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0:03
yes i'm serious today we'll be doing 100 calculus limits and of course we'll be doing them in one take i know finally right and you know the deal the fire is in the description for yoga convenience already so go ahead and download them and work out the questions along the way we will cover everything that you have to know for calculus 1 calculus 2 including how to determine the limits from the graph and then how to do the algebraic limits and also the definition of
0:28
derivatives lopita's rule and we also talk about the epsilon delta definition and also yeah this kind of very long limits at the very end so yep you're going to have tons to learn in this video i don't know how long it's going to take but let's just go ahead and get started here we go so we're given the graph of the function so this right here is just going to be y is equal to f of x and then you can see a better picture on the file so seriously look at the pdf all right so
0:59
first off we are going to find the value for f of three so the way to read this is that x is exactly equal to three this is not a limit question so we go to x is exactly 3. do not look at the open circle but rather you have to look at the closed circle and the value for that is 4. so the answer for this right here is 4 and then we are done then we are going to take a look at the limit as
1:24
x approaching 3 of the function so here's the deal you see how we do not have the plus or minus here we actually have to do both so look at when we have the positive version right here this means we are approaching the three right when x is three from the right so you can use the right hand and then you are approaching and look at the y value you see the y five is approaching two so when x is approaching three plus we get two so keep that in mind and then we also have to do the
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minus meaning that you can use your left hand because we're approaching three from the left-hand side so you can start here because it's approaching here and you see it's like this and the y value is also approaching two so when x is approaching three minus we also get two and two and two are the same so the answer for this right here when there is no plus or minus
2:23
well the answer is just equal to 2 right keep that in mind and you see how these two questions are very different this is the typical look at the graph look at the closed circle question and this right here you really have to start to worry about the calculus part of the limit all right now let's see f of two f of two we go to x is equal to two uh that's the vertical asymptote right there's no
2:51
dots well technically the reason i put down some cut is just because of the vertical asymptote but yeah we do not have the value of the function over there so in this case what we say is undefined
3:04
and there's a small difference between undefined versus does not exist both of them kind of mean that we do not have the answer but undefined is when we are talking about the value of a function or computation so just keep that in mind if the limit doesn't have the answer then we say the limit doesn't exist all right here lastly we have the limit as x approaching two well again we do
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not have the plus or minus we check both if we go to the 2 from the right you see that the curve is going straight up so that's positive infinity and if we approach 2 from the left hand side you see the curve is also going straight up so both are going straight up meaning positive infinity so be sure we check both all right so that's pretty much it however let me just give you guys two more like uh extra practice yeah this is not on the file to make you guys feel special
4:04
because we have the horizontal asymptote as well and that's the case when x is approaching past infinity this means x is all the way to the right and when you have the picture right here you see how it's impossible for us to include everything so you will just have to kind of just assume that okay it's just going to follow this pattern the y-value is going to approaching one all right so you are not going to see like this and then oh no no nobody will do that to you hopefully
4:39
that's not the case especially another test if you have given a graph nobody will do that to you however though if you are using a graphing calculator you are grabbing like a weird function maybe you have to worry about it but yeah that's a different story anyway limit as x approaching negative infinity f of x this case we are approaching left right and then if you look at a better picture from the
5:06
file it's actually approaching one as well so this is more like this right it's more like that
5:14
yeah so here are some terminologies that what is like to go over you see how when x is approaching infinity we get one this means we have a vertical so this means we have a horizontal asymptote
5:31
when y is equal to one right that's the horizontal dashed line right here and likewise we also have one so you don't have to write down the one twice so we have a horizontal asymptote at y is equal to one and you see how when x is approaching two the function this limit gives us infinity this means the function is going straight up sometimes you may have negative infinity meaning go straight down that's the case that we have vertical acetone
6:03
so vertical asymptote at x equals two so that's the vertical dashed line right over there so there you have it this is the first question and we have just a couple more to go and yeah
6:18
so let me just erase this um for the first couple questions just maybe use the time stamp to navigate especially uh because i will have to take some time to graph the
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function so yeah i'll try to i'll also try to give my best graph possible for you guys on the
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board let me see if i can there we go
6:47
okay number two all right number two again look at the file much better that way anyway i'm just gonna give you guys the important values so here we have a two right here of course that will be important and you'll see that the curve looks like this and then we have an open circle here and this is at negative three and then when we have the other piece is up here and then we have a closed circle and then we have this part and of course this is our
7:26
function um couple parts first part f of two let's do this first so this means when x is exactly equal to two go here and look at the closed circle so up here and look at the y value three done f of two equals three and you guessed it part b we will have the limit as x approaching two
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this right here is going to be slightly different than the ones that we did from the first question firstly we will have to again do the plus or minus on the side because you didn't have that so you really have to check both so before we answer this question let me just do this in blue what's the limit as x approaching two let's do the pass d version first
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so in this case we are approaching two from the right so we see okay the wi-fi is approaching three right so just kind of look at this and then approaching three so this is three however if you look at the limit as x approaching two minus meaning from the left hand side
8:39
just left hand starting from here and then just kind of go to two but you see the y value is approaching negative three so this right here will give us negative three and notice that three is not equal to negative three therefore the answer for this
8:58
we do not have the answer and again just like what i said earlier when we are talking about limits the way that we answer it does not exist so just put down d and e i really don't know why i like to put on dots but anyway i'll spell this out at least one time this abbreviates does not exist and the reason that we really say does not exist is because when we ask the limit we like to ask is there a number or the still existed
9:30
number that the function is approaching when x is approaching a certain number if it's a no then your answer it does not exist there is no such a number so i think that's the reason why
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c limit as x approaching infinity
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of the function well this means x is all the way to the right and again don't worry about like this being negative
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when the curve is like this you see that the wi-fi is approaching zero so yeah this is zero and then if the question also asks you for like okay what's the horizontal asylum from here we can say the horizontal asymptote is y equals zero
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right so if the question is asking us for limit we can zero or if the question is asking us find the horizontal asymptote we do that and then we say y is equal to zero all right d
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limit x approaching negative infinity of the function this means we go all the way to the left and the y value is approaching zero and you do not have to write the horizontal twice because they are the same right so yep that is number two and then again we'll just have a couple more to go
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how are you guys doing are you guys taking your calc one with a two classes let me know
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and uh i'm going to check
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yeah periodically because i'm using my phone i want to make sure that
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my phone is recording and uh i'm also charging it hopefully nothing goes wrong yeah
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anyway though number three this right here looks slightly weirder or more weird anyway okay
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so here is the picture firstly i will give you guys the horizontal slopes inside we have a horizontal dashed line here and then likewise here this is going to help me to draw the pictures and then we have a
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curve goes like this and up and then and then
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okay and then flag and then like this and then like so and when let's say here is two and then let me just get rid of that and then we have a closed circle here and the void for the closed circle is negative one and then here is one and then here is
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actually i will just tell you because my picture is going to be really bad so yeah this right here is meant to be 1.5 and this right here is meant to be y is equal to two all right so yeah it's yeah
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i'll try to fix it a little bit
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okay
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maybe slightly better maybe worse uh this is worse
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do you think i'm going to edit the video no i'm not so just be patient and then use the timestamp to navigate as i said anyway this right here is two all right this is horizontal asylum let's finish this question as quickly possible because the picture doesn't look nice anyway here is number three and um first question part a what is the value for f of two that means x is exactly 2 you do get the y value the dot right here right so it's negative 1. so f of 2 is negative 1. done
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b you can guess it what's the limit as x approaching 2 of the function f of x
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yeah well when we're talking about the limit as x approaching two we do not care when x is exactly two right so the answer is actually not negative one but rather you see how the curve is approaching what the wi-fi is approaching 1.5 when we go from the right and then if we go from the left the curve is also approaching the y-value 1.5 it's pretty much the y value of this open circle especially when you have the left and right point
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the left and right end that they're approaching to the same open circle so it's that 1.5
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and in this particular case you see how originally we have a curve but like we're just missing that like that little hole right this right here is actually called the removable discontinuity so removable removable this continuity and the reason that it is removable it means that imagine if you just fill in the open circle guess what the curve right here will be continuous that's why you can remove it meaning that
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because you can just fix it easily you can fill in the circle and then no more discontinuous business
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all right number three limit x approaching infinity
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up up up up up up up and we have a horizontal stop here y is two right so is 2. and again here it tells us horizontal asymptote y equals 2. and then last question limit as x approaching negative infinity
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all the way to the left and the wi-fi here you can look at the picture on the file it's approaching negative 2.
15:36
and guess what if the question is asking us to find the horizontal asymptotes in fact we will have two horizontal asymptotes this is the first one and the other one is y equals negative two the truth is when we have a function it can have at most two horizontal asymptotes in this case you can see that we have one here and one here you may see that hey this picture is just that we just graph it randomly right but in fact inverse tangent also has two horizontal acetone and
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you might be wondering how many fertiliser tops that we have infinitely many because if you look at tangent x it does have infinitely many verticos at all but anyway that's it for this right here
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all right number four it's a funny looking picture kind of it's pretty though okay okay okay like this okay and then the y-value here is two and then open circle here and then we have a wave
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okay someone like this
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yeah and then just kind of continue like this so here's the deal whenever you are given like a picture like this uh just kind of follow along what's the trend of the curve in this case it continues it's like a sine wave cosine it's more like a sine curve because it starts with zero right here right okay part a let's find the value for f of zero this means when x is exactly zero right x value
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exactly zero we look at the close circle go up here the y is two so the answer here is 2. done
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b
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the limit as x approaching 0 f of x
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what do you guys think i'll tell you first the answer does not exist
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why because again we do not have the plus or minus so we actually have to do both and if we look at the limit as x approaching zero plus
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the y value is approaching zero right the y value is approaching zero when we go from the right so this right here is zero however if we look at the limit as x approaching zero minus
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use your left hand you see the y value is approaching two and again when we're talking about the limit we do not care about when x is exactly zero right you see approaching and the y-value here is two when we go from the left-hand side and you see zero and two they are definitely not equal that's why the limit does not exist when x is approaching 0. however though if this is the question then this is the answer if this is the question
19:00
that's the answer so you can have you will have to answer it individually but for this does not exist is the answer because the positive and the minus version they do not equal
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all right limit as x approaching positive infinity do we have any us horizontal no because again if you follow the curve it keeps going up and down up and down right so the value of the function does not approach to a y value as x goes to infinity so here this right here we also say doesn't exist doesn't exist d limit as x approaching negative infinity f of x this means x is up approaching to the left all the way right so ready
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so again this
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if this is the picture then all the way all the way that way then uh do not just say hey it's going to come down no no no nobody is going to do that to you all right if a test is like that then people will be upset but anyway though yeah that's the case so the answer is pass the infinity and because the wi-fi is all the way out all the way up so this is d and this is number four first page
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done cool huh but i know it's just a lot of graphs that's why i know i dropped the microphone hopefully uh it's okay because in fact this is my new microphone because the previous microphone
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yeah i don't know what happened but i it just got bad and uh i i'm pretty sure it's because i dropped a lot now but i dropped a few times yeah let me just use that i don't know how many is the last but anyway number five number five if because i dropped it earlier and then the stun went back and then i broke it double circus i i cannot hear how the sun is recorded right now this is question number five on the 1996 questions to go
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anyway all right number five this is the famous question right so the curve looks like this
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somewhat like that right and in fact here we still have an open circle i would like to tell you guys the function what no not yet let me just give you guys the questions first let's go over them without knowing what the function is firstly what's f of zero
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here if you look at this right here f of zero but when x is zero which is right here if you look up that's just an open circle there's no answer for that right so f of 0 right here is and the find so that's the answer for that okay and then if you look at part b if we do the limit as x approaching zero of the function this right here if you look at the wave over here it's actually one so if you approach from the right the wi-fi is one if we approach
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from the left the y-value is also approaching one so the answer for this right here is one
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all right so that's pretty legit c limit x approaching infinity
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as you can see the curve is like this right and then it's going to get like a more flat than flex in fact this is not like stein curve
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the wi-fi is actually approaching zero so in this case it's zero and again if you look at
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the picture here it's much better to see it right the wi-fi is just approaching zero
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similarly the left-hand side it does not go up like down like that it's not like a wave the wave cannot die down so d limit x approaching negative infinity f of x is also equal to zero okay
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all right so as i said earlier i would like to tell you guess what this function is because we are going to see this again later on this function is rather famous this is actually just f of x equals to sine x over x all right so that's why you can see that when x is equal to zero it's undefined because we actually have a zero on the bottom so does not work and then for number two i wanted to tell you this is what i call the limit because this is perhaps the most
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debatable or controversial limit or what not when we have the limit as x approaching zero
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of this function sine x over x
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per number two you can see that right one for part b you can see that the y value is approaching one so yeah the answer for this right here is one keep this in mind this is one of the sikh weapon and later i will just change the x to theta and you will see uh how we will utilize that all right so this right here i also want to tell you guys that the limit as x approaching infinity zero likewise x approaching negative infinity zero one quick way to see is you can put infinity
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to the top well sine is in between of negative one and one but if you have infinity on the bottom guess what it's like one divided by infinity to zero or negative one divided by infinity which is zero yeah so either way but like if you look at the picture again the wave cannot die down that's why it's zero and then likewise that it's also zero however if we look at the typical sine curve
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which is like the one that we did earlier so this is sine x then the limit as x approaching infinity of sine x this right here doesn't exist right doesn't exist likewise the limit as x approaching negative infinity of sine x this right here also doesn't exist uh but when we divided by x then we end that with zero so um some of the
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crazier cases now you might have nothing before because uh yeah the starting of calculus so maybe you have nothing this kind of function with this kind of curves before especially that one so i want to just mention that but we'll encounter that a lot more later
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cool so the first five questions
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other ones that i gave you guess the graph and then you are going to figure out the limits and then the next five i will just give you guys a function and we are going to see how to get the limits so ladies and gentlemen number six consider we have the function one over x minus two
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you have a couple ways to do the following questions um by a couple way i mean like two ways one is you come up with the graph first two you don't use the graph so kind of up to you i will show you guys both let me show you how we can do it without a graph suppose this is the function part a we want the limit as x approaching 2
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of the function f of x all right here's the deal if we do not use the graph then i will just put a 2 into this x and see what happens and if i do that i'll just do some scratch work right here on the side because this is not like a gentleman computation put a 2 into this x we get 1 over 2 minus 2 which it looks like we have 1 over 0 right what does this mean well let me tell you if we have a number of zero when we're doing limits
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either we get past infinite t or negative infinity what i mean by that is just that we will have to worry about the sign if it's positive or negative when we're doing this but how long by putting two
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no we cannot put the inside two so that's the deal here's the thing again when x is approaching two there's no plus or minus we actually have to do both let's check the limit as x approaching two let's do the positive version first and then the function is one over x minus two yeah so let's see when we took when we put two plus into here we are going to get one over two plus minus two
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two plus is just like a number slightly bigger than two just think about it like that and you can think about it as that two point zero zero zero one so earlier when we do like
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two minus two not so much this is better two plus minus two this right here is approaching
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zero two minus 2 is 0 but this right here is a little bit more than 2. so this 0 is actually a little bit more than 0. so this is technically 0 plus okay and then on the top we still have the 1. so we have 1 over 0 plus and when we have one over zero plus when we're doing limits this right here positive over positive we get positive and then infinity whenever we are doing this kind of like computation if we have a plus or 0 minus on the bottom
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if the top is non-zero then you are going to expect to get either past infinity or negative infinity and again just work out the sign so this right here gives us positive infinity okay and then let's see if we have the limit as x approaching two minus and then one over x minus two same thing put this right here now we are getting one over 2 minus and -2
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this right here is a little bit less than 2 like 1.999 this minus that yes it's like 0 but technically it's less than 0 just a little bit that's why it's true minus on the top is one
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one over zero minus three minus is negative because again just think about this as 1.999 minus two we get negative small number zero minus is a negative so we will end up with again you have infinity but just have to work out the sign positive over negative we have negative so this right here gives us negative infinity so you see this and that they are not equal so that this is like a sad face that's why the answer for this right here is does not exist okay does not exist
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okay part b let's look at the limit as x approaching infinity of the function f of x
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so for this one right here let's see if we put infinity to the function then we are going to get one over right one over infinity minus 2 and let's see if we can reset it out along the way infinity minus 2 is still infinity so we still have the infinity on the bottom and we have a 1 on the top if we have a finite number over infinity guarantee we get zero right one divided by infinity yeah
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approach is zero so the answer for this right here is just zero imagine one dollar yeah share with like one billion people like 100 billion people how much money is each person going to get and that kind of thing anyway limit as x approaching minus infinity of the function put negative infinity into our function there so we are looking at 1 over negative infinity minus 2
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minus infinity minus 2 is still minus infinity and we have 1 over that right but guess what
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it's approaching zero don't say negative zero it's the same as zero yeah so that's why when you have a non-negative number sorry when we have a num when we have a finite number over infinity we get zero so this right here is just zero done there now let me show you guys the graph for this if we have one over x minus two it's just that one over x which is like this like that but because it's a
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x minus two on the bottom we actually have a vertical stop right here at two then we have a vertical asymptote like so and then the curve looks like this like that
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so if you have this picture in your mind earlier this question will be much easier because you can see that the limit as x approaching to 2 from the right this right here is positive infinity as we mentioned earlier and then you get negative infinity when we go from the negative side right so that's that and then they are not approaching to the same waveform you would that's why this doesn't exist and then when x is approaching infinity the y-value is approaching zero
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the y-value is approaching zero when x is approaching negative zero so yeah so this is number six so no dual graphs i'm pretty sure if you are taking your k1 class or maybe pre-calc if you are watching this for your pre-calculus class uh your teacher will would like you to know the graphs of this basic function say the rational functions the square root functions and then the other ones that i am about to mention
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so all right so number seven our function is natural log of x so part a let's see we want the limit as x approaching 0 plus of the function alpha lnx pretty much so let's see what will we get when we have 0 plus inside here well let me tell you it looks like this right if you just want to plug in and then kind of reset the l or do computation
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here's the deal if you have ln of zero plus in fact you will have to know the graph in order to actually see what the answer is i'll tell you though the answer for this right here is negative infinity it approaches negative infinity and notice that how i have been drawing arrows when i do these computations because arrow means approach but when we have the limit and then we put equal because the limit has that approach already we are trying to answer
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limits equal to whatever but anyway though the answer is negative infinity right here and as i said i think it would be much better if i just give you guys a graph right now to make sense of this l and x one of the functions that you definitely will have to know looks like this
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yeah sounding effect for the wing huh anyway one and all that so when x is approaching zero from the right-hand side you see the function is going straight down and uh
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l and x it only makes sense when x is greater than zero this is the domain of the function
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right domain of the ln the inside has to be greater than zero that's why x approaching just zero wouldn't make sense x approaching zero minus wouldn't make sense either
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so that and then let me see what else do i want to mention yeah that's about moving to the next one b
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okay next one limit as x approaching e of the function when x is approaching e we can put e into l and x and guess what l and e is just equal to one so the answer is just one
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in fact e is about like 2.718 ish right so let's say e is about right here when we put e into the ln function you see how the function is nice and continuous and the y value here is one yeah that's it so some of the values you definitely have to know ln one this is not limit it's just a regular computation and one is zero and then ln e is one okay and then this right here is like the limit so just remember that if we have ln and then zero plus
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this goes to negative infinity and you know the next one is the limit as x approaching infinity
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you see how the curve is like this in fact it's going to get bigger and bigger than the y value so this right here gives us infinity so another expression that you should remember is ln if you have infinity inside it approaches positive infinity so this is what we have
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okay so take a look take a look
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okay
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wow number seven already that's that's cool and i think i'm just going to check to see if if the phone is recording every 10 questions so that way i can have a peace of mind and to be very honest with you guys i do not have a peace of mind because i dropped the microphone earlier hopefully the sound is okay if it doesn't i'll record it again don't worry
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anyway number eight aha our function earlier was ln x you expected here we have e to the x
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okay so part a let's look at the limit as x approaching negative infinity of the function
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so again if you don't use graph maybe just want to like plug in and kind of reset it out this is how you can do it put negative infinity to here so we are looking at e to the negative infinity and you see we have negative exponents so maybe if you want to you can just say this right here is uh
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technically i will put equal because it's like computation then the negative right so it's one over e to the infinity and here's the deal e is 2.718 yeah each raised to infinity's power equals to infinity and 1 over infinity is approaching 0. all right so this right here is 0. okay so this right here perhaps i'll tell you guys this thing real quick the deal is that if we have a base right if you have a base that's a bigger than one
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bigger then 1. if we have this raised to the infinity's power this right here will go to infinity let's say 1.001 raised to infinity's power we get infinity all right e is 2.7 inches so that's why if you look at the bottom it goes to infinity um another way to look at this is that if we have a base that's in between between i will say this between zero and one so i would just say
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um oh i'll say this uh between yeah let's use the word between between
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zero and one but not equal to one not
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equal to one and then raise to the infinity's power now say one half or zero or 0.001
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raised to infinity's power we can say that this is approaching zero and you might be wondering what if we have a base that's less than zero well if you're talking about functions when we have a function written this form b to the x power we will have to make sure that the base b is greater than zero yeah technically you can say b greater than or equal to zero but like zero to the x power it's almost always zero right but yeah anyway zero to the x
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power this right here x cannot be negative numbers otherwise we divided by zero but
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yeah when we have exponential function b wants to be greater than zero that's why okay
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so again when the base is bigger than one to the infinities power is infinity we can draw a conclusion if it's between zero and one nine equal to one then we get
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zero okay and then i also want to mention if the base is exactly
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equal to one right just like earlier i said not exactly not equal to one yeah if it's exactly equal to one and then raised to infinity's power this right here is approaching one okay however if you get let's say one plus to the zeroth power or one minus sorry 1 plus 2 infinity's power or 1 minus to the infinite power one with the other this right here right this right here they are the so-called in the determinant form meaning that we do not know what the answers are
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why well it's on my shirt already no jokes aside here this right here are the limits form
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and the crazy part was like the dangerous part of this is that a lot of times we don't even put a plus or minus that's why you see my shirt right here we have a one to the infinities power a lot of times it's misunderstood that hey it shouldn't not just be equal to one no no this right here this one is a limit form meaning that is technically saying one plus or one minus so this right here is a one of the most confusing part and i would like to just
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give you guys an example before we go especially we're talking about e to the x power right so the famous example for this one is this is called the fact well actually no i'm not going to tell you guys the secret weapon yet i'm just going to put down notes
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if we have the limit as x approaching infinity 1 plus 1 over x to the x power
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check this out if we put infinity to here and then just work out like that you know do this kind of thing we are going to get 1 plus 1 over infinity right put this in here with infinity and then to the infinities power now i'm going to kind of argue like we'll resent things out along the way 1 over infinity is 0 so this right here is approaching 1 plus 0 and then to the infinities power right and then what's one plus zero one plus zero is one so this right here is approaching one
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to the infinities power right and now it's like contradicting i'm saying that this right here is exactly one two three no what am i talking about
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let me tell you this particular limit the answer that we get is actually e which is about 2.718 ish okay so i'm just going to write this down right here e is about 2.718
45:11
and again if you look at this why this is a indeterminant form it's because we just write it down as one to the infinity but really it means the limit for either one plus or more minus why because this zero here it's not exactly zero one divided by infinity if the value is not exactly zero the limit is zero so technically you should say this is like zero plus just a little bit bigger than zero
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that kind of thing and then right here we have one to the infinities power this is technically
45:44
one plus and then raised to infinity's power yeah but unfortunately when we're talking about the limit form nobody writes it like this so you just have to be aware of this kind of situation okay and what i mean by exactly equal to one is that if today i have a very redundant example let's say if we have a limit as x approaching infinity if we just have a solid one nothing else and then raise to the x power
46:18
you see i'm not adding anything just one just just one and then to the infinities power the answer for this right here is equal to one why because the base is exactly equal to one and if you put infinity to here you really get one to the infinity but in this case you have to really understand what the base is that's what i'm trying to say all right so keep all these things in mind
46:43
well i'm not looking at my show i'm looking at the question okay that's part a now let's look at part b limit x approaching zero f of x this is nice we can put into the x value here e to the zero power is just one so the answer is just one done and then lastly let me see if i can fit in par c
47:09
okay our fitting part c right here if we have the limit x approaching infinity
47:16
per our discussion if we put infinity into here e to the infinity e is bigger than one right to the infinite power this right here will give us infinity
47:27
so let me just kind of make a quick divider a b c and then these are the notes now i was going over earlier of all when we have a function if we know the graph we can answer a lot of the limits equation without worry so let me give you guys the picture e to the x looks like this so x approaching infinite negative infinity all the way to the left y for u approaching zero x approaching zero this y value is one it's nice and continuous it's just one
48:04
x approaching past infinity all the way to the right y value goes up to my ceiling to the third or the fourth i don't know tenth floor it's infinity so ladies and gentlemen that's number 10 and more importantly please please please know this really really well all right this is like the most asked question in any calculus class especially when we're talking about limits
48:40
so
48:56
all right number nine number 10 i'm going to do like a function and then it's inverse again so let's look at number nine let's look at the function just tension x and um
49:10
let's see i'm going to write down abc and then i'm going to let's put it down here let's say a we have the limit x approaching pi over 2 f of x b limit x approaching infinity f of x c limit x approaching negative infinity f of x
49:38
ready cool inverse tangent sorry regular tangent ladies and gentlemen it looks like this
49:45
we have infinitely many vertical asymptotes the first one is right here if you look at the
49:53
positive exponent first pi over 2 and then the next one is you just add pi and you get three pi over two and so on so the one on the left the first one here is negative pi over two
50:09
and then so and so on and then we are going to get bunch of this
50:13
i'm just going to do this to make it more clear
50:18
right so bunch of this and one of the graphs that you should definitely have seen back in a precalculus 3 class and with that being said let's look at the first limit as x approaching pi over two we do not have the plus we don't have the minus so we do both when x is approaching pi over two from the right we get negative infinity so let me just write this down here
50:43
if we have the limit as x approaching negative so pi over 2 minus
50:50
of our function tangent x this right here is negative infinity
50:55
sorry this should be a plus sorry pi over 2 plus from the right the function goes straight down
51:03
and if we have the limit x approaching pi over 2 minus tangent x
51:10
this right here goes straight up so positive infinity they do not equal
51:18
d and e i preferred with dots you don't have to put down the dots
51:26
anyway limit as x approaching infinity whoa this is crazy just to keep on going all right the y value does not approach to a certain number so this right here d and e i prefer dots similarly x approaching negative infinity d and e when you have the picture in mind or just draw it down right in front of you the limit questions are so much easier isn't it
51:57
cool now we are going to turn this just the middle portion and reflect it about the line y is equal to x and we get the inverse right so
52:29
the aircraft my first my original tangent wrong no i didn't tangent wait hold on so it was like this
52:39
and then i'm going to do a reflection
52:44
all right i'm going to do a reflection so anyway let me do number 10 right here f of x
52:51
equals inverse tangent of x so this was the original tension and we just uh we'll just look at the middle like the the first one because if you do the reflection on everything then the result is not going to be a function anymore so just rotate this not flip this reflect this so inverse tangent
53:15
yeah um we are going to get a picture like this
53:22
ah let me do it
53:31
all right this one's positive pi over two right but if we reflect it this is going to give us
53:51
okay so we have pi over four one and then we will have one pi over four so yeah so yeah what am i talking about anyway so if we reflect it this vertical will become horizontal as though and then this vertical slope which is at negative pi over two will be another horizontal asymptote so this right here is negative pi over 2 and this is pi over 2. and the curve looks like this
54:20
okay so it has that 2 boundary yeah that's horizontal isotopes
54:28
anyway this is number 10 and uh here we go a limit x approaching negative infinity of our function
54:41
if it goes to negative infinity we get negative pi over 2.
54:48
b limit x approaching one f of x we can just simply put one in here inverse tangent of one
55:00
how do we do this we will have to ask ourselves tangent right the original tangent of what angle will give us one look at the first answer and the answer for that is pi over four so this right here is pi over four and again the reason is because regular tangent of pi over 4 gives us 1. and it's nice and continuous so that's why when x is approaching 1 this y value is pi over 4.
55:32
so this is one of the values that you should remember it's going to happen a lot inverse tension is one of the very common function that you have to deal with in your calculus class not just for limits but also later on for the derivative and also the integral anyway last one the limit as x approaching positive infinity we get positive pi over two
55:56
ah so much better when we have a graph in front of us right certainly ten questions let me just check
56:03
good wait how long did i take wow 56 minutes for 10 questions are you serious
56:14
wow i thought it was just like 30 minutes oh my goodness
56:27
okay so that's the first category we determine the limits from a graph and the second category is still kind of similar depends on how you want to do it we will see how to find the limit from piecewise functions okay so number eleven at this this time that we are just doing one question per per question earlier was like three questions per question anyway
56:59
i'm going to give you guys the function here right here so we have our function f of x is a piecewise function you just have to pay attention to the conditions so here we do three x plus one this is when x is not equal to negative two but when x is equal to negative two we get two so
57:24
so just like this and our goal is to find the limit as x approaching negative 2 of course that will be the interesting number of the function so here's the deal when we're talking about the limit as x approaches negative 2 we do not care about x is at exactly negative 2. so we don't care about this we just have to worry about this
57:50
and uh you don't have to worry about the 11 right because left and right is just this anyway so we can just put negative 2 into here so let's just do a computation real quick we will just get 3 times negative 2 plus 1 work that out we get negative five
58:12
and that's it right seriously that's it i will give you guys a picture real quick to illustrate why this is the case firstly let's focus on graphing this which is just the equation of a line the y intercept is that one and then the slope is three meaning that we go up three times one two three and then go to the right one time well you don't have to be like super precise you just have to make sure that your picture makes sense so this is the
58:44
equation of the line 3x plus one however x cannot be equal to negative 2 it means that we will have to go to when x is equal to negative 2 go to the line and then erase it
58:57
and then we'll put an open circle here but now the question is what's the y value for this open circle to do so we put negative two into here just pretend that the point was still there and then we saw that it was negative five so this right here is negative five
59:15
however we still have to grab the second piece when x is negative two the y value is actually a two so when x is negative two we actually go up to two right here this is two let's say this is two and then we have a closed circle so this right here will be the picture for this piecewise function and now you should be clear that again just like what we did in the first category
59:39
when x is approaching negative 2 from the left and from the right both of both of them give us negative 5 the answer is negative 5 done do right so that's number 11.
59:53
i just do one question and erase that's why it takes me a long time okay
1:00:02
number 12. let me give you guys the function first f of x equals again two pieces
1:00:13
first one 1 over x minus 4 squared and this is when x is not equal to 4 and the other one is 3
1:00:24
this is when x is equal to 4 and the question is asking us what's the limit as x approaching 4 of the function same deal do not worry about this because we're talking about limits so just put 4 into here and then let's see it looks like we are going to get 1 over 4 minus 4 and then square ah we get a one over zero and like earlier earlier we were able to just find the answer nicely so that's why we didn't have to do the plus or minus but for this one we
1:01:05
shall make a better observation if we have a four plus here okay that means the x is approaching four from the right-hand side four plus minus four we get zero plus and then square either way it's still positive so yeah one over zero and that's a positive so this right here gives us positive infinity and if we have one over four minus and then minus four
1:01:40
inside gives us zero minus and then we square that
1:01:47
this right here when we square a negative like zero minus this technique this through point zero zero to one it will give us zero by become positive all right so this right here also gives us positive infinity they are the same that's why the answer is past infinity so this is how you can do it without utilizing a graph just kind of write down expressions and then just make arguments or like resent out the answers along the way so that's pretty good
1:02:20
i'll give you guys a picture real quick though oh by the way
1:02:27
see limit this infinity uh if this is part a i'll give you guys like a popular like a note on the side f of four what's the answer for that f of four is three right so just remember all the things that we have done so far and here's the picture for this though
1:02:51
x is four that's the main value here that we care okay vertical isotope because the bottom cannot be zero huh one over x minus 4 squared it looks like a volcano so the picture looks like this i will use red to make this one stand out so like this
1:03:13
like that that's why past infinity however not done yet when x is four the y value is three so when x is four we should actually go here and then say let's say this right here is three
1:03:26
so we actually have a dot right there yeah
1:03:32
okay that's number 12.
1:03:56
all right number
1:04:00
just kidding
1:04:02
okay number 13 our function f of x is equal to a two two pieces again don't worry negative x plus two this is when x is less than zero
1:04:20
and then the other one is negative x squared
1:04:24
plus two and this is if x is greater than or equal to zero
1:04:30
and you can see that zero is the interesting number so yes we look at the limit as x approaching zero of the function so for this one we actually have to do left and right because we have the inequality right x is less than zero and also x is greater than or equal to zero so we have a left and right so here's deal let's go ahead and consider if x is approaching 0 minus let's do that one first if this is the case then the function is this piece that we are
1:05:05
looking at so look at negative x plus 2. and this is not bad at all because if you put 0 in here this is just going to give us 0 right negative 0 is still 0 and then plus 2 just 2. you don't have to worry about the 0 minus because plus 2 is just 2 that's the value that that's the limit that we get and then if we have the limit as x approaching 0 plus that means x is greater than equal to 0 we look at this piece which is negative x squared plus two
1:05:37
guess what when we put this right here that's still zero plus two is still two hey they are the same that means when x is approaching zero of the function we just get you
1:05:51
all right yeah okay and then this case i also like to tell you it happens to be that when we have f of zero how do we do this pay attention to the equality x is equal to zero right here right x is greater than or equal to zero so utilize this one so for we'll have this we technically will have to do negative zero squared plus two and work that out f of zero is two if that was the question
1:06:28
okay a picture real quick for this curve take technically it's a line in the curve
1:06:34
0 is the like endpoint so pay attention to that when x is less than 0 we do this which is just the equation of a line negative x plus 2 2 is the y intercept so we have up here let's say this right here is 2. and then negative x is just
1:06:53
like this right the slope negative 1. well if you would like you can just plug in negative 1 in here negative negative 1 is 1 1 plus 2 is 3 so negative 1 comma 3 and you have another point
1:07:05
however this is x less than 0 so technically i should have put an open circle here i'm just doing this in water so open circle because of this inequality and then for the second part we have this parabola negative x squared is like this right upside down parabola however it's been shipped up two units it happens to be here and just the right hand side because that's only when we have x is greater than or equal to zero
1:07:36
so we started right here because thanks to equality here and then we have a parabola
1:07:44
so this right here is the line this right here is the parabola and you can see that the function is continuous when x is equal to zero that's why the limit and the value of the function they are both equal to 2 because the function is continuous all right
1:08:04
so hopefully this right here makes the piecewise function not so intimidating because in fact you just really have to follow the direction like the condition and you do the piece here and there that's all
1:08:38
all right number 14
1:08:43
our function is this
1:08:48
2x plus 1 and this is if x is less than negative 5 and then the other one is x
1:09:03
squared minus 12. this is if x is greater than or equal to negative 5. all right and you can guess it the question is the limit as x approaching negative 5 of this function
1:09:21
okay because of the inequality that we have so we do the left and right yeah so firstly let's look at the limit as x approaching negative 5 from the negative side
1:09:34
so if x is approaching negative 5 from the left-hand side from the negative side is less than negative 5. so that means we look at this piece so we have 2x plus 1.
1:09:47
how do we do this well just plug in negative 5 in here and we get 2 times negative 5 plus 1 and that's negative 10 plus 1 which is negative 9. okay on the other hand when we have the limit as x approaching negative 5 positive that means we do this right which is x squared minus 12.
1:10:14
and to do so just plugging negative in here and just work that out because this is not like weird indeterminate form so just plugging work out that would be the answer so negative 5 squared minus 12. that's 25 minus 12. 13.
1:10:33
very different so the answer is does not exist
1:10:40
you see if the question doesn't ask you to make graphs then this right here will be it just show this and that and then draw the conclusion for the limit as x approaching negative five
1:10:52
but of course i'll give you guys the graph real quick just to make things super super clear
1:10:57
because this is for practice right if you're on the test if they just want us to figure out the answer i would stop right there okay we care about negative finances right here this is a line with positive slope and let's see when x is a negative 5 we get negative 9 so we should have negative 5 and also let's say this is negative 9 and then we have an open circle like so okay and then
1:11:32
uh two so you have a positive slope so the picture should look like this
1:11:36
okay slope two well if you were like you can just plug in i say negative 6 and then plug into there work that out yeah that's okay what's this this is a parabola yeah this is the parabola x square but you drag it down 12 units so let's say this right here is negative 12 and then you just have a parabola that looks like this but when x is at negative 5 the value is 13 so the picture is going to look like this is crazy it's a metric so i'll try my best
1:12:12
so this is the part that we were talking about here we get negative 9 and then here we get a positive 13. that's it and that's how you do it yeah
1:12:29
okay how are you guys doing how are you guys doing hopefully you guys are doing good
1:12:35
you know just a couple more questions to go right hang there
1:12:41
if i can do it you guys can do it too you don't have to watch this video one time through right take a break if you would like but just make sure that you do the practice all right
1:12:58
number 15
1:13:03
function f of x equals first we have one over x if x is less than zero
1:13:14
if x is greater than 0 we get cosine x also including 0 yeah so let's see
1:13:25
i think for this one let's just do the graph and then
1:13:29
find the answer so if we want the limit as x approaching 0 of the function again we will have to do 11 right let's go to the left right so firstly we do the limit as x approaching 0 minus if it's true minus then the function that we need to use is one over x
1:13:51
so what's the answer for this though well we put zero minus in here we get one over zero minus positive divided by negative and then we have a zero red limit four on the bottom so this right here approaches negative infinity
1:14:10
and if we have the limit as x approaching 0 plus we look at cosine x and for this right here just plugging straight into cosine and then we can actually work that out and we get 1. right so the answer is just one cool but the thing is that they are not equal therefore the answer right here does not exist that does not exist okay and then oh let me give you guys the graph real quick
1:14:45
when it's less than 0 we do 1 over x and 1 over x is like this right this is 1 over x usually it's like this and that but here we just care about when x is less than zero so just this part cosine x cosine starts with zero comma one right so we start right here and then it goes down up and then like this that's it and of course you can also like worry about if the limit is x approaching infinity of the function what's the answer with this ah doesn't exist either right
1:15:21
because it keeps going up and down up and down in between of negative one and one the y value just
1:15:30
bouncing right between negative one and one so no
1:15:34
all right number fifteen okay hopefully i don't need to like erase a question every single time um because oh no for number 16 i still have to make use of the whole board but we will see
1:16:01
number 16 our function is 1 over x minus two
1:16:10
this is when x is less than two and then ln of x minus two this is when x is greater than 2 and notice i cannot put equal sign for the first one for the second one if i put the equal signs right here well we cannot divide it by zero
1:16:32
and if i put the equal sign here that's also no good because l and 0 is undefined so here f of x right the domain for it x cannot be equal to 2 but let's see we want the limit as x approaching
1:16:48
2 of the function and you know it let's do exactly what we did so here take a look at the limit as x approaching 2 from the left-hand side first if it's a left-hand side then we are looking at this right here so that's 1 over x minus 2. okay so put negative put 2 minus in here so we look at 1 over 2 minus and then minus 2. two minus two is zero yes but this is a little bit less than two so this
1:17:20
right here is technically zero minus all right so one over zero minus this is positive this is negative and the limit four on the bottom is zero right so all together we get negative infinity
1:17:33
okay secondly limit as x approaching two plus then we are looking at ln of x minus two well put two plus in here we get ln of two plus minus two two minus two is zero but this is a little bit bigger than two so the inside is technically zero plus what's ln of 0 plus remember the things that we talked about maybe like 14 minutes ago this right here approaches negative infinity
1:18:09
so surprise they are the same so the answer is actually negative infinity
1:18:16
okay all right so just real quick how does the graph of this look when x is two
1:18:24
one over x minus two it has a vertical isotope and then the picture looks like this
1:18:31
okay ln of x minus two ln of x minus two also vertical asymptote and
1:18:41
it actually like this because remember ln of x it actually grows to infinity right so it actually goes up but this piece it has a horizontal double this one does that so it's a pretty cool curve this is the ln x minus two part and this is the rational function one over x minus two just left the function
1:19:13
so
1:19:27
all right number 17
1:19:31
here's the deal we are just going to evaluate the limit as x approaching 0 and i give you absolute value of x over x d i make a mistake this category i'm saying that it's the limits from piecewise functions i don't see no piecewise functions technically there is because when we have absolute value of x it's technically an absolute value it's technically a piecewise function so i would like to write this down right here for you guys
1:20:02
note when we have absolute value of x this actually means either we get negative x or plus dx
1:20:10
we get negative x when the inside namely x
1:20:14
is negative because negative times negative that's how we can produce a past the output
1:20:21
and if the inside is already positive then we just you know have the x back
1:20:29
and the truth is you can put an equal sign here over here doesn't matter because 0 is still negative 0 is just 0. but usually people pull
1:20:40
equal sign there all right so let's see this right here is our p-square's definition but we want absolute value of x over x so how exactly do we do this
1:20:55
well check this out when we have absolute value of x over x this right here
1:21:03
we can just go ahead and do negative x divided by this x and this is going to reduce to negative 1. similarly we do x divided by this x and this right here will give us positive 1. but be careful though this right here yes it's still if x is less than zero however for this right here this is if x is greater than zero no more equal sign we cannot use equal sign anymore because the expression right here has the x on the bottom we can have 0 on the bottom so we lose that equality
1:21:45
right so this is how you write that piecewise function for your absolute five of x over x and this right here actually has a special name sign but like the way you spell it is s i n g because it tells you if it's like a positive or negative sign of the number x so yeah
1:22:08
but anyway though you don't have to know well i don't know it depends on your teacher but yeah but you need to know this and the piecewise function for that anyway what's the answer for that
1:22:20
from the negative side we get negative one from the right hand side we get positive one so they do not equal you can make a graph of this real quick
1:22:32
negative one
1:22:35
and then one now this is the graph for
1:22:39
absolute value of x over x but either way though this right here does not
1:22:46
exist yay i think i can write another question right here without erase number 17. okay number 18. all right let's take a look at the limit as x approaching 0 and then we have
1:23:04
x over absolute value of x plus x squared all right so how can we do this you can use this again you can just rewrite everything again or right here let's just go ahead and break down into like a positive negative and check if we have the limit as x approaching zero minus
1:23:30
the only thing that's going to uh we only have to worry about the following is this this part
1:23:38
so x stays x on the top but after 5 x if x is approaching 0 minus meaning the input is negative this right here you replace that with negative x and then right here you have the plus x squared so that's how you do it okay you could have done the same thing like like this earlier but i think that's more clear anyway once you write this we can simplify it you see on the bottom we can factor out an x so let me do that and then we have negative 1 plus x and then we can see that
1:24:14
this x and that x cancel out and that's okay because when we have x is approaching zero minus x is not exactly equal to zero so cancel it out that's totally fine and then we will have what
1:24:28
we have 1 over negative 1 plus x right so let's see
1:24:36
let's just plug in 0 into here then we are going to get 1 over negative 1 plus 0 and we can draw a conclusion right here the answer is just negative one
1:24:48
so this limit gives us negative one similarly let's do the limit as x approaching zero plus in that case again this is the only part that has to change on the top is still x this guy is just x and then we have plus x squared factor out an x on the bottom
1:25:09
and then cancel out this x and that x and then put a 0 in here then you see we get 1 over 1 plus 0 and that will give us 1. so this limit gives us positive 1.
1:25:24
however 0 minus and 0 plus they give us different limits so the answer for this right here does not
1:25:34
exist okay
1:25:38
cool now another special function for you guys for number 19 namely the flow function but let me erase the ball first i'll give you guys a quick intro and then show you guys how to do this
1:26:11
i'm not sure if you guys can see i'm actually sweating because right now it's like 5 p.m yeah and it's hot and i cannot turn on my ac because one my ac doesn't really work and
1:26:24
i should have told you that hence like reason number one
1:26:29
i cannot really turn on my ac because one if i turn on my ac then it will be super loud that's going to ruin the video and the second reason is that my ac doesn't really work anyway
1:26:39
but you'll be loud that's the main question that's the main issue though anyway number
1:26:48
19. okay here is the limit as x approaching three and then ladies and gentlemen this is the floor function so let me explain to you guys what this is this right here right this right here it's called the floor function where some people idea will write it as flow of x one way with the other so what this means is that you just always run down to the
1:27:20
integer right wrong down so let me give you an example
1:27:27
when we have the flow of let's say
1:27:31
five well you run down so this right here will be zero okay if you have flow of two point
1:27:39
one you run down so this is going to be two right right down to the integer and then if you have the flow of let's say 8.9 man this is almost nine right but no you still run down so this right here is equal to eight and it might be a little bit trickier if we have negative numbers such as negative three point two
1:28:04
what's the answer for this it's not negative three but rather it's negative four perhaps the easiest way to see is that on the number line so you have the zero and then negative one negative two negative three and negative four right i know this is like a horrible picture
1:28:23
much better negative zero point sorry negative 3.2 is right here and you always run down so you always move the point to the left whole number well then you just move the you just run down and then it's the first integer so you run down so that's why it's negative four
1:28:46
you see that prepare ripple there i don't know if i need to drink it but if i need to i will it's been an hour it's hard most likely it's room temperature already maybe i'll get one from the fridge we'll see but anyway hopefully this right here is clear
1:29:03
now what's the limit as x approaching 3 of the flu function of x this question would be so much easier if we look at the picture right so we will look at a picture ladies and gentlemen let's have a look here is zero one two three might as well four and minus will let's put negative one
1:29:26
so use the idea earlier check this out when x is let's say 0.5 the y value will be zero
1:29:33
when x is 0.9 the wafer is just 0. so in fact anything from here to here is just zero so that's the first piece however when we have the abs the flow of one if this is a whole number already then you just get the whole number back so when x is one you jump to one so you have a closed circle here and you have an open circle here
1:29:58
and then you continue when we have one point one flow of that is one likewise anything up to almost 2 because when x is exactly 2 we make another jump to 2. same thing
1:30:14
open and then up 3 and then here and then like this and that's also so and if you would like when x is 0 you get 0 and then anything from negative 1 to here we get negative 1 so you have a piece like this so take a stair i believe some people also call this the stair function but like the flu function is the most common one or maybe the greatest integer function but anyway though check this out when x is approaching 3
1:30:52
from the left we get 2. from the right we get 3. so the answer for this right here is doesn't
1:31:01
exist and let me just indicate that real quick if we have the limit as x approaching 2 sorry
1:31:10
if we have the limit as x approaching 3 from the left hand side so 3 minus of the flow function this right here gives us 2. but if we have the limit as x approaching 3 positive then this right here gives us 3 unfortunately they are not equal so we do not have the limit for it
1:31:39
okay
1:31:55
and depending on your calculus car some people covered a special function that was one of the special functions the flow function um most of the people we do and we also to cover the fractional part of the function but like yeah anyway number 20 yay
1:32:17
here limit as x approaching 1 and the function here is absolute value of x plus the flow of x wow
1:32:34
okay let me just show you guys how we can do this
1:32:38
um yeah now let's just do it without the graphs if x is approaching one from the negative direction guess what because it's one minus so this right here is still just x right so we'll just have x just take out the absolute value it doesn't matter and then
1:33:01
it's not so easy to write this so perhaps i'll just i i think i'm just going to write it like this plus the flow of x yeah and here's the deal one minus is just like one right one and then plus if we have one minus what's the flow of one minus
1:33:23
well if you look at the picture earlier it's like this right zero to one and then just a jump and then like this and then so on so when we have x is approaching one from the left hand side it's zero so it's one plus zero so the answer for this one is one but if we have the limit as x approaching one plus just take out the absolute value it doesn't matter
1:33:56
okay that is going to be still one plus one plus when we put it here you just run down right so just one so it's one plus one two yes but they are not equal therefore the answer for this right here dne all right so yay
1:34:23
20 questions done and now let's have a look
1:34:28
one hour and 34 minutes okay so on page four we are going to just compute the limit so no more graphs so that's good because the graph took a while to do right
1:34:55
all right and also we're just going to be doing algebra so this is going to be the limit as x approaching some number a and for now a is finite uh in this category and i'll show you guys the techniques the things that we will have to do to figure out the limits and i'll also show you guys how to organize our work and all that good stuff first question on page number four number 21 the limit as x approaching eight
1:35:23
x squared minus seven x minus eight over x minus 8. first thing first
1:35:33
when we are just computing the limit it's always a good habit to plug in this number into all the x's and see what happens if we get a nice number then that will be our nice answer if we end up with one of these numbers well i should have said one of these limit indeterminate forms then congratulations we will have to do more work so let's see i will show you guys all the work for now put the 8 into all the excess then we are going to get 8 square minus 7 times 8
1:36:05
and then minus 8 over 8 minus 8 on the bottom we get 0 on the top this is 64 minus 56 which is 8 and a minus 8 which is 0 0 over 0 we cannot show any conclusion why because it's one of the indeterminate forms the cat is scared we are scared of the determinant forms we cannot do any conclusion yet but don't worry i'll show you how to take care of this
1:36:39
here's the deal if we get 0 over 0 and we have that then we can expect to cancel a common factor in this case it will be the x minus 8. to make that happen
1:36:53
notice the top is just a quadratic trinomial we can actually just factor it the usual way and then we will just get the limit as x approaching 8. and please please please when you are doing this kind of limit write down the lim x approaching eight
1:37:11
your teacher your professor will really appreciate that i will also yeah later on you'll see why that this is needed especially once we get to the definition of derivative when there are a lot of variables being votes so it's not we are just being picky is actually necessary
1:37:29
all right anyway i'll write this down and then on the top let's factor it we will get x
1:37:37
plus one times x minus eight because this times that was b negative eight together we do produce the negative seven x over x minus eight have a look x minus eight x minus eight cancel out so nicely and once you can solve this common factor then you wouldn't get the zero over zero
1:38:01
because you can see when we put the 8 in here 8 plus 1 next number that's the nice answer so now just go ahead i'll show you guys the work 8 plus 1 so the answer is 9 and then we are done okay so that's the answer for that all right number 22.
1:38:21
let's see the limit as x approaches 3 and then we have
1:38:29
2x squared minus 5x minus 3 and on the bottom we have x squared minus 2x minus 3. all right so again you should have a good habit that plugging the string to all the x's uh you can work this out on your own let's see three in here is nine times size 18 that's 15 so 18 minus 15 is 3 3 minus three would give zero
1:39:00
nine minus six is three minus three is zero so yeah zero over zero do more work for this trinomial trinomial we can factor it i'll show you guys how to factor this guy on the side 2x squared minus 5x minus 3. it's usually a little bit harder to factor when we have a number that's not 1 in front of the x squared so i prefer to show my students the tic-tac-toe method you just ask yourself what times what gives you 2x squared and the combination is 2x times x and then you come
1:39:34
here and you ask yourself what times will give you negative 3 and the word matters i'm going to put negative 3 here and 1 here this works because now we can do the check and i can confuse you 2x times negative 3 gives us negative 6x 1 times x gives us x in the middle this is where you check negative 6x plus x does give us negative 5x so this is the correct factoring
1:40:03
we have found our correct combination but this is how you are going to write down the answer for the factoring wise anyway here we have the limit x approaching 3.
1:40:14
on the top here read these across so we first have 2x plus 1
1:40:21
and then bottom is x minus 3
1:40:25
right so that's the factoring you cross multiply to check and then you really across like 2x plus 1 x minus 3 that's for the answer for this right here you can factory the usual way right just like
1:40:39
x times x give us x squared and then we need plus one and then minus three to make this happen and you see it x minus three x minus three cancel and then we can just put a 2 sorry we can just put the 3 into this and that and then work out and we will get 2 times 3
1:40:58
plus 1 over 3 plus 1 the seven that's four so all together seven over four
1:41:09
done yay all right now i'm going to erase the ball
1:41:28
number 23 we have the limit as x approaching 127 no just kidding x approaching four
1:41:41
and then we have a square root x plus five and then minus three and the minus three is outside the square root and on the bottom we have x minus four all right have a good habit plugging four into here four plus five is nine take the square root of that is three three minus three we get zero four minus four has zero zero over zero situation we have to do more work
1:42:10
here's the deal when we have a square root it seems that we cannot factor the usual way huh however when we have a square root expression like so we can use the conjugate to help us simplify it and by that we mean to multiply the top by and again let me just put a 0 for 0 all over here
1:42:32
all right i'm going to keep this and that so we have square root of x plus 5 and then 3 but instead of the minus we are going to put plus in the middle and of course we do the same thing on the bottom so square root of x plus 5 and then plus 3. okay and now we can work this out relatively easily when i say that it's because when you multiply the count you get we have an easy way to do it here's the deal when we are multiplying a minus b times a plus b
1:43:13
and this is the purpose of the conjugate just change the sign in the middle
1:43:17
this right here when we multiply we actually just get a difference of two squares namely a squared
1:43:25
minus b squared so coming back here what we can do is just square the first thing and the square squared will cancel perfect huh so we get x plus five and then in the middle we have the minus and then we look at the second thing which is just a three and we square that three squared is a nine good and on the bottom don't multiply the out because the purpose of the conjugate is to fix the top just keep the bottom as how this and you'll see good things will happen
1:44:00
x minus 4 times all that so square root of x plus five and then plus three
1:44:07
now on the top we have x plus five minus nine you don't have to write it down again and i rewrite the whole thing you can just write the result above and you see this is just x minus four just be neat and then this is clear and you see on the top we just have x minus 4 but guess what on the bottom we also have x minus 4 as a factor so this and that can be cancelled it and then when we have that well we can now just plug in the 4 into the x and then work that out
1:44:39
so on the top we have 1 let's cancel this so we just have square root of 4 plus 5 and then plus 3.
1:44:50
so only oh we have what that's 3 plus 3 so we'll have 1 over 6.
1:45:01
all right so that's the answer for that
1:45:04
okay number 24 here we have the limit as x approaching
1:45:14
2. two two all right x squared minus two x over on the bottom we have a square root so square root of x plus two and then minus two how many twos we have here one two three four five
1:45:37
anybody though have the good habit check this out put two in here and here two squares four two times 2 is 4. 4 minus 4 is 0.
1:45:47
2 plus 2 is 4 square root of 4 is 2 2 minus 2 is 0. so this is yes that's through over super case don't worry count you get again fix the square root part first so i'm going to multiply the bottom by square root of x plus 2 and then plus right plus 2 and then do the same thing on the top right here square root of x plus two and then plus two so let's fix the bottom first this is going to be the limit as x approaching two just like what we said earlier we square
1:46:24
the first part which is just going to be the x plus 2 now and then minus square of the second which is 2 squared which is 4. so that's the fast way to multiply all the counts you get
1:46:39
on the top though i'm not going to multiply the l actually
1:46:44
i'm going to factor this because as i said earlier when we get a 0 for 0 situation you're expecting to cancel out x minus 2 because x is approaching 2. that's how we can end up with that's how we can get rid of the 0 over 0. anyway from here i'm going to factor out an x so keep that in black so here we get x minus 2. so this part is from there this part i'll keep it i'll just write in red so that you can see that this is the part that we had earlier
1:47:18
what's this yeah this is just x minus 2 right what can we do next yeah ladies and gentlemen x minus 2 and x minus 2 cancel very much all right so now we can just go ahead plugging 2 into here so we have a 2 here times square root of 2 plus 2 plus 2. who came out with this question so many twos yes i did anyway this is 2. two is four times two eight
1:47:54
all right done so oh look i remember why i came out with this eight
1:48:02
and twenty-four rest in peace kobe bryant
1:48:09
you think i didn't set this up of course i did
1:48:31
it's hot anyway
1:48:37
number 25
1:48:40
hey you know what we're going to do another limit obviously limit as x approaching zero
1:48:50
on the top we have one plus x squared
1:48:55
minus one and on the bottom we have one plus x to the third power and then minus one yes if you plug in zero you get one minus one on top and likewise one minus one on the bottom so this is a very clear zero first real situation do more work what do we do though well on the top technically it's a difference of two squares so if you are a fan of factoring be my guest on the bottom is a difference of two cubes again if you're a fan of factoring be my guess
1:49:29
you can also multiply the out so if you are a fan of multiply things out the binomial part be my guest likewise the bottom be my guess so i'm going to do it the more natural way that i think most of students will do which is actually just multiply the out usually multiplying out is more
1:49:49
not necessarily natural not necessarily easier but
1:49:54
uh more direct i would say so yeah 1 plus x squared just write 1 plus x times 1 plus x multiply the out we get 1 plus 2x plus x squared and then we have that minus 1.
1:50:08
this right here just to impress you guys i'll show you guys what the answer is what we do is we do 1 to the 3rd power which is 1 plus 3 times 1 to the second power which is 3 then times x to the first power next and then use three again as a coefficient three times one to the first power which is still three times x to the second power and lastly x to the third power so that's the
1:50:40
pascal's triangle or the binomial coefficient so one one one and then one two one one three three one that's why i use three and three earlier for these two coefficients and when i did this technically i was doing one times one to the third power nice
1:51:00
and then plus three times one to the second power the power goes down and then the x power goes up and then plus the next coefficient so it's 1 first power x squared and lastly plus 1 and then x to the third power so that's how i did it pretty much and then minus one so you can take a look
1:51:30
all right this is not so bad you see the 1 minus 1 1 minus 1 pretty good and then the beauty is that we can actually just factor out an x and i think for this one let me just yeah let me just write it down right here so let's see when we factor on x on the top we get 2 plus x yes on the bottom when we factor out x we get 3 plus 3x plus x squared and then the x and x cancel so finally we can put 0 into the x here so 2 plus 0 over 0 into here it's just 2 over three
1:52:15
right because this right here will be zero so yeah so the final answer is just two over three
1:52:22
yeah all the x's are zero so just two over three and now we are done yeah so this is for that and yeah as i said this right here i think it's more natural if you do the um
1:52:38
factoring it might be slightly more bizarre
1:52:44
all right anyway number 26 limit x approaching we have one
1:52:55
x square plus 3x minus 4
1:53:00
over 2x plus 10.
1:53:05
okay go ahead and factor and try to cancel no no no why again have the good habit put the 1 into here let's see what we get so we get 1 square plus 3 times 1 and then minus 4 over 2 times 1 and then plus 10 do we get 0 on the bottom no we don't huh it's the 12 on the bottom on the top what do we get 1 plus 3 is four four minus four is zero so we have three over twelve can we draw a conclusion yes when we have a number when
1:53:48
we have a zero over a non-zero number we can draw conclusion and the answer for that is zero
1:53:54
so in fact the way that you can show work for this question just plug in
1:53:59
because it does not give you any determinant form whatever the next answer is that's it zero done
1:54:09
yeah this question might be slightly more difficult if we are talking about x is approaching negative five because that would make the bottom zero in that case you have to investigate the left and also the right limit but that wouldn't be a 0 for 0 either so yeah so for number 26 it's just like a reminder that hey don't forget to have the good habit that always plug in the value because sometimes when you do a lot of algebraic simplifications it's like huh what do
1:54:38
i have to do with that sometimes simply plug in and then work that out maybe you are done already
1:54:53
27 here we have the limit as x approaching negative three and we have x to the third power plus 27
1:55:05
over x plus three so let's see if we plug in negative three raise that to a third power we get negative 27 we get zero right plus twenty seven zero zero on the bottom we also get zero so zero for zero how do we do it this is a sum of two cubes so i will remind you guys how to factor that when we have a to the third power plus b to the third power the way to factor this is a plus b
1:55:39
times a squared minus a b and then plus b squared okay and in this particular case you can also do polynomial long diffusion or synthetic diffusion i think both of them will work out okay but let me show you guys how to factor this because i have the synthetic diffusion equation for number 28 so for this one that's factory so notice 27 is the same as 3 to the third power so
1:56:13
i'm just perhaps going to write it right here so we have x to the third power plus 3 to the third power and then we can just drawing but let me put on the result right here so this is the limit as x approaching negative 3. so this right here just look we will have x plus 3 that's the first
1:56:34
factor so x plus 3.
1:56:37
to continue we will have x squared so let me just write that down x square
1:56:44
minus a b in this case minus x times three so that's minus three x
1:56:51
lastly plus b squared and that's plus three squared so that will give us nine
1:56:59
so that's the factoring for x to the third power plus 27 on the bottom we still have the x plus three guess what zero over zero no more
1:57:11
then put the negative two into all the axes we are going to get negative three square minus three times negative three and then plus nine this is nine and not nine another nine twenty seven
1:57:31
yeah called barry is a pure coincidence that you get twenty seven all right for number 28 i'm going to show you guys the synthetic division and also the long division together so i'm just going to erase this so that we have the whole board to to show you guys both
1:58:10
all right for number 28 we have the limit as x approaching negative 2.
1:58:18
and we have x to the third power minus 5x minus 2
1:58:25
over x plus 2. as you can see on the top we have a trinomial however it's a cubic it's not quadratic so it's not that easy to factor in the usual way but again let's have a good habit plug in negative 2 into here negative 2 to the power we get negative 8 plugging into here negative 5 times negative 2 is 10. negative 8 plus 10 is negative negative 8 plus 10 is 2 minus 2 we get 0. likewise put the negative 2 here we also get 0 so 0 for 0. we have to do more work so the first
1:59:01
thing is that we can just do polynomial division and to do so i'm just going to
1:59:08
put it inside here x to the third power and then we see we do not have the quadratic term right so i'm just going to say we have plus 0x squared and then minus 5x and then minus 2.
1:59:22
and then here we have the x plus 2 on the outside all right so now we have to ask ourselves what times x will give us x to the third power the answer is x squared so i'll put the result line up right here x squared times that gives us x to the third power take this times that we get plus 2x squared and when we do long division we subtract so this right here we have this this minus r 0 0 minus that we get negative 2 x squared bring down the negative 5x
1:59:57
what do we do next here we have x we need to get negative 2x squared so we need negative 2 x take that multiplied by this we get negative 2x squared take that multiplied by that we get negative 4x
2:00:14
all right subtract negative 5 minus minus become plus so negative 5 plus 4 is negative and then we have the x and then we have that minus two this is x plus two we need that so we need negative one so put a negative one multiply we get negative x minus two subtract we get zero cool so with this being done this right here tells us we just have to compute the limit as x approaching negative two
2:00:48
the whole thing reduces to that and that's x squared minus two x minus one
2:00:58
so what we can do at the end is just go ahead and plug in negative two so negative two in here we get four and then plus another 4 is 8 and then minus 1 is 7. so the answer for this is 7. all right so last point just you plugging then you evaluate so this is another way to cancel the 0 for 0 right because the x minus 2 is gone
2:01:23
yeah okay so how do we do the so-called synthetic division here's the deal
2:01:31
look at the coefficients so starting with one and we do not have the x squared term so just zero and then minus five and then minus two right the same right here one zero negative five negative two now
2:01:49
here we have here we have x plus 2. ask yourself how can you make this equal to 0 x has to be negative 2 right so we are going to do the outputs of this number so just put negative 2 right here and then for the synthetic division we are going to add check this out this is how we do it first bring down the one next we do this times that and then uh let me put it like this we do this times that negative 2 times 1 we get negative 2
2:02:23
and then we are going to add so we get 0 plus negative 2 so that will be negative 2 and then we continue negative 2 times negative 2 is positive 4 and we put it here negative 5 plus 4 is negative 1. next this times that is positive 2 and we put it here negative 2 plus 2 is 0. this number is the remainder very nice why remain the zero
2:02:53
just like this earlier so like normal fraction part so how do we read the answer from here though
2:02:59
originally this is x to the third power right so keep that in mind x plus the power and x square you don't have to write it down by that just keep that in mind when we do the synthetic division these are the coefficients for our answers and the power goes down by 1. so we have the x square
2:03:17
x and also no x so as you can see this part becomes x squared right minus 2x minus 1. so that and of course once you put this back here we do the same thing and then you can compute it and get 7. all right so again just a way to kind of get rid of the rational expression case
2:03:57
you know the hard questions are at the end epsilon delta definitions of the limit and also the riemann sum integrals pretty much definition of integrating right almost like that yeah
2:04:13
okay
2:04:19
oh we'll also be doing the derivative the definition of the property though it's coming uh yeah anyway a couple more to go number 29 here we have the limit and then we have x approaching five
2:04:39
1 over x plus 2 minus 1 over 7 and x minus 5. here we have a complex fraction so how do we simplify this the key is that we don't like complex fractions so i would recommend you guys to multiply the top and bottom by this and that just like the common denominator so i'll put it as 7 times x plus two likewise the bottom check this out this is going to give us the limit oh by the way it will be zero derived i'm not gonna say that
2:05:16
one over seven minus one over seven right and then 5 minus 5 is 0 0. anyway limit as x approaching 5.
2:05:26
take this multiplied by that we will just have the 7 left
2:05:30
then take this multiplied by that the 7 will cancel so we have this minus
2:05:36
and then we have the x plus two but be sure we put parentheses around it because we will have to distribute the negative in a second on the bottom i'm just going to write it here we have the 7 goes first do not multiply out anything so 7 and then x minus 5 and then x plus 2.
2:05:57
okay here's a very common mistake a lot of students once they get to here they see the x plus 2x plus 2 then in parenthesis they will just cancel it this is wrong the reason is because we must fix the top first the word of operations so let's see for the top we will just get seven and then distribute we get negative x and then minus two right
2:06:22
so on all we just get negative x and then this is uh plus five
2:06:35
hey this is x minus 5 this is negative x plus 5. they're almost the same don't worry let's factor out an x so let's factor out negative so let's put negative like so and then we can change this to be x and then minus 5. check negative x negative negative we get the positive so the top right here is negative and then x minus five
2:07:03
now cancel cancel this is the right way to do it on the top we have negative one
2:07:13
on the bottom we're plugging the five here so we see here we have the seven times five plus two which is just seven seven times seven is 49 all together negative 1 over 49
2:07:33
okay
2:07:36
all right number 30. here we have the limit as x approaching zero and then we have one minus cosine x over sine x what how do we even do this wow
2:08:02
how do we factor it no we are not supposed to find it check this out 1 minus cosine squared don't know too much about however we do know that let me just put it down right here
2:08:14
if we have 1 minus cosine squared x this right here is nicely equal to what sine square x yeah so how can we go from here to here well this is precisely 1 minus cosine x times 1 plus cosine x yeah so let's again just do the conjugate business take the top multiply by 1 plus cosine x
2:08:49
and then of course we will do the same on the bottom 1 plus cosine x all right and then for our discussion here when we do this times that we will just get sine squared x that's very nice so this right here is the limit as x approaching zero oh no the top is sine squared x i'm just going to put that down in red
2:09:16
and then we have what here we have the sine x
2:09:20
and then here we have the one plus cosine x oh by the way i forgot to mention the reason that we have to do this is because when we put zero into cosine x is one one minus one is zero one plus two in here sine of zero is zero so yeah it was a zero plus two situation that's why i bought it into this have a good habit you guys anyway once we get to here what we can do is just cancel the sign and cancel that so we have sine of x on the top over that on the bottom
2:09:53
then we can go ahead and put this roon here so we are looking at sine of zero over one plus cosine of zero but you know what sine of zero is zero cosine of zero is one on the bottom is just one plus one which is two zero over two is just zero
2:10:17
so the answer for this is zero and hopefully right now you can see that when we have three over zero the answer is imaginable like you don't know what the answer is yet you can get a negative number not imaginary numbers all by the way i shouldn't use that word and then when we have three over zero it's possible to end up with zero so many possibilities yeah okay
2:10:43
i'm going to take the ripple from my fridge because that ripple is not
2:10:51
cold anymore so i'm i don't know why i took it out yeah anyway
2:11:05
yeah it's the summer that's why especially when you have no ac anyway i'm going to put this down here and also check
2:11:58
all right
2:12:04
okay number 31 you can see that the marker is drying out relatively faster than usual because it's hot anyway here we have the limit as x approaching one and then we have
2:12:21
inverse sine
2:12:24
of the inside is 1 minus square root of x over 1 plus square root of x yeah you'll see why i left a big gap like that okay here's the deal if we put one inside of the x here and here
2:12:47
oh sorry 1 minus x 1 1 minus square root of x
2:12:52
i didn't mean to change the q no i wrote it down wrong anyway if we put 1 to here and here we just zero plus zero and that's just the inside so what we will have to do is the following all right technically we actually do this right this is the question first
2:13:13
let me just redo this right okay here we go this is number 31 we have the limit as x approaching one inverse sine of one minus square root of x no no no no no what am i talking about okay
2:13:27
maybe i shouldn't drink the red bull anyway number 31 the limit as x approaching 1 and we have the inverse sine of 1 minus square root of x over 1 minus x there we go all right if we put 1 to all the x's then we get 0 over zero but that's inside the deal is that we can actually look at the limit first do the inside out what i mean by that is the following this is the same as saying inverse sine of the limit as x approaching 1 and then look at that first 1 minus square root of x
2:14:03
over 1 minus x and the reason i can do this is because inverse sine is a continuous function when x is approaching one in this particular case so this is actually legitimate but anyway though what i'm saying is that you can just do a limiting cell that's pretty much it and for the inside we have done this many times already like the conjugate business let's just go ahead
2:14:32
no i'm not going to do the kanji get you can multiply 1 plus square root of x and also 1 minus square root of x you can do that but i want to showcase something different just so that way you have more tools that you can possibly use later on here's the deal when we have 1 minus x this right here is the same as saying 1 squared minus square root of x squared hostile now this is a difference of two squares so we can say 1 minus square root of x times 1 plus square root of x
2:15:15
yep so in fact this guy is just that which is 1 minus square root of x and then 1 plus square root of x if you do the conjugate you will end up with the same thing like this after the cancellation anyway this and that cancel out very nicely so what we are going to get is the inverse sign
2:15:41
and then we just have to put a one in here and that's on the bottom so don't forget we still have the one and then the one would have been a one plus square root of one and this is just inverse sine of one half now i know this is the hardest part how do we do this the way to do it is you have to ask yourself the original sign of what in goal will give us one half
2:16:09
and that's still hard huh i know sometimes it's hard for me too so i will use the
2:16:14
special triangle sine is the opposite over hypotenuse so here's the deal here is the triangle
2:16:23
one over two we see we have a special right triangle and the angles are what 30 60
2:16:31
this is 30 degrees this is the angle 30 degrees is the same as pi over six so ladies and gentlemen inverse sine of one over two is pi over six
2:16:48
next one number 32
2:16:52
all right here we have the limit as x approaching
2:16:57
one and then here i am going to give you guys inverse tangent of 2 over x minus 1.
2:17:09
okay can we do the same that we did earlier not not really in the following sense because you see if we put one inside here we get two over zero you see plus infinity was a negative infinity
2:17:26
we will have to do what yeah we will have to consider cases right so let's break it down this is not zero plus zero this was zero over zero let me just put it down somewhere but this is not zero plus zero this is two over 0. so we have to do what we did like 2 hours ago anyway x approaching 1 minus first
2:17:55
so what exactly do we get if we do one minus we put it here so we're looking at inverse tangent of 2 over 1 minus and then minus 1. so this is approaching what 1 minus 1 is 0 but this is one minus so it's zero minus right so this is inverse tangent of two over zero minus two is positive zero minus is negative so the inside will approach
2:18:27
negative infinity so inverse tension of negative infinity is what this will approach negative
2:18:34
pi over 2. so this right here gives us negative pi over two all right okay and let's see if we have the limit as x approaching one plus inverse tangent of two over x minus one
2:18:54
well you can see i'm just going to change this right here this is going to be one plus one plus minus one is zero plus two over zero plus is positive infinity and then inverse tangent of positive infinity is positive pi over 2. so this right here gives us positive pi over 2. unfortunately they are not the same therefore this limit does not
2:19:21
exist and in fact in this particular case i don't remember i forgot to mention this earlier in this particular case we actually have a so-called a jump
2:19:38
this continuity at x equal to 1 why because as you can see when x is approaching 1 from the left-hand side it does have a limit but that limit is not the same as when we approach the
2:19:55
one from the right-hand side these two limits have to be finite and they have to be different to inverter for us to have a jump this kind of continuity the picture of this should look like the following i'll just give you guys a picture real quick one is right here when x is approaching one negative is negative pi over two right so you will get
2:20:20
and then when when x is negative infinity we get zero so yeah you'll get like this and then negative pi over two and then like this yeah so there there's a jump it's very similar to
2:20:41
oh man yes it's actually number question number two this is a jump right here too it's a jump
2:20:48
when when x is at two right there so yeah i forgot to mentioned it
2:20:55
anyway
2:20:57
that's number 32
2:21:02
so sometimes when x is approaching to some number uh you don't have to do you may not get 2 over 0 but if you get like two over zero like five over zero then break down two cases like the things that we did earlier so those will be helpful
2:21:35
oh
2:21:43
number 33
2:21:46
here limit as x approaching 5 and then we have 1 over x squared minus 25
2:21:56
so 1 over x squared minus 1 over 25 i forgot what i said and then on the bottom we have x minus sorry we have x squared minus 25. cool do not cancel this and that they are very different fix the complex fraction first and yes this is a zero over zero situation okay check it real quick by the way though let's multiply the top and bottom by this and now which is 25 x squared
2:22:30
so we are going to get the limit as x approaching five this times that the x squared will cancel so we get 25 minus this times data 25 will cancel so we have x we have 25 minus x squared
2:22:49
on the bottom you guessed it we are not going to cancel things we are not going to multiply things out let's just keep it as how it is 25 x squared times x squared minus 25
2:23:06
cool now can we cancel this and that no not yet
2:23:13
this is kind of out of water we like to have the x square goes first right and we can do so by just changing right factor out negative and then we can say this is x squared minus 25 so this and now the same now we can cancel the x squared minus 25
2:23:30
yeah so what do we have on the top we still have the negative and it's a negative one on the bottom we have this 25 times 5 squared which is just 25. so what's 25 times 25
2:23:50
i'll show you guys the fast way to do it whenever you are multiplying two numbers if they end with a five yeah you can do the following five times 5 is what 25 okay you do 2 times not 2 but times the next number which is 3 2 times 3 is 6. so 625 so it's negative one over 625
2:24:17
okay i have a video on that
2:24:24
pretty cool stuff you guys can go and check that out on my just algebra channel
2:24:40
all right ladies and gentlemen number 34 limit as x approaching 8 and then we have the cube root of x minus 2 over x minus 8. wow how do we do this
2:25:01
we can multiply by the count the count you get to kind of complete the the to get rid of the
2:25:09
cube root that's that's possible
2:25:14
but let's look at the following all right firstly we can factor this as a difference of two cubes i will show you guys that on the side when we have a to the third power minus b to the third power this right here is equal to a minus b times a squared plus a b and then plus b squared
2:25:45
what's x x is the same as the cube root of x to the third power what's 8 it's the same as 2 squared i mean 2 to 3 power so i'm just going to put on the factoring right here just i don't want to write this down just put it here so x minus 8. apply the formula we will first get cube root of x minus 2 and you see the cancel already huh and we just have to figure out the rest which is
2:26:18
okay we have to square this so i'm going to just put it down like this the cube root of x yeah and then we have to square it
2:26:28
and then plus a b so we have to multiply these two together so that's 2 and that so it's 2 and the cube root of x and lastly we have to put a b squared which is 2 squared which is just a plus 4.
2:26:46
so right here we have the plus four very nice huh cancel cancel
2:26:56
all right now let's put the eight into here here and
2:27:03
we have one on the top over here cube root of eight and then we square that and then plus two times the cube root of eight and then we add the four
2:27:19
this is two inside square that we get four this is two times two is four
2:27:29
1 over 12.
2:27:33
done ladies and gentlemen that is number 34.
2:27:40
all right i'm going to do one more question with snap with this marker and i'm going to change to a better marker so that way you guys can see uh you guys can see this better i have a brand new marker over here and then later on when i do like smaller questions
2:27:59
and sometimes if you just let the marker sit you know it's going to come back alive
2:28:05
if you mean if you just put a marker like this for a while then yeah it will come by life and you can solve like two or three integrals before it tries out again
2:28:19
but anyway number 35 this is called nasty square root because you have a square root inside the square root limit x approaching 9 and we have
2:28:35
square root 1 plus square root of x and a minus two over x minus nine here's the deal
2:28:45
plugging nine into all the x's yes you guessed zero over zero
2:28:51
i don't remember if i checked earlier yeah you it was obvious you are zero plus zero for number 34. all right for number 35 here let's go ahead and multiply by a conjugate first to get rid of this square root so i'm going to multiply the top and bottom by square root of 1 plus the square root of x and change this minus to a plus and then the two stays and then let's do the same on the bottom
2:29:18
cool now this right here is the limit as x approaching nine
2:29:26
how do we multiply out the conjugate just square the first thing so the the big square root will cancel so we have one plus square root of x what's next minus 2 square that which is 4 good on the bottom don't do anything just write it down so we have x minus 9 and then square root of one plus square root of x and then plus two good on the top we can actually just do one minus four which is square root of
2:30:03
square root of x well one minus it's just this square root of x minus three
2:30:11
yeah what we'll do next
2:30:15
do you want to multiply the conjugate again sure be my guest but you know it this guy
2:30:23
x minus 9 is the same as square root of x squared minus 3 squared isn't it
2:30:33
so we can factor this as square root of x minus 3 times square root of x plus 3.
2:30:44
cancel the top what's that no more zero over zero yay finally
2:30:51
one over don't forget this we have a square root of x is nine just plug in and then plus 3. so this is the first factor and then square root 1 plus square root of 9 and then plus 2.
2:31:12
just work this out this is three plus three is six times this is three plus one is four square root of four is two two plus two is four so we have one times four we have six times four and one over twenty-four
2:31:36
yes i did it right i believe so
2:31:47
yeah yeah by the way nice number 35 when we have double square roots you can do the conjugate twice or just take advantage of like how we can factor it like the waist i show you guys
2:31:59
use radicals and then the difference of two squares what may be the difference of two cubes
2:32:14
page four done oh this sheet of paper is done moving on to page five all right for now we'll be talking about the limit as x approaching infinity okay
2:32:46
all right number 36 here we have the limit and then we have x approaching infinity
2:32:54
and then we have cosine of inverse tangent of x
2:33:01
okay so for this kind of questions um it really depends on your teacher your professor your ta to see how they want you to show work uh this is how i will love my students for how i will show work to help students navigate the computation you can just plug infinity into the x and just kind of recently along the way so we are going to get cosine of inverse tangent of infinity
2:33:29
and just have to remember that we are taking a limit so then that will be okay and then just draw arrows because you are taking a limit so inside out inverse tension of infinity meaning the limit as x approaching infinity of inverse tangent the inside here gives us pi over two so we have cosine of pi over two and this is nice because cosine of pi over two nice and sweet and nice right shorter than three what's cosine of pi over two zero yeah
2:34:05
so the answer for this right here is just zero and then we are done just inside out inside l and of course for this kind of questions you will have to remember the limit as x approaching infinity like the things i told you guys on the first page use the graph or the spatial function and all that number 37 uh switch this a little bit
2:34:28
limit x approaching infinity inverse tangent of cosine of x
2:34:38
well can we plug infinity into cosine x what's this if we uh cosine of infinity
2:34:51
what does this mean this right here does not exist
2:34:57
why because remember cosine looks like what right so the limit as x approaching infinity of cosine x this right here is just dna all right done if the inside is dna already you don't need to worry about what's after this it's not possible to do so this right here is dne
2:35:23
it's not because the inverse tension of the n is dnu kind of like that but don't do that this is enough much better that way
2:35:31
number 38 everybody will love this kind of question
2:35:35
limit as x approaching infinity we have 2x to the third power
2:35:45
minus 5x minus 3 over oh x squared minus 2x minus 3. sorry this right here meant to be it's a square
2:36:00
does this look familiar it should huh is that yeah it's number 22 isn't it except for the fact that for number 22 it was x approaching 3 but right now it's x approaching infinity when we have x approaching infinity much easier we just have to care about the dominating part on the top and also the dominating part on the bottom in this case we just have to worry about the biggest power of x on the top which is 2x squared on the bottom we just have to worry about the x squared
2:36:34
reduce we get two and then we are done this is the limit that you do not want to miss
2:36:43
all right why does this work i'm going to show you guys
2:36:49
uh i just want to make a remark first though if you are taking my class if you do this i'm totally cool with that but if you are taking a class uh with a different professor if your professor
2:37:01
wants you to do the following then just do it this way here's the deal when we are taking the limit as x approaching infinity and keep in mind this is x approaching infinity that's why we didn't even need to bother to factor whatsoever we just had to pick out the dominating part when we have x approaching infinity let me write this down again we have 2x squared minus 5x minus 3 over x squared minus 2x minus 3. when we have this look at the biggest power of x and that is x squared
2:37:37
and what you want to do is divide everybody by x squared divide everybody by x squared
2:37:47
and once we do that we are looking at the limit as x approaching infinity this is out it's just a 2 but then we will have minus 5 over x and then minus 3 over x squared and then that would be just 1 and the minus that is 2 over x and then lastly minus 3 over x squared as x goes to infinity this has no xo it has to be 2 but 5 over x this will approach 0 because we discussed this before 5 over infinity will approach zero you can just a minus one
2:38:29
it's just zero similarly this will also approach zero likewise zero likewise zero in another word these parts they don't really matter it's just a 2 and also the 1. reduce that we get 2. done
2:38:50
however if the power on top let's say we have x plus 3r power let me just
2:38:57
probably give you guys like a small bonus on the side
2:39:07
before we continue that's number 38 so number 38 with a star next to it just like a smooth variation if we have the limit as x approaching infinity if we have 2x to the third power and minus 5x minus 3 over x squared minus 2x minus 3.
2:39:31
the dominating part on top is 2x to the third power and the dominating part of the bottom is x squared the highest power of x pretty much you see that degree right here the power here is bigger than the bottom if you reduce you just have x as x goes to infinity this right here will go to infinity if the degree of time is bigger the whole thing goes to infinity
2:39:52
however number 38 starts the heart if we have the limit as x approaching infinity 2x that's it's the power minus 5x minus 3 by the bottom is x to the 6th power and the minus 2x minus 3.
2:40:11
you see now the degree on the top is 3 the degree on the bottom here is 6. this is bigger than that if the bottom is bigger you get zero because you can reduce it right you get one over x to the third power after you reduce you of two over x plus zero power as x goes to infinity you get zero so here are the three situations you just compare the degree aka the highest power of x on the top and also on the bottom if they are the same reduce the coefficient if the top is bigger
2:40:49
then the answer will be infinity if the bottom power right here is bigger then you get zero that's it these are the limit questions that you do not want to miss on your calculus test
2:41:25
all right so that was 38 now let's look at 30 9
2:41:32
here we have the limit as x approaching infinity we have inverse cosine and then we have square root of 3 x plus three over two all x minus nine
2:41:51
so let's do this study fast way x is approaching infinity we can do this inside out so just look at the inside first so because x is approaching infinity that's exactly what we talked about earlier so just look at this because that's x first power likewise that so in fact they cancel just square root of 3 over 2. so this right here is just inverse cosine of square root of 3 over 2.
2:42:19
now the hard question is what's the inverse cosine square root of 3 over 2 picture time we have a right triangle square root of 3 is adjacent and then the 2 is the hypotenuse because we are talking about cosine yeah so this is the angle
2:42:38
it's again 1 2 square root of 3 so this right here is pi over 6 30 degrees
2:42:46
this is pi over 6. done
2:42:58
all right next we have the limit
2:43:02
as x approaching infinity and then we have let's do this one ln of 2x to the third power plus 9
2:43:16
over x to the fifth power minus 12.
2:43:24
again we can do this inside out and this is x approaching infinity so here let's just focus on 2x plus 3 power and then x to the fifth power the degree on the bottom is bigger so we get zero so this right here is uh i i should write it on this just marker here inside here we get zero right and then we have ln but l one zero is not really a thing but technically this inside is positive because when we have x approaching past infinity
2:43:56
so this zero is technically positive infinity so what is l n of zero plus what did i say earlier
2:44:09
this is true plus and then what's the ln of zero plus yeah there we go this divided by the zero plus sorry and then what's the l naught of zero plus negative infinity good
2:44:21
so yeah you can just reason out this kind of questions in no time that's why i said if you want to show work for this it might be harder than it has to be if you do this kind of things i think i will be happier usually i like to ask this kind of questions in the multi-purchase format so that way you guys don't have to struggle and you guys don't have to worry about how to show work you just have to determine the answer
2:44:50
yeah i know it's just little thing right but my head my arm it's still very tiring 41 have a look at the limit as x approaching infinity sine of x over x what's the answer for this
2:45:11
we did this for question number five sine of x is seen between negative one and one is finite as x goes to infinity the bottom is infinity so if you have a finite number divided by infinity yeah you get zero for sure but if you want to do this a slightly more legitimate way you do you do it like this this is called the squeeze theorem the idea is that you first have to mention that okay sine x is in between of negative 1 and 1. so we can write this down and this is our inequality
2:45:48
cool and now we can divide everybody by x so that's what we are looking at right so divide everybody by x like so good and then we will take the limit as x goes to infinity
2:46:06
for everybody and if you do that you can see that this is the same we have
2:46:12
zero here and then one over one over x x goes to infinity is also zero and then in the middle is exactly what we're looking at the limit as x approaching infinity
2:46:25
sine x over x and as you can see this limit is in between of zero and zero therefore it has to be zero done with that squeeze it has to be zero done
2:46:41
all right number 43 i'm going to do 43 44 45 hopefully
2:46:49
not possible i don't think it's possible from those three questions on the same board
2:46:56
maybe we'll try
2:47:04
number 43 we have the limit as x approaching infinity
2:47:10
and we have square root of x
2:47:15
plus 3 minus square root of x plus so x minus 1. i'm not going to close the parentheses for this one let's see when we put infinity into here and here square root of infinity plus 2 just infinity square root of infinity is infinity and the reason is because if we have infinity for the base raised to a positive exponent this right here will give us infinity and the square root is the same
2:47:46
as saying one half power so that's why square root of infinity gives us infinity and then minus same idea infinity so we are talking about a new kind of indeterminate form for this video infinity of infinity the answer is not necessarily zero what is infinity minus infinity in determinant form
2:48:10
it's right here anything that you see right here is indeterminate meaning that we cannot draw any conclusion yet
2:48:20
well how do we do it just like what we did before when we see square roots use the conjugate and give it a try let's multiply the top and bottom that's why i didn't close the parenthesis let's multiply the top boundary pattern by square root of x plus three change this to a plus and the square root of x minus one and then let's do the same thing on the bottom square root of x plus 3 plus square root of x plus 1. cool now this is the limit as x approaching infinity
2:48:58
the bottom stays the same huh so square root of x plus three plus square root of x plus one
2:49:06
on the top you just square the first term which is just x plus three and then minus square the second time which is x minus one but you should put parentheses the first parenthesis doesn't matter but second parenthesis doesn't matter all right and then we can simplify this this is the limit as x approaching infinity x minus x is gone three minus minus which is four so it's four over square root of x plus three plus square root of x plus one what's the answer for this zero
2:49:47
why because the top as you can see is just a four it's finite but if you have the infinity on the bottom you will get infinity plus infinity which is infinity four over infinity we get zero
2:50:03
okay so the x and minus x comes so yeah cool so in this case yeah my goodness infinity minus infinity we did get zero do you have anything wrong don't worry have a look let's look at number 44.
2:50:22
oh i'm going to change a little bit huh i'm just going to change a little bit let's look at the limit as x approaching infinity square root of 2x plus 3 and then minus square root of x minus 1. right so i'm just going to change this a little bit well i have a 2 right here instead okay
2:50:45
and it's the same idea we also have infinity minus infinity so i'm not
2:50:53
going to say too much about it we'll just go ahead and get work so multiply the top and bottom by conjugate which is square root of 2x plus 3 plus square root of x minus 1 and then same thing on the bottom square root of 2x plus 3 and then plus square root of x minus one cool same thing under like earlier here we have the limit as x approaching infinity all right so here we have square root of two x plus three and then plus square root of x minus 1 what's on the top
2:51:36
square this minus square of that so we have 2x plus 3 and then minus x minus 1. okay let's see what exactly we are going to get this time this is the limit as x approaching infinity x sorry 2x minus x is x so we have x on the top now huh 3 minus negative 1 is 4 and that's a positive 4 so x plus 4 over those things square root of 2x plus 3 plus square root of x minus 1.
2:52:15
what's the answer this time i'll tell you the answer for this one right here is infinity
2:52:26
all right again x is approaching infinity so pick the dominating part on the top which is just x to the first power on the bottom pick out the dominating part the plus 3 doesn't matter so we have square root of 2x and then from this we also have the square root of x yeah so technically we can look at this as square root of 2 times square root of x plus square root of x and we can combine them as square root of 2 plus 1 times square root of x this is just a coefficient
2:53:03
but the main thing is we have x over that this is x to one power this is x to the first power
2:53:14
this is x to the one half power which one's bigger top is bigger whole thing is infinity told you and as you can see infinity minus infinity a we end up with infinity all right quick summary
2:53:32
the number 32 we get 0 the answer for number 33 is 0. this tells us this expression and that expression are about the same size as x goes to infinity okay however
2:53:48
the answer for number 44 is equal to infinity it means this expression is also so much bigger than that as x goes to infinity so that's why when you subtract this infinity is way bigger than that
2:54:04
so not all infinities are the same size yeah
2:54:11
all right i'm also going to show you guys number 45. it's another infinity minus infinity situation and we will actually get a very interesting value it's a finite value it's not like a super cool number like e to the square root of 17 for something no it's just and you'll see
2:54:41
number 45
2:54:44
here we have the limit as x approaching infinity and then square root of
2:54:55
x squared plus six x plus one and then minus square root of x squared plus 1. and notice this time the inside here they are both quadratic earlier they are just both linear
2:55:12
yeah anyway if you put infinity into this and that we get infinitely minus infinity all right so just want to mention that real quick but you know the deal is that we will just multiply by the conjugate so let's go ahead and just get work so square root x squared plus 6x plus 1 and then plus square root of x squared plus 1 over the same thing so square root of x squared plus six x plus one and then plus square root of x squared plus one
2:55:51
okay this right here gives us the limit as x approaching infinity on the bottom we have square root of x squared plus six x plus one plus another square root of x squared plus one
2:56:11
you know why calculus is hard it's because a lot of times we just have to write the same thing over and over and over yeah but anyway though bring this down now multiply this out we just have to square this which is just x squared plus 6x plus 1. you don't need a parenthesis for that by the minus you do need a parenthesis for this x squared plus 1. all right so now check this out x squared x squared the cancel we have 6x on the top let me just put on the result here
2:56:44
and then 1 minus 1. i i i i'm just going to rewrite everything again to make it super clear so if we cancel the top this is the limit as x approaching infinity we will just get six x on the top one minus one is also gone and then all this
2:57:09
okay picked up the highest power of x sometimes it's just this 6x to the first power but guess what on the bottom this is the highest power right here and square root of x squared
2:57:25
what's square root of x squared yes we can cancel them
2:57:31
but if you just cancel them you actually have to put an absolute value because the output of a square root is always positive quick example when we have square root of negative 4 squared you do this inside out this is 16 right and then take the square root you get 4. so you see if you have square root of negative 4 squared if you just cancel them and then just say the answer is negative 4 this is wrong you have to include the absolute value and then you see this
2:58:04
is in d equal to four now okay however you don't need the absolute value if you know the inside if you know the x is positive already x is approaching positive infinity so this part is just equal to x and that's legitimate just x no square root needed likewise here this is also equal to x so guess what on the bottom we actually just have x plus x wow
2:58:36
so technically to get down the bottom which is of 2x
2:58:40
so 6x over 2x the x cancel 6 over 2 the answer is three ladies and gentlemen i told you
2:58:49
infinity minus infinity a we also give nice number three can you get pie yeah if you manufacture a good example yeah you can get pi can you get e to the 17 power minus 2 yeah if you would like you can try it manufacture equation it's possible but again infinity minus infinity is not always equal to zero we cannot draw any conclusion unless we do more work because that is an indeterminate form
2:59:26
right number 45 let me just take a look at uh
2:59:32
wow three hours already
2:59:41
definition of derivatives so
2:59:46
i wanted to tell you guys that for number 45 sorry for number 46 to number 50 we'll use the definition of derivative to calculate the derivative of a function as some number so i will write this down for you guys if we have f prime of a this right here i call this definition one
3:00:07
oh i'm just going to write it like this yeah so i call this definition one just put one over here this right here is the limit as x approaching a and then we have f of x minus f of a
3:00:23
over x minus a all right so it's pretty much the difference of quotient and then you let x approaching a so that's how you can find the slope of the tangent line
3:00:35
secondly this is why i call the definition of two they are equivalent is the one with h the limit as h approaching zero and for this one we get f of i forgot
3:01:07
now i remember
3:01:11
okay i know my merch yeah i designed this i have a t-shirt and also a canvas print yeah so i'll put it on the side
3:01:26
okay all right anyway f of a plus h minus f of a yeah all over h okay so definition one definition two let me just talk about them briefly this right here it's easier if you like factoring this right here is easier if you like expanding so that's the advantage and also yeah i have another video that kind of comparing both definitions so you guys can
3:02:04
take a look at that video on my another channel just calculus
3:02:12
anyway what are we doing okay number 46 let's say our function f of x is equal to a 4x to the third power 4x to the second power plus 3x we want f of negative 1 f prime of negative 2.
3:02:35
okay ladies and gentlemen number 46 f prime of negative 2 let's use the first definition
3:02:42
this is just the limit as x approaching negative 2 and you see f of x is just that so we can just put that here 4x squared plus 3x and then minus we need f of negative 2. so let's put it on the side let me that's what you should do here f of negative two
3:03:02
means we're plugging negative two into this x and not x so we get four times negative two squared plus three times negative two this is four times that is sixteen
3:03:18
yes sixteen minus six is ten
3:03:25
yes so this right here is minus 10. so f of negative d2 is 10 and then on the bottom we have the x minus a is negative d2 so minus minus becomes plus all right cool and now if you take a look right here doesn't this remind you of the algebraic limits that we did yes is there a fast way to do this yes but let's do the things that we did earlier okay
3:03:55
i'll tell you if you put negative 2 into all the axes for sure you get 0 over 0 why because we stated that to be 0 for 0. when we put n into this x and x it was f of a minus f of a zero a minus a zero likewise here zero for zero if you don't get zero over zero then that means you study or wrong
3:04:17
anyway though how do we simplify this we can actually just factor it so let me see let me put it down on the side four x squared plus three x minus ten
3:04:33
uh no common factor so we'll just do a tic-tac-toe right away so you know the deal we need to have x and 2 right here 4x
3:04:45
2 times 4 will be negative 10 is negative 5. why does this work we need to get rid of that right so that's it but check it out check this out x times negative 5 is negative 5x and then 2 times 4x is 8x together we do get a 3x that we want
3:05:05
okay so this is the limit as x approaching negative 2 on the top is just that namely x minus so x plus 2 times 4x minus five on the bottom is still x plus two over although we see that this and that can cancel finally put a negative two into here and we will get
3:05:33
four times negative two and then minus five negative eight minus five is negative thirteen done cool but wow that's a lot of work
3:05:48
i wonder if there's a way for me to know if we did this right or not
3:05:54
yes of course we can do this question in like five seconds check this out
3:06:01
right this is the definition of derivative this is the so-called power rule i will have to mention this along the way because coming up next we will be using the laptop's rule which we need the derivative
3:06:17
but if you are just learning derivative for the very first time then please use the definition anyway this is how it goes start with f of x equals 4x squared plus 3x and 3x is like the 3x to the first power i'll show you how it goes to get f prime of x whenever we have
3:06:40
x to some power right here now on the bottom not just right here huh if you are multiplying by some number that's good that's cool you can put the num put the power to the front and multiply this
3:06:52
multiply the two with that and then minus the power so my minus 1 to the power so this right here will give us 4 times 2 is 8 x to the first power can we do the same thing for this one yes put a 1 to the front 3 times 1 is 3 and then don't forget to minus 1 to the power 1 minus 1 is 0 x plus 0 power is 1 so this is our f prime of x it's so cool because now we can just plug in negative 2 that we wanted
3:07:27
so f prime of negative 2 is 8 times negative 2 plus 3. you bet it is negative 13. done
3:07:38
power rule but don't be too happy about it because knowing the power rule is not knowing all the calculus this is a nice result after the
3:07:49
definition of the referee i'll say
3:07:54
all right i want to make sure that i didn't i have to make sure that i am not erasing the definitions because i would like to have them on the board until the end of the category until this category
3:08:16
if you guys need help with derivatives yes you know it we have the 100 we have had 100 derivative took me a long time to do that as well anyway number 47
3:08:36
okay function x over x plus two we want f prime of three hey can we do the power rule
3:08:51
if we put a one one bring it to a one one minus one zero x plus two is one so we just get one and then one plus two the answer is just one third no no no no no no don't do that that's why
3:09:11
the definition of the robot is the key all right especially to take care of us when we're trying to do something new the limit as x approaches 3 right we're following the first definition
3:09:26
we get x minus 3 and then on the top f of x is that so it's just x over x plus 2 and then minus we will have to put f of 3 yeah so i'll just indicate that right here f of 3 is just three
3:09:45
over three plus two so it's just three over five so it's minus three over five
3:09:53
okay as we can see we have a complex fraction situation therefore we are going to do the things that we did right that's why we talk about the algebraic ways to 2d limits i'm going to multiply the top and bottom by this and that so it's 5 times x plus 2 and likewise here 5 times x plus 2.
3:10:21
all right here we are going to get the limit as x approaching 3. let me just keep this how they are so 5 and then x minus 3 x plus 2. now let's multiply the top this times that x plus 2 cancel so we have 5x
3:10:42
minus 5 and 5 cancel but we have the 3 and also the x plus 2. am i right let me just double check
3:10:59
okay cool now i had to leave space on the top so i'm just going to kind of do it on the side so you see this is just 5x minus 3x and a minus 6 yeah so just 2x hold on oh no no 2x minus 6. so this right here is 2x minus 6 and then we can factor out the two so 2 times x minus 3. okay so the top one so the top part is that so let me just rewrite this
3:11:37
so we get 2 times x minus 3 and see we can cancel things out on the bottom so 5 times x minus 3 x plus 2 very nice huh x minus 3 cancel now put a 3 into here so we are going to get 2 over 5 times 3 plus two that's five times five is twenty-five so all together we get two over twenty five done
3:12:14
it was not like power rule yeah you have used the so-called quotient rule but we're not going to get into that in this video we have the 100 we have the 100 derivative for that already so go ahead and check that out
3:12:45
all right number 48
3:12:48
let's talk let's take a look at this squirrel function f of x equals square root of 2x plus 5.
3:13:01
and then we want f prime of 2. and let's just use the first definition um i think it's more straightforward more direct this way yeah but this one also works don't worry we will have a chance to practice this data on ufc anyway this is the limit as x approaching 2 and then here we have our function square root of 2x plus five minus okay we need the f of two so let me write it down here
3:13:33
f of two equals square root of two times that two and then plus five that's nine take the square root is three so it's two right here and over x minus two aha once we set this up it becomes
3:13:52
what's that called it becomes that algebraic limit question right so let's go ahead and just do the count you get multiply the top and bottom by square root of 2x plus 5 plus 3 and then let's also do the same on the bottom square root of 2x plus 5 plus 3
3:14:10
okay now this is the limit as x approaching 2.
3:14:18
okay square this so we just have 2x plus 5 and then minus three times three squared which is nine and then the bottom stays so x minus two times all that
3:14:36
all right the top right here let's just do it on the side real quick that's 2x minus 4 and then we can factor out a 2 right so this this so i'm just going to squeeze out the result here 2 times x minus 2. okay and then i'm just going to cancel the x minus two x minus two perfect all right we have this two right here though be really careful so two over
3:15:03
put this two in here so we have square root of two times two plus five and then plus three
3:15:16
that's nine square root of nine is three two plus three is six two over six is one third is it
3:15:25
yeah okay all right so now
3:15:31
that's yeah that's okay let's just do this with actual numbers uh for the derivative for the uh derivative technique uh it's coming anyway so don't worry that's practice oh what two more questions okay anyway though
3:15:54
number 49.
3:16:06
if you are still watching leave your time stamp right now
3:16:11
appreciate it you're working hard i am working hard too okay let's see number 49
3:16:21
f of x equals x to the third power plus 4 and then we want f prime of one so use the power rule for this later on but anyway though if we want f prime of one
3:16:36
as i mentioned here a couple minutes ago or so if you use this definition it's easier to you have to factor it if you like factoring and you use this then you will have to uh expand it now let's just do this one so this right here is the limit as x approaching one function is x to the third power plus four and then we need to know the f of one right so f of one is just equal to one to the third power plus four which is 5. so this means we have x 3 power plus 4 and then minus 5.
3:17:22
yeah and then on the bottom we have x minus one yeah it's not bad to factor this as a whole so this is good limit x approaching one on the top is just x to the third power minus one and on the bottom is x minus one okay this is the same as one to the third power so factor it difference of two cubes we get x minus one times x squared plus 1 times x which is x and then plus 1 squared which is just 1 and then on the bottom is x minus 1 and then x minus
3:18:06
one cancel ladies and gentlemen put a one here here one plus one plus one we get three yay cool
3:18:17
now let's see if we can finish these questions in three seconds no like that's five seconds i'm talking about the question also it doesn't count ready
3:18:29
okay f prime of x bring the power to the front minus one we get three x square well this is just a 4 but if you use the power rule you can look at this as x to the 0th power put the 0 to the front
3:18:45
0 times 4 is 0 so it's no more so just like this f prime of one three times one square and we get three wow wow powerful power rule so powerful isn't it okay
3:19:27
oh
3:19:34
all right number 50.
3:19:39
here we have our function is just one over square root of and we want f prime of 9 so there we go f prime of 9. this is my 9 this is my a here we go limit x approaching 9 our function is one over the square root of x and we have minus f of nine which is just one over square root of nine and that's just one third okay not bad at all so minus one third
3:20:26
let me just write it down better for you guys what's on the bottom x minus nine yay
3:20:34
okay here we have two things to worry one we have a complex fraction a small fraction instead of a big fraction second thing is we have square root yeah by the way let's take care of the complex fraction first multiply the top and bottom by this and that is 3 square root of x
3:20:52
okay so we see this is the limit as x approaching 9. this times that square root of x will cancel so we have 3 and then minus 3 cancels so we have square root of x
3:21:09
i'm just going to keep them as how they are so we have 3 square root of x times x minus 9.
3:21:19
this is square root of x my bad all right so for the top i'm going to change the order of subtraction this is negative square root of x minus 3. i switch to water and put a negative on the outside for this let's not do the conjugate anymore so just look at this in the factory we get square root of x minus 3 times square root of x plus 3.
3:21:48
all right
3:21:50
so now you see this and that cancel out completely very nice so ladies and gentlemen on the top we have negative one over here we have three and here we have square roots of x x is nine
3:22:08
that's gone and then here we have square root of nine plus three yeah and now just have to work this out this is three times three times three plus two is six
3:22:23
nine times six is fifty-four so all together we get negative one over fifty-four
3:22:31
done if we get a small bonus right here in fact we can also use the power rule for this question check this out when we have our function this is 1 over square root of x in calculus a lot of times we look at the square root as the one half power so this right here is the same as saying
3:22:53
x to the one half power but we can bring that up so this right here is actually just x to the negative one half power okay and the reason that we want to do this is because we can use the power rule check this out put the power to the front and then minus one so we get f prime of x and we get negative one over two at the front and then x this minus 9 is negative 3 over 2.
3:23:22
yeah what's next well we want f prime of 9 yeah so f prime of 9 this right here gives us negative
3:23:34
now let me just write it down negative one half and then we have 9 and then to the negative 3 half power the main question is how do we simplify this if we can do it then we are done yeah so usual thing put the power right if it's a negative power then put it to the bottom so this right here is negative 1 on the top over two
3:23:59
and then once we bring this down to the bottom we have the over two right here right let's change that to a square root and then we have the nine and then raise that to the third power so that's pretty much how you do it again this negative bring this down to the bottom the over two power change to the square root and then the third power stays and you see square root of nine is three three to the third power is 27 times 2 is 54 and that's on the bottom oh no negative 1 over 54
3:24:34
how cool is that huh
3:24:44
yes no i'm not dying yet i know my lights
3:24:54
it's not funny yeah still have like 15 more to go 51 more to go
3:25:04
actually i don't know if it's 50 or 51 i'm not ready
3:25:10
go see you'll see
3:25:16
all right next we want to do this 51 we have given the function f of x and this is just a regular polynomial it's quadratic we have three x squared minus five x plus seven instead of saying f prime of like 2 f of prime of negative 7 we want f prime of just x
3:25:42
so now here's the trouble can we still use definition 1 kind of if you put this x
3:25:51
you if you make this x red then you are talking about the limit as this x approaching this x right this axis let's say it's red then we can just replace the a with the red x
3:26:05
but you know this does not it doesn't look good at all right it does not look good at all so no no no that's why we have definition too if we want to do f prime of x then we look at this right here and just replace the a with x and it's still a jet
3:26:28
yeah see it's the official definition of the derivative
3:26:35
okay so i am just going to
3:26:44
now i'm just going to write this right here here we go we're looking at definition 2 and just change the a to x that's all you have to remember so here we have the limit as h approaching zero and this right here it's not easy
3:26:58
this is perhaps the one of the first things that people dislike about the calculus part is pretty demanding also anyway we want to have f of x plus h so what that means is we want to look at all the x's here and then plug in x plus h i'll show you we will first get 3 times x plus h and then square that and then you continue minus f of x so minus 5 of
3:27:30
minus 5 times x plus h and then plus 7 so that's the f of x plus h part and then we want to minus f of x which is just the original so just minus and then we just put this down 3x squared minus 5x plus 7. and you can see that this expression is pretty long already and um perhaps it's the longest especially if you just come from pre-count and yeah pretty intimidating so that's the idea but for now i'm just going to erase both of this so that way we can
3:28:11
actually just put down the definition of derivative for f prime of x and then we'll focus on that so let me do that for you guys right here so ladies and gentlemen f prime of x this right here equals the limit as h approaching zero we have f of x plus h
3:28:35
minus f of x and then over h and this is definitely a definition that you have to remember for your calculus class so how do we take of this well we will have to expand it that's why i told you guys earlier i just used definition one because now we will just be doing this right here
3:28:58
just do it all right here we go this right here we have the limit as h approaching zero open this we get a string in the front and then we have let's just put this down in black x square plus 2 x h and then plus h squared and then distribute we get minus 5x minus 5h and then this is the plus 7. distribute we get negative 3x squared minus minus becomes plus 5x and then minus 7. and over h but don't forget we still have this two to be distributed let
3:29:38
me just put it on the top here we have three x square plus six x h and then plus three h square
3:29:49
now here's the best part we can cancel things hopefully let's see
3:29:55
3x squared negative 3x squared cancel good negative 5x positive 5x cancel good 7 minus 7 no more good this this and that remains on the top and notice everybody has an h we can factor that out so we will here we have the limit as h approaching 0. factor out the h we get first thing 6x second thing plus 3h and the last thing right here we have the minus five don't forget to have this remainder and then all divided by h now cancel out the h and this is the
3:30:43
reason why i told you guys you should keep writing down the limit as h approaching 0 when i acquire versus approaching however because now you see how we have both x and h in the expression
3:30:54
but it's only h is approaching 0 so you're plugging zero into h so this is like a reminder what you have to do you put h to be zero in here so you just have 6x
3:31:09
no more and then minus five and here you have it that's the answer can we do this in five seconds yes f of x equals three x squared minus five x plus seven derivative in action power rule bring the power to the front minus one six x to the first power second part minus five x to the first power yeah bring the power to the front minus one so just minus five x plus zero just one
3:31:44
this right here no x so it's just a constant the derivative of this is just zero so done
3:31:51
i know i know i know why are we doing this you'll see let's uh practice one more and then
3:32:01
you'll see for 53 54 55 then you will really see why we need the definition of derivative
3:32:20
we'll just check very good
3:32:28
52.
3:32:32
it's a very simple function it's just x to the third power i didn't want to go too crazy we want the f prime of x so here we go f prime of x
3:32:43
okay let me just put it down like this so here we go we want the limit as x as h approaching zero and then this part means we're just plugging x plus h into here so we have x plus h
3:32:58
third power and then minus f of x which is just that original and then all over h open this we get limit as h approaching zero and let's use the particle triangle one three three one so first we will have one times x to the third power so is x to the third power plus three times x squared and h to the first power and then plus three times x to the first power h squared plus h to the third power so that's the expansion for that and then minus x to the third power over h
3:33:37
we can only cancel one thing
3:33:40
but the good thing is that you have h you can factor that out and we get 3 x squared plus 3x
3:33:50
h and then plus h squared okay and then you see that this h and that h cancel finally put the zero into this h and h so they are gone and we get three x square
3:34:09
done
3:34:11
yeah so pretty much if you do this if the question is asking you the definition of derivative for extra third power if you do this then you get full credit and you'll be happy you know you can yeah but you can of course check the answer in your head or so but just don't let your professor your teacher your ta sees two seconds one two three x squared done yeah
3:34:45
okay so now let's see we have the power rule already why do we need the definition of derivative right well of course the reason is because the power rule only works for x to some power right x to some power if we don't have x to some power then oops
3:35:08
i don't know what other sound effect i can give you but anyway let's look at number 53
3:35:15
f of x equals sine x how do we do the power rule technically you can still do the power rule but you will have to take calculus 2 because you have to write this as a power series and use the power rule but let's not go there yet anyway f prime of x limit as h approaching zero right if we cannot use the power rule use the definition of derivative first we put x plus h into here so we have sine of x plus h and then minus the origin not just x and then over h
3:36:01
well as i told you whenever we use this definition we will think about how to expand it especially the bigger problem which is this and we do have a way to expand this this is just a angle some formula for sine and i'll tell you guys what the formula is this right here is still limit as h approaching zero we will first get sine of the first thing we choose x times cosine of the second which is h and then we add sine of the second
3:36:32
so sine of h times cosine of the first so it's like this and then minus sine of x and over h cool
3:36:45
impressive but what do we do next well keep this in mind this is the limit as h approaching zero so this is like in the h world meaning that if you don't have h like for example the science you can actually put it on outside of the limit and don't worry about it so have a look this term has sine x this term also have this x we can put them together and we can factor out the sine x and the kick outside of the limit so let's do two things right now i'm going to pair this
3:37:22
and that right here okay and then factor out the sine x and then put it on the outside of the limit so we have the sine x right here and then let's look at the limit as h approaching zero
3:37:36
here we'll just have cosine of h and then minus one and then over h we still have that denominator all right and of course we have other stuff too but for the other stuff is this over h and then we can take the limit of that so let's break it apart let's also put the cosine x outside so i'm looking at this right now for the second part i'm going to add cosine x on the outside same reason because
3:38:08
it doesn't have the h and then we just focus on the limit of sine h over h as h approaching zero
3:38:19
all right so the reason that we can do this is limit of first sum it's the sum of the limits so you can break them apart and also if you don't have the h then you can put the expression outside of the limit so that's why we are doing that's why we can do this okay this right here we get 0 for zero likewise that but you know what this right here we have seen it question number five so let me just remind you guys that note if we have
3:38:52
the limit earlier use x right now we have h i know i'm going to be using theta right now
3:39:03
i think it's like a more universal
3:39:06
variable for this limit in particular by the way the sine of an angle sine of theta over theta this right here will give you one and then you can see question number five for the graph for now this video i will just tell you guys look at the graph that we have
3:39:25
uh well you can also just plug into calculator i didn't do anything with the calculator in this video because i just plug in the calculator right so you should be able to just plug into the calculator but this right here you please do not use lovely house rule what's lopita's rule just wait for it anyway what i'm trying to say is that this right here per our discussion earlier is one so we have this times cosine x and then what's this then
3:40:01
in fact you can try it multiply the top and bottom by conjugate and use this to help you out you can try it but i will tell you the answer right here you get zero or you can use the graph just use the graphing calculator use your calculator plug in the numbers that's not a discussion
3:40:19
for here but like that that's equal to zero so we get what zero cosine x hey the answer is just nicely equal to cosine x sine and cosine yeah they're always together in uh precalc trig and also calculus of course yeah they don't run away from each other yeah and let me just make sure that i did this right aha derivative of psi x is equal to cosine x told you
3:40:58
okay anyway i want you guys to pay attention to that hey we run into this limit right here when we are trying to get the derivative of sine x this is gonna show up later on like an hour or two i don't know by the way
3:41:31
all right so now let's take a look at 54. here we have the sorry uh we're still doing the definition of derivative number 54 we have the function f of x this is 1 over 1 plus x squared and then we want f prime of x can we use the power rule right here and say the answer is f prime of x equals to the power rule and then we get 1 over 1 plus 2x can we bring the 2 to the front minus 1
3:42:08
no no
3:42:15
i will show you we can actually use the power rule along with something called the chain rule and then we can actually still formulate the easy way the aka the slow cool easy way to do this kind of questions but let's do the definition of the probability first the limit as h approaching 0 f of x plus h so we put that here 1 over 1 plus parentheses x plus h and then we square that and then minus 1 over x squared and all over h wow
3:42:48
it's a complex fraction with a lot of things anyway multiply the top and bottom by this and that so oh sorry it's not just x squares one plus x squared wow crazy
3:43:02
uh hope you guys can see this okay let me just fix this a little bit
3:43:09
1 over 1 plus x squared ready multiply the top and bottom by this which is 1 plus x squared also that and perhaps i'll just multiply this out right here at the same time so we have 1 plus open that we get x squared plus 2xh plus h squared and then of course let's also put that down on the bottom here 1 plus x squared plus 2xh plus h squared all right so here we will have the limit as h approaching
3:43:44
0. take this times that this and that cancel so we have the first part 1 plus x squared and the minus
3:43:55
this times that this and that can so so we just have this so i'm just going to keep it in the parentheses we get 1 plus x squared plus 2xh plus h squared all right cool and then the bottom maintain it h times 1 plus x squared times 1 plus x squared plus 2 x h plus h squared wow
3:44:26
i wish i could just copy and paste maybe it's easier anyway
3:44:32
so we're just going to see what we can do check this out 1 minus 1 this and 9 cancel yeah 1 minus 1 because you have to distribute the 1 and x square x squared cancel so you see that we just have this and time left and they both have h we can factor that out and it has a negative in the front let's keep the negative in front as well so we have the limit as h approaching 0
3:44:58
negative h on outside and then we have 2x plus another h like so
3:45:09
all right and then all that stuff h times 1 plus x squared times 1 plus x squared plus 2 x h plus h squared
3:45:25
okay let's do this in our head plus zero into h right plus two into h so we have negative two x on top over here we have what one plus x all right and then put zero into h so this and that cancel we have one plus x squared whoa look at that it's a pretty clean result this is now the same though so we can just put it as power 2 right so negative 2 x over 1 plus x squared and then square and then we are done
3:46:02
cool okay now as promised earlier i told you guys we can use the power rule and you see this is totally wrong but how can we really avoid doing the definition of derivative check this out it's all about rewriting and then also pay attention to the so-called chain rule
3:46:25
f of x is that 1 over 1 plus x squared let's first purposely rewrite this as parenthesis with 1 plus x squared inside but then because we bring up we we brought this up right so it will be the negative one power and the reason that we want to do this is now we can use the power rule bring the power to the front and then minus one what do we get f prime of x we have the negative right here
3:46:57
and then the base you keep it which is one plus x squared and then 1 minus 1 negative 1 minus 1 is negative 2. okay wow look i can bring this down to the bottom and this is negative 1 over this is 1 plus x squared and n squared
3:47:20
wait
3:47:25
i thought we can use i thought we could use the power rule but again negative one on top right here earlier we have negative 2x what's the deal
3:47:37
is that wrong that's wrong huh it must be wrong the eastward is always right huh no
3:47:44
definition is much better stole it so here's the deal how can we come up with the 2x well when calculus so we have to think derivative here's the deal when we put the negative 1 to the front and the minus 1 we just do the power on the outside and the idea is you also have to look inside to see what the function is and multiply the derivative of that so you see the derivative of this the derivative of 1 is zero but derivative of a to it the derivative
3:48:16
of x squared is 2x so we have to multiply this by 2x so this is the so-called chain rule part which is needed so you see if we multiply by 2x all together we have the answer that we have earlier
3:48:38
so hopefully you guys can see why we need the definition of derivative it tells you what has to be right if you want to use the shortcut well earlier without seeing that then we'll miss the 2x right
3:48:55
so that's the reason for the definition of the robot and then let's do it again for number 55.
3:49:08
and maybe you guys can count how many times have i said have i said here's the deal i don't know i don't know why i kept saying here's the deal here's the deal
3:49:20
especially for this video i have no idea
3:49:30
number 55
3:49:34
f of x equals square root and we have 3x plus 1. wow yes we can use the definition of derivative and we can also use the power rule and also the chain rule you will see first let's do the legitimate way f prime of x equals the limit as h approaching zero f of x plus h you change the x to f you change the x to x plus h so you have the square root three parenthesis x plus h and then plus one and then minus the original function which is just
3:50:17
that let's just just copy that down square root of three x plus h and then all over h okay
3:50:28
square root business use the conjugate right so let's multiply the top and bottom by square root and let's multiply this also 3 x plus 3 h and then plus 1 and then plus square root of that which is 3x plus 1. and let's do the same on the top on the bottom 3x plus 3h plus 1 plus square root of 3x plus 1. cool all right here limit as h approaching zero
3:51:04
open the top right multiply the top we just get this square which is just inside here three x plus three h plus one minus this thing square which is just 3x
3:51:17
plus 1 and then the bottom stays the same huh so we have h times square root
3:51:26
3x plus 3h plus 1 and then plus square root of three x plus one ladies and gentlemen
3:51:36
three x minus three x less than negative so one minus one cancel perfect on the top we have 3 times h now on the bottom here we have h times all that so this h and h cancel
3:51:51
cool therefore we have a 3 on the top over put 0 into here so we have square root of 3x plus 1. that's the first square root and now we actually have another one huh
3:52:08
wow and they happen to be the same and we're adding therefore it becomes two of them so on all we have three over two square root of three x plus one
3:52:24
done now the moment that you have been waiting for f of x equals okay we have the square root of three x plus one but as i said earlier let's write the square root as one half power that way we can use the power rule so three x plus one raised to the one-half power
3:52:49
once we do this we can go ahead and use the power rule meaning that put the power to the front minus one so we have the one half at the front and then the inside is three x plus one one half minus one is negative one-half negative one-half is the same as bring this down to the bottom and then give it a square root to it so we have one over two square root of three x plus one ah man
3:53:26
this is what we have now that's what we got earlier three on the top we all have one where are the other two one plus three one plus two three not funny i know but you know the deal what you have to do when you're doing this process is that once you do the power rule on outside you will have to look at the inside and ask yourself what's the derivative of the inside function the derivative of 3x plus 1 is just a 3 right so you have to use the power rule inside again
3:53:57
so you have to multiply by 3. again this part is the so-called the chain rule part and for more practice with derivative you know where to go 100 derivatives so multiply by 3 multiplied by 3 and we get that all right so that's that yay done with the page five now you think we're done with derivative no we can never run waveform derivative we have the l'hopital's rule
3:54:34
and let's see i'm going to give you guys a quick quick quick explanation on what the lobby toss rule is and then if you haven't studied derivative yet then a lot of this wouldn't make sense like what's the derivative of natural log of x what's the derivative of tangent x or things like that so be sure you watch my other video for the derivatives all right
3:55:01
maybe uh you're just taking cal one maybe like the first three weeks or so this is like a good place for you to uh maybe just stop and then later on yeah you come back starting at 56. so usually the question went to question 55 is like the first month was throwing your car one class like first three four weeks okay i talked too much
3:55:34
no no no no i did 54 already no
3:55:41
ladies and gentlemen let's have a look this time we want the limit and we just want
3:55:48
the limit as x approaching 2 and
3:55:53
we have x squared over so x minus x squared minus four over ln x minus l and two
3:56:08
how do we do this
3:56:13
factor it sure we can factor the top easily but how can we factor the bottom i want x minus l12
3:56:22
is this just the same as ln and then x minus two
3:56:26
no no don't do that it's not at one time no this is not even funny no don't do that no
3:56:37
wow so here is the time that i'll show you guys the so-called lobby toss rule keep in mind the l'hopital's rule says so i just abbreviated la petite rule this right here it's only good when we have three over zero or infinity over infinity and all you have to do is the following check first we do have zero for zero so you get to use login toss rule and then you just go ahead and do the derivative on the top and then you do the derivative on the bottom and that's the process
3:57:12
if it works you have the answer everybody happy if it doesn't work too bad we just have to think about it if i want to do it anyway let's see this is the limit as x approaching 2. let me just get the answer for you guys first and i'll explain why this is the case
3:57:31
differentiating x squared minus 4 we get 2x and then take the derivative of l1x we get 1 over x
3:57:39
okay so again it's just the derivative of the top and then you do the derivative of the bottom and the derivative of beyond x is one over x this is a number that's why it's zero so just that all right so seriously you will have to know your derivative really well in order to use lopital's rule okay now we can put the 2 into the x's if you would like or you can multiply x and x just simplify first so this is the same as limit x approaching 2 and just 2x squared
3:58:14
yeah and then just worked out right so it's two times two squared four times two which is eight now what circle it i usually box it sorry all right cool done
3:58:33
now here is the reason why this works first i want to tell you guys that a lot of mathematicians they do not like the l'hopital's rule sometimes i hate it too because it costs the circular reasoning i'll explain that in the next category
3:58:55
but let's see explain why this works it's just that you will have to do this the long way and then i'll just tell you if we look at the limit
3:59:05
let me just rewrite this again so let me just rewrite this right here right if we have the limit as x approaching 2 and then on the top we have x squared minus 4 yeah and on the bottom we have ln x minus ln 2 right looks good not really because it's zero of zero
3:59:27
but let's do the following i'm going to divide the top by x minus two likewise we'll do the same on the bottom as well and then i'm going to write this as do the limit on the top and then divide the limit on the bottom and we can do so because the limit of a quotient is a quotient of the limit assuming they both exist so this is the same as saying the limit as x approaching 2
3:59:56
x squared minus 4 over x minus 2 and over limit x approaching 2 and then
4:00:06
so again the limit of a quotient is a quotient of the limits provided they both just they do so that's why we can do this now if you look here at the top right here
4:00:19
what does this represent i know we can solve this by using the algebraic limits but if you remember the definition of derivative and you know it i'm going to bring this up right here this thing
4:00:33
doesn't just remind you of a derivative it does huh yes and i would like to tell you you see how we have x is approaching two so i will tell you this right here represents i'll just spell that out for you guys represents f prime of 2 because x is approaching 2. so it's the derivative of some function when x is 2. but what's the function though where the function
4:01:03
i'll just write it as f of x it's just that just x squared right yes similarly we can also talk about the bottom right here this right here represents
4:01:19
again f prime of 2 as well but i use f right here already so just to reduce confusion i'm going to use another letter let's say g just another function right because we're working on the same question so let's use g it's the derivative of some function g at two and then what the function is is just l and x so where g of x equals l and x so in fact the type is a derivative equation likewise
4:01:49
the bottom is also a derivative equation and you see this right here is really just f prime of 2
4:01:58
over g prime of 2.
4:02:03
so just the derivative of the top divided d by the derivative of the bottom assuming that we get a nice answer for this right it does so yeah so what's f prime of 2 though well if you look at this carefully you see that
4:02:21
we know f of x is x squared so this means f prime of x is just 2x so this means f prime of 2 is just 2 times 2 which is 4. so on the top here we actually just get 4.
4:02:38
on the top here we just get 4. and on the bottom as we saw earlier this thing right here represents g prime of 2 and yes i'm changing the color to make it prettier for you guys
4:02:54
and now you see this is g of x is l and x so we have to know the derivative g prime for x is just one over x and that's exactly how we did right and then we plugged in two
4:03:07
so just one over two so the bottom here is one half
4:03:12
simplify the fraction multiply two and two so ladies and gentlemen we end up with eight
4:03:18
yeah so this is the idea behind the the so-called lopitos rule and again we use it's useful when we have three over zero and also infinity we can only use laptops rule when we have zero plus zero or infinity of infinity however there might be some cases that lobby does rule does not help us uh get the answer in that case you just have to try other things but if you don't have this intervention for
4:03:50
then can i use lopitas rule and the reason why that a lot of mathematicians don't like the lobby does rule because it costs circular reasoning is because you see how when we are doing limits we are using the derivative but when we are trying to get the derivative we will have to use limit
4:04:14
see earlier when i show you guys this i can actually tell you hey let's just do the derivative no you have to do it algebraically first yeah so yeah be be really careful i will tell you though the best time for us to use lopitas rule is infinity over infinity but then still a lot of people will still not like it too much but it's convenient i would say yeah that's enough explanation because i want to move on and usually i will just say i want to go home but
4:04:53
i'm home already so it's not like when i was doing that 100 integral 100 derivative questions
4:05:06
okay number 56 no number 57
4:05:14
limit give you guys a tough one x approaching zero and then we have one over x
4:05:23
minus one over inverse tangent of x okay just trust me for this one
4:05:32
technically we do the zero plus plus zero minus but if we have zero plus we get infinity minus infinity if we have a minus here we still get infinity minus infinity so indeterminate point either way in water first you sloppy does rule you see that we have to have zero plus zero over infinity of infinity namely a fraction here we have two fractions not good idea therefore let's just combine them and to do so i'm just going to write this as the limit as x approaching
4:06:05
zero right we need this and not together right so it's x times inverse tangent of x and then pretty much multiply inverse tangent of x here so we need the inverse tangent of x here and a minus just x there much better and you can check if we have zero into all the x's we get 0 over 0. and because of that let me just indicate that we have 0 over 0 we are using the logitos rule i'm just going to put ddx and ddx just to show you guys that we are using a so-called beetles rule
4:06:42
i forgot to put this earlier so just go ahead and do it on your own go back to your notes and do it as i told you you had to know your derivatives really well
4:06:53
what's the derivative of inverse tangent of x do i need to go to the board sure but i don't need to because i know the derivative that is one over one plus x squared okay derivative of that's the derivative of this is 1 so it's minus 1. this right here we need the so-called product rule
4:07:14
we are going to keep the first function which is x and then times the derivative of the second so that is 1 over 1 plus x squared and then we add the second function which is inverse tangent of x times the derivative of the first so it's 1.
4:07:30
so these are the things that you would need to be able to do for your derivatives
4:07:38
before we proceed let's fix the complex fraction situation and i is to multiply the top and bottom by 1 plus x squared and likewise here so let's see we will have limit as x approaching zero this times that is just one this times that does distribute the negative so we will have negative one and then minus x squared that's cool we have the one minus one so far so good i don't know anyway take this times that the one plus x cancel so we just have x this times ah okay plus
4:08:26
let's let's distribute so this times that is inverse tangent of x and then this times that so it's plus x squared inverse tangent of x i told you this is the hard one but anyway one minus one zero
4:08:44
plugging zero into all the excess i tell you is zero over zero so yes do lopital's rule again try it it is making us progress so yeah so here we go lopito's rule on the top
4:08:59
differentiating on the top and then take the roof on the top on the bottom
4:09:04
limit as x approaching zero derivative of this guy is negative two x derivative x is one the root of this is 1 over 1 plus x squared derivative this product rule again the first function is x squared
4:09:25
times the derivative of the second which is 1 over 1 plus x squared and then we add the second function which is inverse tangent of x times the derivative of the first which is 2x
4:09:39
i know sometimes you want to do the derivatives
4:09:43
love it might make you cry it makes you want to cry right so yeah
4:09:55
if ripple can sponsor me one day while i'll be really happy
4:10:00
or any kind of any like a beverage it doesn't have to be energy drink anyway
4:10:08
what do we do next man okay let's see
4:10:15
this right here shouldn't be too hard because i did it before let's multiply the top and bottom by one plus x squared again oh okay so how am i going to
4:10:34
this doesn't make sense does it make sense
4:10:39
okay let's see here we have the limit as known x h
4:10:48
x approaching zero perhaps let's multiply the bottom here and see what happens right so before we go there let's see this times that we get 1 plus x squared this times that we just get plus 1. this times that they cancel so we just get plus x squared here right and this times that oh my god
4:11:16
oh i don't think we end up with i don't think we end up with zero first rule anymore check this out
4:11:23
here this times this is just i'm just going to keep it one plus x squared and then
4:11:31
let's put a 2x here at the front first 2x and then 1 plus x squared and the inverse tangent of x all right so just multiply out the reason now i say it's not 0 for 0 is because when x is approaching zero this is zero this is zero this is zero but we have one plus one which is two on the bottom
4:11:58
but this is upsetting because when we put zero on the top
4:12:07
we get zero so on o we just get zero right zero over two
4:12:17
what does that mean the answer is just equal to zero
4:12:22
yes let's double check
4:12:31
yeah nothing wrong for sure he has to be right zero why not huh
4:12:37
yeah you guys can pause the video and just verify this with me verify this for me and then just do it on your own whatnot you should be sure all right all right so 57 check
4:12:53
good four hours and 12 minutes so far so good but the 57 is numbers is zero
4:13:18
my hand because that's pretty dirty anyway
4:13:24
all right moving on to a slightly more gentle question
4:13:31
limit as x approaching 0 plus and then we have x times ln x oh okay i can talk about another indeterminate form check this out when we put zero into here we get zero and then when we do zero plus into l and x we get negative infinity right so yeah it's like this so we have a zero times infinity right here the negative doesn't matter we care about this times that that determinant form and that's no goal you cannot draw any conclusion why don't
4:14:07
you use btos rule how check this out pick one of the functions it will be easier if you pick x
4:14:16
just trust me on this and then you want to bring this down down i'll show you this right here is the same as the limit as x approaching zero plus keep the l and x on the top but then change the x to one over x on the new on the denominator because now you can see if you multiply the top and bottom by x then you get the original back cool huh now if you put zero or zero plus here in here you get infinity over infinity so we are able
4:14:50
we can actually unlock we are allowed to use lupita's root now so i will pull this down ddx
4:14:58
ddx
4:15:02
all right so here we have the limit as x approaching zero plus the derivative of this is one over x derivative of this write it as x to the negative one power bring the negative one to the front minus one so it's negative
4:15:20
x to the negative 2 power which is just negative 1 over x squared before we plugging 0 let's just simplify it multiply the top and bottom by x square and then we will see this is the limit as x approaching 0 plus we have a negative they cancel and then they reduce to just x put 0 in here we get zero yes oh my goodness another zero not so satisfying but that's just how he is okay
4:15:56
please do not say zero times infinity with zero times negative infinity is equal to zero all the time sometimes it is but not all the time okay let's try another one number 59
4:16:09
limit as x approaching another 0 plus
4:16:18
1 over x
4:16:23
plus ln x now here's the deal if you put 0 plus in here and here you get infinity and minus infinity so this is the infinity minus infinity situation again let's just combine the fractions and let's see how this goes right so this right here equals the limit as x approaching 0 plus multiply x and x right so we just have x here and then we have 1 plus and we get x times l1 x but check this out
4:16:59
if we put zero plus into all the x's what do we get should we just do l'hopital's rule right now can we just do lopitas rule and say
4:17:14
no do not do lapido's ruling right now why because if you put zero plus in here on the bottom we get zero on the top hey what's x times ln x we did it earlier and then it's the same thing but we have a one so in fact we end up with one over zero this is one over zero
4:17:41
so what do we do next well you don't have to worry about doing l'hopital's rule in this case we kind of did for this part right here but the truth is the truth is we can actually kind of argue what the answer should be already so i would just have to say
4:18:02
this part the limit as x approaching 0 plus x l and x this right here is equal to zero we did that for 58 so i can say that okay use it and then you can see this is just going to be 1 plus this is zero right because what we did earlier and then we have
4:18:27
x is approaching zero plus so the bottom here is zero plus
4:18:32
and you might be wondering that is this true plus with zero minus doesn't matter because you're adding one so the top is for sure positive for sure it's a positive a non-zero number and divided by zero plus so the answer for this is what positive infinity that's it yeah that's it
4:18:58
if you don't think this is like 100 legit then you can also do it this way you can factor out the 1 over x this right here is the limit as x approaching 0 plus and then factor out the 1 over x and you will get 1 plus if you factor one of x is like x times l and x right and um
4:19:22
it's pretty much like this but like another look and you will see if you put zero plus in here this part is positive infinity but this is zero it's just like one plus zero so infinity times one is infinity yeah so another way to look at it but thanks to this part imagine if this director turns out to be negative 1 then this question will be harder but it's not and you see infinity minus infinity gives us infinity
4:19:54
yeah you never know what you are going to get yeah doing limits is just like a box of chocolate
4:20:18
number 60
4:20:22
i want to say 67 but i no not yet
4:20:28
limit as x approaching
4:20:33
zero plus
4:20:36
yeah of course who came out so this question
4:20:43
all right if we put 0 plus into here and here check this out we will end up with a
4:20:49
0 times 1 over 0 plus infinity e to infinity infinity so this is zero times infinity situation
4:20:58
so we are going to do what we did earlier here's the x bring that down down this right here is the limit as x approaching 0 plus e to the 1 over x stays on the top but then we write this as 1 over x on the denominator and we will use the laptops rule because now we have infinity over infinity so it's not beta's rule ready and then we can just say ddx
4:21:30
ddx
4:21:34
cool now we have the limit as x approaching 0 plus what's the derivative of e to the something
4:21:43
the answer is you first repeat that e to the 1 over x and then you multiply by the derivative of the inside because of the chain rule and the derivative of that is negative 1 over x squared
4:21:56
the bottom when we differentiate that we also get negative 1 over x squared
4:22:02
let's cancel each other out completely oh my god check this out when we put zero plus into here now we get what we get this
4:22:13
this is now e to the one over zero plus which is e to the infinity which is infinity told you zero times infinity is not always equal to zero sometimes infinity can win two with infinity
4:22:36
all right number 61 limit as x approaching infinity x times two to the one over x minus one
4:22:51
all right similar flavor this time now if we put infinity to here we get infinity times one over infinity 2 to the 0 is 1 1 minus 1 is 0 so infinity times 0. so do the same thing put this x down down so we're looking at this as limit as x approaching infinity here we get two to the one over x minus one over one over x and you know it we can use l'hopital's rule because this time we get zero over zero which is not beta's rule
4:23:27
so ladies and gentlemen d dx here ddx just means that we are going to take the derivative
4:23:37
so here we have the limit as x approaching infinity the bottom here we get negative 1 over x squared how about the top what's the derivative of 2 to the some power
4:23:51
let me tell you you first repeat the 2 to the 1 over x
4:23:55
but because the base right here is the 2 it's not e so we will multiply by ln of the base which is 2 and then we multiply by the derivative of that because of the chain rule which is negative 1 over x squared derivative minus 1 0 so no more and then we see that this and that cancel and when we put infinity to here 1 over infinity is 0 0 2 to 0 is 1 so
4:24:26
0 times infinity gives us ln 2.
4:24:31
told you 0 times infinity you can get a lot of different things indeterminate forms crazy huh
4:25:04
number 62 limit as x approaching infinity
4:25:10
and then we have log base 2 of x plus 2
4:25:18
over log base 3 of x plus 1. wow
4:25:26
okay we get infinity over infinity so let's just get it out of the way and let's just go ahead and use l'hopital's rule because this is infinity over infinity situation
4:25:41
so everybody's favorite ddx the dx here is the limit as x approaching infinity
4:25:56
so how do we differentiate this though well i'll tell you when we have ln so when we have log of 2 right log base 2 you just put this on the bottom first so you have 1 over x plus 2
4:26:10
but you also multiply by ln of that base so you multiply by l and 2 okay and then you check the derivative of the inside but the derivative of x plus 2 is just 1. so if you forget the chain rule you know you're lucky just because multiply by 1 doesn't matter all right let's practice this again
4:26:30
derivative of log base 3 of x plus 1 we put this on the bottom first so it's 1 over x plus 1
4:26:39
times ln 3 okay and then multiplied by the derivative of this which is also one cool now
4:26:49
we want the x to be going to infinity but let's just put this on the top so on all we are looking at the limit as x approaching infinity and then we have x plus 1 times let's distribute using y and let's see how do we write it let's do this x times tau and three plus ln three distribute and l minus three is the constants the coefficient is the coefficient of x
4:27:19
and then this right here same thing x l one two plus lm 2 but guess what this is everybody's favorite derivative this is everybody's favorite limited question calculus question because x is approaching infinity just care about the x to the first power the dominating x right yeah yeah they have the same decreasing power so they cancel and then it's just l and three over l one two isn't it good and now icing on the cake we can use the change of base formula
4:27:59
and write this as log base 2 of 3.
4:28:06
oh my goodness log base two of three not kidding this right here is the answer for that so you see infinity over infinity we get like a irrational number yeah
4:28:20
okay number 63 one of the classic calculus version
4:28:27
limit as x approaching 0 plus x to the x power check this out if we put 0 into here and here we are getting zero to the zeroth power let me tell you if you are just talking about computation zero to the zero okay i'm just going to do this like more legitimately i'll say four limits
4:28:54
and again for limit means that the limit for meaning i just put this number into here and here and just get the form of it tanning colleagues has zero plus plus zero plus right but usually we just write zero to the zero but again
4:29:07
it's really zero plus to the zero plus right like that okay but anyway when we have this limit form this right here is in the in the
4:29:25
determinant okay why inside one here is right here okay but anyway if today we're just doing computation
4:29:38
and by computations i mean that you know when you have 2 to the 3rd power what do we do you just do 2 times two times two three times that's computation but anyway
4:29:49
when we're just talking about computation like no calculus with no limits involved zero to the zero this right here is the most debatable number or anything i don't know a lot of people will say this right here is undefined i used to say that and uh a lot of people will also say this right here should be equal to one i would like to tell you for computation this right here i think the best answer that i can tell you is that this right here has no agreement
4:30:26
i think this is the best answer that i can say and i know i think people will agree that there's no agreement on this if you really want to say this is one then you can make a facebook post for like a youtube video or like a twitter and then you can see how many people will like debate and all that if you say this isn't defined you know same thing will happen like you say this right here has no agreement
4:30:51
hopefully you guys leave me alone but anyway this right here is zero over this this is right here zero to zero but we're doing limits so it's indeterminate meaning we have to do more work this is how we don't like base x we like to have base e this is the fast way to do it change the base to e
4:31:17
and we can achieve that by doing this write x as e to the ln x power because e and l and cancel still get x back and then you see we will have to raise this to the x power right so perhaps because this is the first time we're doing it so i will show you guys a step by step like this i'm not going to skip anything right so this is x raised to the x power okay
4:31:45
and then you might be wondering now in the parenthesis y sentences just don't worry too much about it yeah i think this might be bad okay anyway now i'm just going to write this as the limit as x approaching 0 plus we can multiply the powers and then we get e to the x times ln x power
4:32:05
does this look familiar yes it does number 58 yes but the way that you should take care of this is the following this is e and then you do the limit of that which is the limit as x approaching 0 plus and i just focus on x l and x but i should say
4:32:33
use question 58 all right so in fact we get e and earlier we saw that this is equal to zero so on o we get one done and of course if you are doing this on the test do not just say look here 50 look at question 58 nobody will know what you're talking about just redo everything from question 58 on your test so that people will be happy
4:33:03
okay if my student if you go from here to here and say use question 58
4:33:12
i will give you focus no i will not give you for credit
4:33:16
maybe i should because i know that way you are watching my video i don't know
4:33:23
but the deal is that will not ask you guys that question is too easy or something more difficult
4:33:39
all right
4:33:41
all right number 64. wow that's eight square
4:33:57
so limit as x approaching 0 plus and here we have x to the 1 over 1 plus ln x power and check this out when we have 0 plus here put it here basis zero and let's do this in our head zero plus into ln l and zero plus is negative infinity and then we have the one right one plus one plus negative infinity is just negative infinity yeah and it's 1 over 1 over infinity is 0. so this is actually a 0 over this is actually a 0 to the 0 indeterminant form let's see if the answer is 1 or not
4:34:42
do the same thing that we did earlier rewrite the x as e to the ln x so this right here is the limit as x approaching 0 plus e to the ln x power and then raised to the 1 over 1 plus ln x power now multiply the powers so we are getting limit as x approaching 0 plus and then we have uh just do it like this e and then l and x over 1 plus 1 x power
4:35:16
now just focus on doing the limits right here so this right here is equal to e and then let's just focus on the limit as x approaching zero plus l and x over one plus l one x
4:35:31
but if you put zero plus into here and here we get negative infinity over infinity we get negative infinity over negative infinity so you can use log beta's rule well in fact
4:35:43
you can just get rid of the one because the one doesn't matter when we have infinity but perhaps this is slightly more legitimate anyway though this right here we get e for the base and then again the blue part is negative infinity over infinity so that p talks rule
4:36:06
all right let's do this in black so this right here is the limit as x approaching zero plus the top is one over x the bottom is also one over x the one doesn't matter they cancel just one so on all the answers e to the one which is just e
4:36:29
zero to the zeros power we end up with e this is the limit right so limiting determinant 1 we can get some other numbers besides 1. it's not like there's no agreement this question everybody will agree that the answer is e 2.7 ish right so there will be no argument for this one right here everybody will be happy if you go to people's birthday party do not show them zero to the student's power show them this limit and you are being invited that's the idea
4:37:03
number 65
4:37:09
the limit has x approaching zero
4:37:12
finally it's just zero anyway x e to the x over e to the x and then minus one
4:37:23
so we're plugging zero well no that will give us zero of zero no good what do we do though hey lovely does rule right now you could but let's make it slightly easier let's divide the top and bottom back into that because they'll have e to the x so i'm just going to divide this by e to the x likewise divide this by e to the x and you'll see we get limit as x approaching zero they cancel reduce is one one over e to the x let's write it as e to the negative x
4:38:00
do we really need to use laplace rule no because now we can see this is just zero over
4:38:08
no no we still have to sloppy those rules
4:38:12
we still end up with zero plus zero right here so yeah we're still going to use slopital's rule so here we go dtx
4:38:20
ddx the laptop's rule is much easier this way though zero over zero poppy does rule
4:38:28
so limit as x approaching zero derivative of x is one derivative one is zero derivative of negative e to the negative x is negative e to the negative dx but we are using the chain rule multiplied by this derivative which is negative 1. so it's actually positive e to the negative dx but then if you put zero into here we will get one over e to the negative zero just zero let's final answer
4:39:03
is one yeah zero plus zero yes it can also be one all right it can also be one
4:39:15
so whole modi you guys are getting like all the things that you have to worry about and you can see that all the things that you have to be careful and all the techniques and these are not the hardest part yet okay
4:39:41
number 66
4:39:46
limit as x approaching zero
4:39:52
sine of two x and then e to the x minus cosine of x all right so just do a quick check zero over zero so go ahead do l'hopital's rule
4:40:12
so d dx d dx all right here we have the limit as x approaching zero
4:40:25
derivative of sine is cosine we did that today cool and then multiplied by the derivative of 2x which is just a 2 so let's put that in the front derivative of e to the x is e to the x the derivative of negative cosine is negative sine so negative t times negative becomes positive and then we have the sine x cool now put zero into all the axis then we get to try it cosine of zero is one so on the top we have two put the zero in here equal to zero is
4:40:56
1 sine of 0 is 1 sine of 0 is 0 so just 2 over 1 which is 2 done so you can verify that on your own
4:41:09
67 we have the limit as x approaching pi over 2 and then we have secant x minus tangent x
4:41:24
wow i'll tell you it's a infinity minus infinity case depending if you are talking about the plus or minus but no let's combine the fraction wait where are the fractions
4:41:38
this right here it's the same as 1 over cosine x this right here it's the same as x over cosine x isn't it so all in all this is just the limit as x approaching pi over two and then we have
4:41:58
cosine x on the bottom and then one minus psi x on the top yeah and in fact yes we get zero over zero but do we really need to use l'hopital's rule for this
4:42:18
we don't because earlier today we do something similar to this question number 30 right similar
4:42:29
to question 30 you guys can look back that's like i don't know how many hours ago but we can multiply the top and bottom by one plus sine x you can do it that way but since this is the logitech's rule category so let's just use l'hopital's rule just want to mention that we do not have to just stop because rule if you don't want to but anyway this is a zero over zero case we're using lobitos rule so this is the limit as x approaching pi over two
4:43:03
derivative once you wrote the rule of negative sign is negative cosine
4:43:09
derivative of cosine x is negative sine
4:43:14
so cancel pi over plugging to cosines zero sine of pi over two is one zero over one is zero
4:43:26
very cool all right oh this is what i like i can do like multiple questions on the same board so i don't have to erase erasers okay number 68 another
4:43:40
big one or final the x is approaching one that's well that was an interesting number too by the way
4:43:49
x over x minus 1 and then minus 1 over ln x you'll know it combine the fractions before we continue so number 68 here we go take the limit as x approaching one this and that on the bottom i will put it as x minus one parenthesis times l and x on the top though i will do this times that right so it's x times l and x and then minus this which is just so i'm going to distribute so it's minus
4:44:24
right minus x minus minus become plus and then one
4:44:32
all right and then verify it you get zero over zero so again that's just two l'hopital's rule wow d dx
4:44:49
okay here is the limit as x approaching one okay product rule here we need to keep the first function which is x times the derivative of the second and derivative of l on x is 1 over x and then we add the second function which is ln x times the derivative of the first which is one continue the root of the minus x is just minus one derivative one is done continue right here um we distribute it now let's let's just do it like this keep the keep this as the first function and
4:45:29
this is the second function so i'm going to keep it as x minus 1 times the derivative of that plus the ln x times the derivative okay oh my god check this out
4:45:46
x over x is one one minus one is zero so in fact this and that cancel out completely perfect here though we can distribute it we get 1 minus 1 over x
4:46:02
ah that's still going to give us 0 over 0 oh my goodness right so perhaps i'm not going to distribute because it doesn't do us any good or if i distribute i'm just going to multiply the top and bottom by x so that we can get rid of that detail x for the complex fraction case so here we have the limit as x approaching 1.
4:46:26
this times that is just x times l and x then this times that they cancel which is just x minus one and then this times that which is plus x l and x
4:46:40
yes it's zero over zero but we can use l'hopital's rule again do this do that
4:46:49
so zero over zero l h this is the limit as x approaching one
4:46:58
okay part of in action keep the first function which is x times the derivative of the second plus the second function which is l1x times the derivative of the first
4:47:12
now the derivative of x is one derivative of one is zero minus one is zero same thing here point rule plus first function times the term of the second plus the second function times the derivative first oh my god check this out this is one this is one when x is approaching one this is zero that's zero on the top we have one wow this is one one plus one is two
4:47:48
ladies and gentlemen for all the work we just get one half yeah all this work we get one half wow number 68 oh my god my hands are so dirty so look at this yeah cool
4:48:08
you see i'm doing so many questions for you guys right yeah if you are taking kelvin for the very first time how many questions are you practicing that's the key right if you want to really learn calculus well depending on if you really want to nano material or if you just want to get over calculus practice i'm doing this for you guys with you guys right
4:48:38
so maybe you are not like using maybe you are not even though you are not
4:48:48
maybe you are just studying for a calculus or math test in general you are not really needing like this material hopefully this can somehow motivate you in any way anyway
4:49:01
this question
4:49:04
number 69 and we have the limit as x approaching one plus and then we have natural log i would like to tell you this is a very common test type of question okay ln on the top and then inside here i picked uh x to the third power minus one on the bottom i have ln of x to the eighth power
4:49:30
minus one very common very popular kaokwang cow2 test question
4:49:40
okay you will get negative infinity over infinity by like yeah it just tells you we can do la pito's rule because we get ln of zero plus or no zero plus that's why anyway
4:50:00
the
4:50:03
dx ddx okay here we go limit x approaching 1 plus when we differentiate 1 we just put that on the bottom so it's 1 over x to the third power minus 1. you don't have to multiply by l1 of the base technically you do ln is log base e when you have l and e is just one so it doesn't really matter anyway multiplied by the inside derivative so that's the chain rule so multiply by 3x squared
4:50:34
do the same thing on the bottom we get 1 over x to the 8th power minus 1 and then multiply the derivative on the bottom which is 8 x to the seventh power wow
4:50:48
okay so what i will do next is i'm just going to kind of um well let's reduce this a little bit yeah this and that will cancel will become the fifth power only
4:51:01
cool and then we are going to get the limit as x approaching one plus and uh to make things more clear i think i'm just going to do the stasi
4:51:21
so this this is out and this is out this is also x25
4:51:30
okay
4:51:34
so i have a 3 on the top yeah i have a string on top and then i'm going to bring this up so it's 3 times x to the 8 minus 1 because this right here goes to the top so it's x to the 8th minus 1 over
4:51:51
this is on the bottom right x to the 3rd power minus 1 and this is also on the bottom so i will multiply this with 8 x to the fifth power okay so kind of verify this on your own just to multiply this bring that to the top and there's 3 with that and then 8 x to the 5th is still on the bottom along with that for the denominator cool x is approaching 1 so what we can do next is just multiply out and then you can do lot better through again let's just lobby those rules again
4:52:28
because why not but let's multiply this out first this is the limit as x approaching 1 plus we get 3 x to the eight minus three and then eight x to the you add the exponent which is eight minus five
4:52:46
minus eight x to the fifth okay and then this time we get zero over zero yeah so
4:53:01
a lot better rule here right well dtx here right dtx here we're using nothing to follow this so limit as x approaching one plus derivative 24 x to the seven yeah zero bring the a to the front is 64 x to the seven minus bring the five to the five is 40 and x to the fourth cool now we're just going to be plugging one to all the x and then you don't have to reduce what not because it doesn't really
4:53:39
matter that much but if you look at the bottom when x is equal to one on the bottom here is just equal to 24 right yeah so it's like this you get 24 over 64 minus 40 so only all we get 24 for 24.
4:54:00
why am i doing this to myself that the answer has equal to one
4:54:06
yeah the answer to this right here is equal to one okay i have a question for you guys just to make sure that you guys are actually uh paying attention so go ahead and leave a question right here right now leave the
4:54:22
timestamp and then leave the answer i want you guys to try this i want to keep the function the same by instead of x approaching 1 plus go ahead and figure out what if we have the limit as x approaching infinity and then we have ln of x to the 3rd power minus 1 ln of x to the 8th power minus 1. okay you just do it the work is almost identical but you have to be really careful of one little place and then you will be able to figure this out i am
4:54:55
going to squeeze in number 70 here because i want to i really want to limit x approaching infinity
4:55:04
we have ln x this is ln x and squared so do not bring the two to the front as like the
4:55:14
log power rule don't do that all right
4:55:18
this is x but anyway though
4:55:23
we will use the power rule for derivative but not the power rule for the logarithm so pay attention to the following
4:55:33
holley anyway
4:55:38
invented into here you get infinity over infinity so lobby does rule last question of love resource rule man yeah hey because it just has a lot of things to write anyway bring this to the front and minus one here we are looking at limit as x approaching infinity 2 to the front l and x and then multiply the derivative inside which is 1 over x the derivative of x is just 1. but the truth is this right here requires l'hopital's rule again
4:56:17
so this right here is the limit as x approaching infinity 2x or 2l and x and over x right put infinity we get infinity over infinity
4:56:31
so not beta is true again and please pay attention to this question because you will see that we have infinity over infinity right this is ln x compared with x the two doesn't really matter but let's choose to log plus rule here d dx dx here we get the limit as x approaching infinity on the bottom we get one on the top we get two times the derivative of the or just one of x
4:57:03
when we put infinity to here we get 2 over infinity just 0.
4:57:10
so what does this mean if you look at this the limit of the ratio of two l and x over x is zero this means l and s is actually so so small even though it goes to infinity as well but this infinity is nothing compared to when x is going to infinity that kind of thing all right in fact you can have your own x to whatever power that you want that will not be bigger than x to some power yeah like a power function so keep that in mind because with that
4:57:45
being said we are going to get into the secret weapon if you know the secret weapon we will be able to finish a lot of the questions in no time and congratulations to myself i finished page six
4:58:04
five hours oh my god
4:58:32
all right i feel like drinking some soda so just
4:58:38
give me 20 seconds was all
4:59:07
hmm oh
4:59:25
i have this haha
4:59:49
right
4:59:56
oh it says real but not instagram real
5:00:07
anyway i should have bought some gatorade i forgot
5:00:14
it's hot
5:00:17
yeah that's why i have to drink something but anyway though
5:00:21
30 more questions secret weapon number one i call this the list
5:00:30
what's the list well this is to compare the kind of infinity i'm using n right here because this way i can also bring the factorial with us and then
5:00:45
it doesn't really matter of course you can use t you can use x doesn't matter here we go among all the functions now go to infinity ln well any kind of log any kind of log functions is the smallest so small that i'm going to put two less than symbols like this and then you will see why next we have n to the power like the power function will power sequence but anyway though continue next we have b to the n and next here is d factorial and then next the biggest one n to the n
5:01:19
power tower however in order for this to go to infinity well you see n goes to infinity right and we talked about this earlier like two three hours ago the power right here has to be positive so let me just indicate that p has to be greater than zero here and then in order for exponential to go to infinity the base has to be greater than one so b has to be greater than one so this right here
5:01:48
is the list if you know it you can solve a lot of limit questions in no time i tell my students that you can just tell me what the answers are if the limits think to this situation how do we use it let me show you number 71 i'm going to give you guys two questions if today we have the limit as x approaching infinity and then we have an x over the cube root of x okay well our next it's a small kind of infinity yeah and the reason i put on two less than symbol is that this is so small
5:02:33
such that when you do the ratio if you have small over something that's bigger because this is the same as x to the one third power which belongs to here when you divide them and take the limit as x goes to infinity let me tell you the answer is just equal to zero
5:02:52
yeah yes that's it that's it for my students you can say the list if you don't quote uh and if you don't quote the reason that's fine too but if you're not my student if you are
5:03:07
taking your calculus 2 class then do not say this is equal to zero by the list your professor is not going to understand what you're talking about so yeah by the way this is just that you can keep in mind so it's a magic when you see this kind of limit you can put on zero right away the proof for this is to use lopita's rule which is very similar to number 70. that's why i told you guys to pay attention to the question earlier all right okay
5:03:34
this is that but what if now this is just first first this is like versus so i'm going to put it as 71 star to make it slightly more clear but what if we have the equation being the limit as x approaching infinity we have the cube root of x on the top and then l and x on the bottom well you can read it backwards this right here is so big compared to that big enough so that when you divide them and then take the limit as x goes to infinity you get infinity
5:04:09
yeah so when you're computing this kind of function for sequence either you get zero or infinity when you're using a list so cool so we'll have a lot of questions that we can actually just finish in no time by the way number 72 number 72 tells you the
5:04:29
the purpose of number 72 is to show you the um how to pick out the dominating part but anyways let's see number 72 we have the limit as x approaching infinity and then we have x to the sixth power
5:04:44
over x squared plus 4 to the x
5:04:56
okay on the top we have x to the sixth power okay on the bottom we have 4 to the x so look at this don't look at that why because if you compare the bottom x squared versus 4 to the x 4 to the x is bigger it's dominating the exponential part right a number to the exponent power right x 1 yeah
5:05:23
and now if you compare this and that 4 to the x is bigger but on the bottom so that means when we do the limit of this we get zero done
5:05:36
72 star if i change this a little bit if i change this to the limit as x approaching infinity 6 to the x power and then 2 to the x and then x to the fourth power what do you think what the answer is
5:05:57
compare the bottom 2 to the x versus x plus 4 this time 2 to the x is bigger than that it dominates exponential part it dominates the x to the p right so x to the fourth
5:06:12
and then if you compare this and that well the top is bigger than that even they are in the same category so of course you look at which bigger which base is bigger and you can actually just see this as 6 over 2 to the next power and then just 3 to the x power when x goes to infinity is infinity
5:06:33
but yeah that's it that's how you do it okay number 73 oh man this is the last question no i have a couple more for the list i have five questions for the list like that limit as n goes to infinity n factorial over n to the 2022 and this is the reason why i put n to be the list right for the list because uh that's how we can say n factorial for positive whole number factorial or maybe zero factorial you can do like fractional factorial but that's a different story
5:07:12
that will be later on not now anyway when n goes to infinity uh n factorial is so much bigger than
5:07:24
n to the p power and to a power so this is bigger than that therefore the answer is one infinity
5:07:31
done you can give me one thousand of these questions i can finish it i don't know i don't want to do actually you can give me 100 of this question i can do it in like two hours or one or one hour maybe by the way 10 seconds wow okay next 74 if we have the limit as n goes to infinity no i use x now most of the time we are talking about x x is much better because we can do laptops rule with like ddx and all that stuff but anyway l1 x minus square root of x wow so this is
5:08:10
slightly different than these because we have a subtraction here but just think about the idea
5:08:16
if you have infinity put it here here we get infinity minus infinity but one infinity is actually much bigger than the other which one's bigger this is bigger because square root of x is the same as x to the one-half power which is here it's bigger than natural log so if this is bigger small minus big we get negative infinity yeah just like that so quick so fast so good
5:08:49
all right let's try another one oh man this is the one i'm going to miss because after this no more the list e to the x minus x to the e what's the answer exponential is much bigger than power this is x to the power e right so this is infinity minus infinity we get positive infinity all right so you can see with the list we can finish a lot of these questions in no time
5:09:45
next i would like to tell you guys
5:09:50
the fact
5:09:53
again i gave this formulas these names why because it's much easier for us to refer them yeah and then once you have a name with something that you tend to like it more understanding more and you want to keep it with you more anyway
5:10:15
the fact i'm just going to write here is the limit and
5:10:19
i don't remember if i use x what he or what not doesn't really matter i'll just use x x approaching infinity you've got 1 plus some number a over x raised to the b x power
5:10:34
this right here gives us e to the a b power this is it and this right here is actually nothing nothing special you actually have seen it before you just never recognize it from your precalculus class with like the calculus questions that you have been doing so here's the deal we define
5:10:58
e to be the limit as n x y naught doesn't really matter seriously let me use n
5:11:11
and this is just one of the definitions so you can take this for the definition of the number e which is about 2.7 each all right and
5:11:19
if you guys recall you are compound interest formula so this is the compound
5:11:29
interest formula you have the amount it's equal to the principal times 1 plus r which is the interest rate divided by however many times you compound a year and then raised to the nt's power t is the number of years so this is the compound interest formula i will have to say that compound n times
5:11:54
in a year
5:11:57
right so that should be clear and you guys should have seen this back in precalc well if you will just continues if you alright let me give you an example
5:12:09
if you put money let's say one thousand dollars into the bank that will give you interest right then the way that they do it i believe most of them do it is they will compound every day so they will compute the interest that you earn every day okay but they will pay you every month so the idea is that if you have one thousand dollar in there if you have i'm saying regular banking
5:12:30
if you want to withdraw tomorrow guess what you still have one thousand dollars if you withdraw in two days you still have one thousand dollars okay that regular checking account while saving account
5:12:43
you have to wait until a month later and they will just kind of just keep track of all the interest that you have earned throughout the month and you get paid per month that's pretty much what works anyway imagine if you can get continuous compound
5:13:03
well in real life that's not possible but in math the idea is just you take the limit as n goes to infinity so you have the continuous compound formula all right
5:13:19
and continuous common formula is like a continuous growth and that's how you can get the exponential growth or decay depends on how you look at it but you will see this is the famous pert formula among is equal to p e r t pert formula where's the e well e is precisely because we are talking about continuous that means n goes to infinity you pretty much compound infinitely many times in a year and you see precisely this as n goes to infinity this part
5:13:53
gives you the e so we have the e and you see it right here as x goes to infinity this and that is the exponent of e just like this so the so-called continuous compound formula r t is the power of e right here and this is just like a coefficient so you actually have seen this before yeah anyway remember this formula then you can use it and i'll show you guys how number 76 take a look at the limit as x approaching infinity and we have 1 plus 1 over 2x
5:14:33
raised to the x power cool all right i tell my students to do this
5:14:41
identify the a and b and then pretty much use it what's the addo pay attention right here is that a it's actually one over two right it's one over two and then b is one so i would write it right here a is one over two and b is equal to one so this right here is just e to the 1 over 2 times 1 which is just e to the 1 over 2 power which is just a square root of
5:15:13
e all right okay so that's the answer for that nice and quick now try another one 77 limit as x approaching infinity and then we have one plus one minus
5:15:34
this to the x power so what's the a value do i hear negative one good yeah it's negative one and then b is again one so here a is negative one and b is one so it's e to the negative one times one which is negative one so this is just one over e if you want to write it as a fraction
5:16:00
and you want to write it as a rational form one over e
5:16:05
okay cool all right i'm going to give you guys a question to see if you guys are really paying attention or not question for you guys all right so leave the timestamp and also tell me what the answer is you see how here is just x to the first power what if we have the limit as x approaching infinity 1 minus 1 over x and then raised to the x square power go ahead think about this and then let me know what the answer is in the comment down below all right
5:16:48
okay
5:17:01
okay all right number 78
5:17:15
limit as x approaching infinity and then we have oh this is different x over x plus 2 raised to the x power okay so how do we do this well again we are going to utilize the fact and this is how first we see that we have two terms on the bottom we don't like that because we cannot split the fraction easy this way right so why don't we kind of just flip it can we do that yes we choose how to negate the power so let's go ahead and write this as the limit as x approaching infinity and
5:17:53
then for the inside i'm going to put it as x plus 2 over x and then raised to the negative x power
5:18:02
and now we can split the fractions so we are looking at the limit as x approaching infinity and then this is 1 plus this is 2 of x and then minus x
5:18:17
so a is equal to 2 b is equal to negative 1 all together is e to the this times that which is negative 2 and if you would like you can write it as 1 over e squared and then we are done
5:18:38
all right so this is the fact for that and now i will show you guys the limit which is the one that caused a lot of argument too for limits lopito's rule will not lobby those rule were or not
5:18:58
all right so this right here i call this the limit
5:19:09
so you see a lot of time we have and one more 100 something limits already and uh
5:19:18
why is this one the one i call the limit right
5:19:23
yeah anyway i'm going to use theta word this is the limit as theta approaching zero and we are going to get sine of theta over theta and this is equal to one this is the limit why because earlier we talked about the lobby does rule and you see that if theta is equal to 0 then we get 0 for 0 right so it looks like hey to do this it's super easy because when we have the limit as theta approaching 0 we can just go ahead and do l'hopital's rule like go ahead and differentiate
5:19:58
this with respect to theta and then do the same thing as well wow and you see the derivative of this what's that what's the derivative of sine theta yes you can tell me cosine but how do we do it well if you recall the way that we did the derivative of sine theta sine of x
5:20:22
you will recognize that when we are doing the derivative of
5:20:28
sine x all right you use a we will use the definition of derivative the limit as x sub h approaching zero all the stuff all the stuff right and then somewhere we have what the limit as h approaching 0 sine of h over h and here we are using the h for the variable you cannot just change your variable and then all of a sudden you know how to take the derivative no you see in order for us to use inverter for us to get the derivatives of x
5:20:59
we will have to find the uh we will have to find the we will have to find this limit first
5:21:08
but how can we use the different how can we use the derivative right now if i try to use if i were trying to find this limit how can we find it how can use the derivative we cannot so this is called circular reasoning yeah so a couple ways to fix this one you can use numerical method meaning i just use the the table of values plug-in you can see the answer you know is equal to one or you can do a geometric proof which i didn't do yeah but i give you guys a graphical proof i graph
5:21:41
sine x over x and we saw that the limit of this is equal to one but here i have something better whenever you encounter this countdown let's just say by the limit that um the answer is one and we are done here number 79 let me show you how we can work this out
5:22:04
limit as x approaching zero and then we have sine of three x over x
5:22:17
okay here the input is 3x what we can do is just do a nice substitution cos theta to be 3x and you see when x is approaching zero you can put it here theta is approaching zero as well good and perhaps we can also solve for x we can divide this three on both sides this means x is equal to theta over three so you can see this is going to be the limit x approaching 0 implies theta
5:22:48
will go to 0 and then here we have sine and then this right here is theta inside now and then the x is theta over three but you know the overstream in the denominator we can put it up right here so all we can say this right here is three times the limit as theta going to zero and then here we just have sine theta over theta and ladies and gentlemen this is just equal to one why
5:23:23
think for the we have this secret weapon to use so it's just one so we have three times one and the answer to that is just three and then we are done yeah much better please do not use sloppy toss rule for this one seriously just use the limit all right much better this way much better this way
5:23:54
all right and let me put it down right here
5:24:01
okay another one limit as x approaching infinity and here we have x times sine of 2 over x
5:24:12
wow this is infinity situation same thing we can do it don't worry let's call theta to be the input which is 2 over x perhaps let's solve for x yeah so multiply x on both sides d by theta on both sides this means x is equal to 2 over theta check this out
5:24:32
here we have the limit when x is approaching infinity put it here two of infinity is zero so theta is approaching zero good what's x that's two over theta and then we have sine and the input is 2 over x which is theta great 2 can be brought into the front yes and then we have the limit and then theta approaching 0 and then what's this what's this sine theta over theta which is one oh no two times one two
5:25:12
done
5:25:15
right am i happy that the secret weapon isn't over no am i happy now what's coming next not really but
5:25:26
i don't know why i put a hardest part at the head when my energy level is the lowest
5:25:36
but let's see we can handle it no worry so by now i will say uh this will cover enough of your car 1 car 2 limits and then uh for epsilon delta definitions depends if you are teacher your professor goes over it or not if you are at the unit if you are at a 4-year university then most likely you will have to know how to do the abstract data definition but don't worry i'll show you guys four words on how you can keep track and then organize everything
5:26:11
5 hours and 26 minutes all right so here is number 81
5:26:19
we have the limit as x approaching one you are going to be like wow seriously this is the question yes this is the question x approaching 1 and we have 6x minus 2.
5:26:34
that's number 81.
5:26:38
so easy right yes we can just put a 1 here i'll show you guys the work if you would like six times one minus two that's six minus two is four hey we are done yay
5:26:52
but how do we know that this is actually true yes we will have to prove it how though just like what i mentioned earlier the upside down delta definition i will remind you guys on the side and really depends if you have done it or not all right so i'm going to actually give myself more space i'm going to put down this right here it's four
5:27:18
okay here's the deal if we have the limit as x approaching some number a
5:27:27
of some function f of x if we do end up with some limits that's called it to be l this right here means
5:27:36
you ready it's ming i know but it means for all epsilon greater than zero there exists a delta greater than zero such that if the distance between x and a is small enough and for distance it means we have the absolute value and we just subtract them so we put absolute value of x minus a if it's small enough how small is more less than delta okay and then we do not need x we do not want x to be a so we want to say this is greater than zero
5:28:12
if this is true then we must have the absolute value of the well it's pretty much the distance between f of x and the limit l it's not enough as well but in this case we write it as f of x minus l to be less than epsilon yeah so intuitively it means that you can get as close to l as you would like then that's this is just like a full movie to say it because again f of x minus l is the distance between the function and the limit you can make it as small as possible
5:28:50
that's why it's for all epsilon greater than zero and we can make it as much as possible it's because you can always find a delta greater than zero so when this happens this happens that's the idea i have a very detailed video going over this so you should definitely try that but for now we will see how to just write the proof write the proof try the proof for 10 of them okay here we go how do we write proof write down pierre first everybody will be happy forwards
5:29:24
forwards that's going to make everything much much much much much clearer systematic easy to follow first word given why when you see for all right just start the proof by saying given and you can just say given epsilon greater than zero and some people might say you should say let absolute
5:29:48
uh be greater than zero like this and then b r b cherry but that's cool too you know uh if you say something like this you sound more like a mathematician and if you write given epsilon greater than zero that's true and consensus and people wouldn't get mad at you so either way but given nothing is okay so you always start given that some greater than zero if i ever ask my students uh epsilon delta definition on the test if you put this down i'll give you a point
5:30:20
among like 10 points anyway anyway next you see that right here it says there exists so to show that there exists you are going to pick you are going to select you are going to choose uh i like to use the word choose so that means you can show that hey we do have such a delta we are going to choose delta but the other is what we don't know yet just leave it blank for now a lot of students get stuck with epsilon delta definitions because they just don't know what
5:30:52
delta is and then they're stuck i'm telling you don't worry just write it down leave it blank for now and then we'll come back to you later okay okay such that we are going to assume that we have this so i'm just going to say suppose if you would like and also say assume but suppose
5:31:13
suppose this is true meaning that the absolute value and here we have x minus 1 x minus 1 because this is x minus a so it's x minus one suppose the distance between x and one is small meaning less than delta but we don't want x to be one okay so we put this greater than zero when you read it backwards because of this that's why five hours ago when we saw the graph you see we don't care when x is exactly 2 or 3 we don't care because we have this inequality
5:31:57
anyway we assume we are supposed we have this condition with that we hope to show this is true and we hope to show that means we better do the check so we are going to check or you can also say we see who we see here and then just go ahead and do the computations and all that but anyway we check check what the absolute value our function is 6x minus 2 and then minus the l which is four
5:32:34
okay function minus l and then we hope to show that this right here is going to be less than epsilon okay so this is the general framework now this is the part that we will have to do the most amount of work and then just do some algebra first minus two minus four is minus six and um
5:32:56
we can also factor the six on outside so perhaps i'll show you guys like this 6x minus 6 and then factor out the 6 and then we get x minus 1. okay and this is great because you see how we have the x minus 1 in the absolute value we know that right here we have this
5:33:17
absolute value of x minus 1 is less than delta so we can say that this right here is less than delta we replace this absolute value with less than delta but we still have the 6 at the front
5:33:32
so you see we create this inequality the less than that we want very nice but keep in mind though at the end we want to end up with what we want to end up with just epsilon we just want to end up with epsilon so what do we need well check this out this right here is 6 6 times what will give us epsilon well it's just epsilon over six isn't it because when we have that check this out six and six cancel and then we'll get the epsilon
5:34:04
so once we figure this out you see delta is just epsilon over six we rush back to here and then that's right bound let's write down epsilon over six and you see because epsilon is greater than zero and over six is still greater than zero so if you would like an indicator but nobody really does that i just want to say delta is greater than zero okay and a few things though you see how we pick delta to be exactly we pick delta to be equal to epsilon over six
5:34:38
so right here i'm going to put on equal sign delta is equal to epsilon over six and i cancel cancel so after the simplification this right here is still equal but the way that you read this is that we have this inequality which is equal to this which is equal to that and then this is less than this and this is equal to that is equal to that in another word this inequality sorry this absolute value is less than epsilon so we are done put box yay 20 more questions
5:35:18
it might be confusing the first few times that we're doing it that's what we're doing ten times
5:35:23
okay so let's try another one i'm going to erase everything not that all right okay let's let's erase this as well why i want to show you guys the more you write it down the better that we remember it okay let's try another one number 82.
5:35:44
here we have the limit and then we have x approaching negative two
5:35:53
and then two x plus seven
5:35:56
okay and then to do this you can just do this in your head negative two here two times negative 2 is negative 4 plus 7 is 3 so this right here is equal to 3 done now epsilon delta definition means if we have the limit and if x is approaching a if we have a function f of x
5:36:20
if this right here gives us a limit that's called l this right here means
5:36:28
no no where's the n this right here means
5:36:35
here we go for all epsilon greater than zero there exists a delta greater than zero such that if the
5:36:45
distance between x and a is less than delta but bigger than zero so radius absolute value of x minus a is less than delta but we don't want x to be 8 so we put greater than 0. this right here implies that the symbol for implies implies that the distance between f x and l is less than epsilon so we write f of x minus l is less than epsilon so just go ahead and write down a few more times and then watch the video where i
5:37:23
showed you guys an actual example with picture with number and you will see that it's like the
5:37:32
epsilon stands for error that's why it's e epsilon it's kind of like squeeze the just like kind of make the function as close to the limit as possible that kind of thing
5:37:49
all right so now here we go what's the first thing yes pf okay for o go ahead and write down given epsilon greater than zero next they exist so we are going to write down choose delta delta is what i don't know here don't worry leave it we'll come back to you later then we are going to assume we have this condition so i'm going to suppose
5:38:17
so given suppose even choose suppose and then check yeah suppose the absolute value here we have x and then minus a which is negative 2 so it becomes plus 2
5:38:31
suppose this right here is less than delta which also don't want x to be negative 2 so that's why
5:38:40
you have the greater than 0 and then check
5:38:46
check what absolute value of the function minus the limit so this minus that in the absolute value
5:38:54
two x plus seven and then minus three all right given the given choose suppose check given choose suppose check now focus on this and do the algebra 7 minus 3 is 4 and then we can actually factor out 2 so let's just go ahead and factor that out it should be okay and then we end up with what x plus 2. absolute value of x plus 2 so good because look back this right here we know is less than delta and we have a 2 in the front yeah now keep in mind we want to end up with epsilon so
5:39:34
we have 2 already 2 times 4 will give us epsilon we better have epsilon over 2 because this way this 2 and that 2 cancel so we know we should choose delta to be epsilon over 2. and of course epsilon is positive over 2 is still positive so satisfy it and because delta is epsilon over 2 so we put equal likewise this is just simplification we put equal sign and we're done
5:40:06
just like that okay
5:40:10
okay cool all right let's do another linear situation usually i'm just saying usually
5:40:20
you will see a linear situation on your exam you usually if not then a quadratic yeah but i'm going to cover both so don't worry however you're in on a class and that's a different story you might see like a rational or like uh exponential or like logarithm but yeah okay number 83
5:40:47
limits
5:40:49
x approaching 3
5:40:54
2 over 3 x minus five
5:41:00
okay let's do this in our head
5:41:03
pull string here cancels two two minus five r is negative history let me see
5:41:14
yeah okay cool here proof first given
5:41:23
epsilon greater than zero next we are going to choose choose what choose delta to be what don't know leave it then suppose
5:41:35
suppose the distance between x and three so we say x minus three we want this to be less than delta and we don't want x to be three so put a greater than zero by looking backwards then we check
5:41:50
absolute value of the function minus the limit so we are looking at two over three x minus five minus negative three so it becomes plus three okay here we go we have a fraction so let's see how it goes okay negative five plus three is negative two so here we have two over three x minus two e okay can we factor it yes we can still factor out the 2 over 3. check this out
5:42:29
so i bring this up so it's x and then we have the minus but our original d here is uh minus two we factor out the two over three so what should we do we should look at this and divide it by
5:42:46
we should just look at this divided by that dividing by 2 over 2 is the same as multiplied by 3 over 2. they cancel so we just have just a string i'm not kidding
5:43:00
check this out distribute back we get this this time starts three cancels two minus yeah what's this less than delta so we know this right here is two over three times delta and at the end we want this to be equal to epsilon right
5:43:23
so you know it delta should be equal to what well this right here we will have to kind of work it out a little bit two over two times okay we want to have a epsilon here and then perhaps we just need a reciprocal right namely three over two because this way
5:43:41
two and two cancel three and 3 cancel perfect so that means delta is equal to 2 over 2 epsilon
5:43:50
yeah this is equal to that put it there done yeah and as i said you don't necessarily need to put a like greater than zero uh this is okay yeah because in my previous video i don't think i put down like delta is less than greater than zero usually you can just write down formula and that's it so here are the three examples for epsilon definition when the function is linear
5:44:47
all right number 84. let's look at a first situation that the function is not in here and then why do i look at the okay it's um it's it's it's going all right anyway take a limit as x approaching 4 and then we have the square root of x man this if it's a computational question super easy huh because the answer is just two yeah but for proof let's just go ahead and do it pf here we go given
5:45:18
epsilon greater than zero next we are going to choose delta i don't know just leave it then suppose suppose what absolute value of x minus 4. to be less than delta and then greater than zero we'll go backwards and then lastly we check
5:45:43
absolute value function minus the
5:45:47
limit okay that's all we have all right now uh we cannot factor this anymore huh but don't worry we did a lot of this example before we have square roots what did we do today the count you get so let's go ahead and do that i'm going to just write it down right here i'm going to have square root of x minus let me write it down like this
5:46:14
square root of x minus 2 right and i'm going to multiply the top and bottom by square root of x plus 2 and then over square root of x plus 2. aha yeah and then you see this right here will give us just x on the top and then minus 4 right so it's that and then
5:46:39
over square root of x plus 2. all right so so far so good take a look hopefully it's okay so far
5:46:52
it's hot it's warm here but it's okay almost done what do we do next though well here's the thing when we have an absolute value of a quotient this right here is the same as the quotient of the absolute value meaning that we can look at this and say this is just the absolute value of x minus four and then over absolute value of square root of x plus 2. but the thing is that the output of square root is always positive likewise when we add a 2
5:47:23
to it it's always positive so this square root on the bottom doesn't matter you can erase that
5:47:30
good and then you see right here this is the place that we can use the less than the outer which is yay but what exactly do we do with the button though we have square root of x plus 2.
5:47:45
well i'm actually just going to get rid of it yeah i'm just going to get rid of it but you have to say a few things for this though so i'm going to
5:47:58
do like this i think it's clear if i write it down right here this right here is actually just equal to so this right here is actually just less than absolute value of x minus 4. just get rid of it why i'm going to give you guys a explanation because square root of x plus 2 is always greater than 1.
5:48:21
i'm not saying the bottom is always positive i'm saying that the bottom is always greater than one that's why when i get rid of this right this right here is smaller than just the top
5:48:33
let me give you an example to convince you guys so suppose you have eight versus eight of course they are the same but if today you have eight and then divided by two then becomes what this right here becomes smaller than that right yeah however if you have eight divided by
5:48:57
let's say a number that's um positive but it's 0.5 compared to 8. which one's bigger
5:49:08
this is bigger now yeah if you divided by 1 they are equal if you divide it by a number that's bigger than 1 then this right here will be smaller than this that's why i'm mentioning square root of x plus two is bigger than one so i can legitimately get rid of the bottom and then just say this is less than just the top so this is the idea so that should be clear right so that's really really good because now when we have just the absolute value of
5:49:41
x minus 4 we can say this right here is less than delta now we can say this is less than delta yeah and then what exactly do you want at the end we just want this to be less than epsilon right so we want the whole thing right here to be like epsilon
5:50:03
so what do we want we want delta to be equal to absolute so we rush back here and just say delta is the same as epsilon and we are done
5:50:15
so this right here is less than delta which is the same as epsilon
5:50:24
and we are done
5:50:27
all right so this right here might be a slightly um difficult case especially if you have not seen these kind of things before but again when we're working with inequalities or absolute values uh just be really careful this is one of the things that we can do
5:50:48
this is the key that simplifies this equation tremendously draw tremendously if you don't use this you can still do it other ways like the ones that we are going to do later but this is so much better thanks to absolute 72 square root of x is always positive and then add 2 to it we know it's always greater than 1. all right so that's 84.
5:51:36
all right 85 another square root example slightly trickier or slightly more complicated but we can handle it limit as x approaching 6 square root of 3x minus two okay put six in here we get square root of eighteen minus two eighty minus two is sixteen square root of that is four
5:52:04
now prove it p f given epsilon greater than zero next we are going to choose
5:52:16
delta i don't know yet so leave it then we are going to suppose
5:52:23
okay and then we have x minus six so just go ahead and put down the absolute value
5:52:30
and then this is less than delta and then greater than zero then we are going to check
5:52:37
absolute value function minus the limit so we have square root of 3x minus 2 minus 4. cool
5:52:48
now for this right here do the conjugate business again so this right here equals square root of three x minus two minus four multiply the top and bottom by square root of three x minus two plus four and then do the same on the bottom as well square root of 3x minus 2 and then plus 4. cool
5:53:12
now multiply out the top we just get absolute value is just 3 x minus 2 and then minus 4 squared which is 16. and then the bottom we just keep it so it's over square root ah
5:53:33
it's just over square roots there we go i like to have a space between a square root and so the fraction bar three x minus two and then plus four
5:53:45
cool now let's do algebra first though we see that minus 2 minus 16 is minus 18 and then we can factor out 3. so on the top here we can have 3 absolute value and then it's x minus 6 okay and then same idea like earlier it's absolute value of the top and then absolute value of the bottom but the bottom is always positive so we don't need the absolute value so we can just say square root of three x minus two and then plus four and what can we do what can we say next
5:54:21
well we can just get rid of the bottom part and say this right here is less than three times the absolute value of x minus six why because we will mention that square root of three x minus two plus four is greater than one right that's why you have to mention you don't mention this is greater than zero you mentioned this is greater than one from here to here we mentioned it's greater than zero because we can carry out the absolute value but from here to here we mentioned this is
5:54:54
greater than one because we want to have this inequality and the carry of the whole thing
5:54:59
cool now we have this which is that which is excellent so this right here is again another less than thanks to that and we have this three right here and then we have a delta and we know it at the end we want to have just epsilon so now what's the delta epsilon over 3 yes this is 3 times epsilon over 3 and then we have this and that bye bye so we come back here and then say delta is equal to epsilon over three this is so cool it's symmetric if you go to 180 degrees
5:55:38
yeah anyway this is equal delta is equal to epsilon over three and then we are done
5:55:48
we're done done yeah
5:55:54
we're done yeah no we have a couple more to go of course
5:56:16
all right quadratic cases so two quadratic cases and then one cubic and then two rational
5:56:25
and then we'll move on to riemann sum and it will be done
5:56:32
all right number 86 limit as x approaching we have
5:56:41
three two x square plus one let's just work this out real quick 3 squared is 9
5:56:54
times 2 is 18 plus 1 is 19.
5:56:59
all right here's the deal if somehow you computed this round let me tell you later on the proof is going to horrendous if the proof does not go as smooth as the example that i've been showing you guys earlier maybe we just added wrong one and did something wrong in the at the beginning
5:57:17
i don't know anyway p f proof given epsilon greater than zero and then we are going to choose
5:57:29
delta i don't know yet but you know i'm not worried about it suppose absolute value x minus 3
5:57:40
is less than delta and then no greater than zero like this lastly we are going to check and hopefully everything goes smoothly absolute value of the function 2x squared plus 1 minus the limit which is 19. right here we go this minus star is minus 18 and then we can factor out two so this right here is two all right
5:58:09
let's see all right two x squared
5:58:14
okay
5:58:19
2x squared i'll show you guys all the steps minus 18.
5:58:24
and then we factor out 2 and then we get x squared minus 9 all right and perhaps our yeah yeah this is okay and then okay and then this is 2 times x minus 3 times x plus three okay so show all the steps one legitimate i was going to go from here to here but now yeah i'll show you guys everything and from here to here we use the fact that the absolute favorite product is the part of the absolute value so it's pretty clear so you can skip that's another step
5:58:57
but anyway though this is nice because we can say that we can have less than two times delta but now here's a troublesome that's the value of x minus 3 how can we get rid of it well how can we get bound meaning that we want to show this right here is less than some number as well so that we can kind of maintain this inequality so it can go smoothly
5:59:24
how do we do it though okay this is where the magic is we are going to come back to the author come back here right now and then put down minimum
5:59:37
and then like this notation like a set notation and then just use one key might just use one oh i heard can you use two can you use 17 sure up to you if you want to have a smooth experience then i highly recommend you to use one if you want to have some at
5:59:57
adventures use pi use one over e just use one please no no use one over pi sometimes two yeah i don't know anyway just put on one why because the moment that we have the minimum and also the one the moment that we have the one right here we are allowed to come here and say because delta we have this one that we can use so look at this inequality just this side and we can say absolute value of x minus three is less
6:00:30
than this number that we put down earlier we put down one so we can use one we put down one over e you use one over e you want to put on square root of 17 square root of 17 up to you now from here we are going to somehow get an expression for absolute value of x plus three to do so one systematic way is just you first open the absolute value so we will know that
6:00:57
when we do that we get absolute value we get x minus 3 this right here will be in between of 1 and negative 1. right that's how absolute value works right if this is less than 1 then that means the insides in between of negative 1 and 1. and now you see we want to have x plus 3 this is x minus 3. so let's just go ahead and add 6 to everybody and this is of course legitimate because this is just a regular inequality and then you see this is 5 and this is x plus 3 and then this is 7.
6:01:32
okay now question for you guys i want to go from here back to the absolute value how do we do it
6:01:41
well here's the idea just compare the absolute value of this number number which one is bigger seven has a bigger absolute value so it's just absolute value of x minus x plus three absolute value of x plus three is less than this number the bigger absolute value why because when we have this of course x plus 3
6:02:07
if it's greater than 5 it has to be greater than negative 7 so you see that we have negative 7 also 7. so of course we can write this as absolute value of x plus 3 less than 7. yeah and later you are going to see a slightly strange case but for now just remember look at which number has a bigger absolute value so you know what now you see we have this absolute value of x plus 3 is less than 7 this is great because this is less than we have less than already we don't
6:02:39
have to write it again this is not a list you can just say that this right here is less than seven
6:02:47
yay
6:02:50
yay seriously okay now question what's your delta b though because again at the very end we want to have this to be just delta so we just want to have this as just epsilon so what should delta b
6:03:09
2 times 7 is i want it to say 17. 2 times 7 is 14. so we better have delta to be epsilon over 14. so that you see 2 times 7 this this and that will cancel
6:03:29
now the outer we said is epsilon over 14. we come back here and then we put down epsilon over 14. close this right here that we do not put on equal sign because we say delta is the minimum of 1 or epsilon over 3 so it's a minimum we want this right here to be less than or equal to because delta is less than this what that okay we're done we're done
6:04:15
please do not ask me to do 100 delta definition please no no 10 is it's good 10 is good
6:04:26
please because this right here are still like to be honest like introductory level when we're talking about real analysis there are crazy stuff such as proof a function is uniformly continuous or dance you have like just so many that's not good let's not talk about it okay
6:04:48
87 limit and we have x approaching negative 2. x squared minus 3x is that right cool all right put that in here we get 4 minus 3
6:05:09
6 so it's 10. right
6:05:17
good
6:05:21
all right so now again just go ahead and prove it this is a quadratic decay so it will be very similar to what we did earlier so here we go pf let's go ahead and say given epsilon greater than zero and then suppose delta is equal to something then
6:05:44
no no no no even choose sorry i'm tired choose
6:05:51
suppose here we go absolute value of x minus negative d2 so it becomes plus 2
6:06:00
is less than delta and then this is greater than 0 and then we want to check
6:06:07
absolute value function minus the limit so x squared minus 3x and then minus 10.
6:06:17
oh my god look at this how beautiful is this this is a quadratic trinomial we can factor it the usual way okay we need x minus five times x plus two so that's the factoring so here we go we have absolute value of x minus five let's let's put on x plus 2 first
6:06:36
because that way we can put on the delta first so x plus 2 times x minus 5.
6:06:43
okay this right here is less than delta that's nice now we have to get a bound for absolute value of x minus five you know the deal come here and write down minimal put on one just use one please
6:07:01
all right come here just this side that's just this part so we can say this right here gives us absolute value of x plus 2 is less than 1 and we want to somehow go from here to here so let's open the absolute value that means x plus two is in between of negative one and one we want to get to negative five so that's minus seven minus seven here minus seven here minus 7 here right so in the middle we have x minus 5. this right here is negative 8.
6:07:38
and then i'm just subtracting sort of inequalities maintaining the same and then less than 1 minus 7 is negative six hey this is crazy because if we have the absolute value of x minus five this is negative that's negative i cannot say this is less than negative six that doesn't make sense this doesn't make sense but the deal is that remember what i told you guys earlier
6:08:08
we are looking at which number has a bigger absolute value i'm not talking about which number is bigger so i'm telling you guys that hey just do absolute value of x minus 5 put as less than the absolute value of negative 8 which is just an 8.
6:08:26
why is this true because here if you open this we get what x minus 5 is in between of negative 8 and positive 8. negative 6 is right here of course this covers that situation right
6:08:46
so here we have it this blue part is this we can just say hey multiply that by eat that's my delta that's my eight at the end we want to end up with epsilon so delta should be equal to what epsilon over eight yes let me just write it down nicer for you guys i'm 8
6:09:11
cancel cancel come back here epsilon over 8. close that make it legitimate and then right here less than or equal to because we use the minimum
6:09:26
so this is how you write the proof for it yeah could there be another choice for delta yes because if you use let's say 2 then this delta this expression will be different and you will still work theoretically speaking yeah but to really understand what's going on for absentee definition i would highly recommend you guys to watch my video because i use actual example my actual numbers and also the graph and i explained it when i'm like when i was like 100
6:10:02
energy right now i don't know i use 78 of my life already so i have like 13 more
6:10:17
okay
6:10:23
all right now a cupid example i also picked this one on purpose number 88
6:10:32
what's a limit as x approaching two of x to the third power you know it's eight why look eight eighteen and eight okay all right proof
6:10:45
this is like the extension of the quadratic case if you have understood
6:10:53
the quadratic cases then this right here not so bad by the way
6:10:59
given epsilon greater than zero choose delta i don't know yet suppose
6:11:09
absolute value of x minus two
6:11:14
is less than delta and put this check absolute value function minus the limit
6:11:26
all right here we go algebra work this is the difference of two squares this is the same as two to the third power so we can factor this as absolute value of x minus two times
6:11:42
absolute value of x minus 2
6:11:45
times x squared plus 2 times x which is just 2x and lastly we are 2 squared which is 4.
6:11:54
cool what's this this is good yeah less than delta everybody likes it what's that
6:12:05
don't worry come back to the delta part and then we just say minimum let's use one okay starting right here let's draw this a little bit longer because i have that okay so we're looking at absolute value of x minus two is less than one for here we want to get to x square part so uh it's a little bit tricky so pay your attention right here um but the idea is the same if you can get a bound for this and you are in good shape anyway open this so we
6:12:42
get x minus 2 is in between of negative 1 and 1. and for this particular case i'm going to just
6:12:51
because i cannot get to x square unless i square everybody no let's not do that let's just add 2 to everybody you'll see why so they cancel and then we see this is 1 this is x and this is three
6:13:08
and then just like what we did earlier we can say absolute value of x is less than three so you pick the bigger absolute value which is this now here is the deo we have this condition thanks to that and we want to see this so i will just tell you guys that notice from here and there we are going to make a connection when we have the absolute value of x squared plus 2x plus 3 plus 4.
6:13:43
you see how we have the absolute value of the sum
6:13:49
this right here is going to be guaranteed less than or equal to the absolute value of the first
6:13:56
plus the absolute value of the second plus the absolute value of the suit yeah so this is the extent version of the so-called triangle inequality so let me just write it down somewhere
6:14:10
by triangle e e quality let me give you guys an example real quick suppose you have absolute value
6:14:21
let's say we have five plus three well is this less than or equal to absolute value of five plus f to five three yes in fact they are equal because it's eight it's actually equal to eight
6:14:35
but if you have let's say this because you don't know if x is negative or not i don't know right so let's say if we have five plus negative three i'm going to put the three right here and make it negative three you see here we actually just get two but here we have eight the right hand side speaker so again when we have absolute value for sum this right here is less than or equal to the sum of the absolute values this is the so-called triangle inequality
6:15:11
so that's another thing that you'll keep in mind because once you have this check this out i'm going to use equal sign because this is equal to absolute value of x
6:15:23
this right here is equal to absolute 5x squared we can put a square on the outside and the reason for that is absolute 5x squared is equal to absolute value of x times x and the per hour discussion earlier you have the value of a product it's a product of the absolute value so this is the same sms it depends if you guys have seen this kind of proof or not in your class hopefully you can just go from here to here but anyway it's this equal sign and then you can put a
6:15:52
2 on the outside so it's plus 2 absolute value of x and then absolute value of 4 is just 4. so from here to here is just straightforward equal sign now we know absolute value of x is less than 3 so we can replace the real absolute value with less than and then three squared plus two times three and then plus four and then from here to here it's going to be equal sign because it's just computation let's see this is nine plus six plus four so nineteen so all we are saying that this
6:16:32
absolute first expression is less than 19 which is 7.1 this is less than 19. let's have a look
6:16:50
because you know once you have that we're almost done at the end we want epsilon
6:16:57
delta should be epsilon over 90 that way when we multiply by 19 they cancel so we can come back here and say we also need it epsilon over 19 and close this less than or equal to finally don't you ever forget the box shading ah cool absolutely i mean um that's some delta definition it's a lot of fun when you know how to work out when you know how to write down a good proof for it it's it feels very satisfying
6:17:51
it feels very very satisfying yeah i think this is the reason why a lot of people like real analysis uh only after they understand how to solve it because the process of like understanding is super hard but once you understand it yeah the feeling is super satisfying
6:18:10
yeah it's like running a miracle the running process is not that enjoyable but after work after you finish the miracle you know it feels good just like why i like to do 100 questions
6:18:26
all right russian no example number 89
6:18:38
all right this is uh limit as x approaching two yeah this is my
6:18:45
uh what's up my baby calculus versus the do calculus video question right it's super easy huh one over two let's take a look at the proof of it here we go p f okay given epsilon greater than zero choose i don't know yet leave it suppose we have the absolute value of x minus 2
6:19:15
is less than delta and then greater than 0 then hopefully we can get the absolute value to work for this we want the function minus the limit now let's do the algebra let's get a common denominator we get absolute value of 2 minus x over 2x okay flip this subtraction so we can use that legitimately so on the top this right here is the same as negative x minus 2 right and then i can take the absolute value on the top and then the absolute value of 2x
6:19:54
yeah and then let's see i'm going to write this down right here the negative in the absolute value becomes positive doesn't matter so we just have absolute value of x minus 2
6:20:07
times 1 over put a 2 on the outside and then we have the absolute value of x this right here we know is less than delta excellent but we don't know what this is 1 over 2 absolute 5x so you know the deal we're going to come here and say delta is equal to the minimum and then we say 1. that way we can utilize this inequality just this part absolute value of x minus 2 is less than one and
6:20:42
want to get this expression huh so hopefully you can find this right here to be less than something that's the goal i want this to be less than something okay open this absolute value of x minus
6:20:54
this means x minus 2 is in between of negative 1 and 1 and then just add 2 to everybody
6:21:03
we know x is in between of one and three uh this means absolute phi of x is less than three right that's what we have done so far and in fact i picked the same number like the previous one on purpose because that's what we did earlier and now we have absolute 5 x is less than 3. we want to get 1 over absolute value of x so should we just do the reciprocal right now well if we do that keep in mind both sides are positive so when we do the reciprocal on both sides
6:21:38
we have one over absolute five of x and then one over three
6:21:43
but we will have to switch the inequality yeah for example we know two is less than three
6:21:52
but if we do the reciprocal on both sides in fact one half is actually bigger than one over three
6:21:59
ah man this is so bad because we cannot proceed we really want to see this right here is less than something but now we have one over absolute value of x is greater than something no good so what do we do i'm so stuck no don't worry the trick is that you do not do the absolute fault right here right now in fact we actually do the reciprocal first and everything will go smoothly check this out everything's positive so do the reciprocal one over one still one and here
6:22:36
we have one over x and this is one over three and then flip the inequalities right this this
6:22:44
do the reciprocals first and then take the absolute value absolute value of one over x is just one over epsilon of x that should be clear and now which one has a bigger absolute value one so in fact this right here is less than one yeah and then minus will put a two right here as well as it's two on the bottom though so let's multiply both sides by two so one over two absolute five x this right here is less than one half
6:23:15
this and we have the less than already very good uh we know this right here is stands down one half so what do we need for the delta this one is rather special huh we want the delta to be 2 epsilon because this way when we multiply by 1 over two the two and two will cancel so hurry up let's put delta to be two epsilon here and then close this
6:23:48
and then the minimum zone then finally draw the box shake the scene
6:24:04
hey if you feel good with epsom delta definition don't worry there's one more
6:24:10
and for the last one i'm going to erase that as well right so i'm going to see if we can record everything i think we have been doing a good job recording all the definitions
6:24:22
right all the steps so you see after 10 epsilon delta definition in a row it's not so better
6:24:40
right limit as x approaching one as i said another
6:24:50
rational case and for this one put one in there work that out with your monster so simple but yet it's going to be also pretty difficult don't worry though i'll take up you guys pf first given given delta sorry given epsilon greater than zero then choose delta i don't know yet then suppose
6:25:20
absolute value of this minus that so x minus 1 is less than delta then greater than zero then we want to check okay absolute value function minus the limit so we have one over two x minus one so two x plus one minus one over three right now let's just get the common denominator all the good stuff so just put it down right here maybe we need more space
6:25:54
okay so three right here minus right and now i'll just distribute so minus 2x and then minus 1. okay
6:26:05
and then on the bottom i'm just going to write it as 3 times two x plus one yeah something like this hopefully you guys are okay with it absolute value all right okay three minus one is two and then we can factor out negative two so this right here becomes absolute value negative 2 and then
6:26:30
this is negative 2x plus 2y so when we factor out a negative 2 this becomes x minus 1.
6:26:41
yeah so far so good we see that and then um
6:26:49
the bottom is just that i can't really do too much so i'm just going to keep it as that so 3 2 x plus 1. now break this apart absolute value of the top and the absolute value of 2 just negative 2 just 2 right so we have 2 and the absolute value and then x minus one okay and then we have this on the bottom put the string on the outside so one over three and an absolute value of this two x plus one all right so that's the algebra that we have so far
6:27:26
you know the deal come here minimal
6:27:30
1.7 no no please don't don't do that of course you should do one over pi okay okay okay
6:27:44
one
6:27:47
all right here let's see look at absolute value of x minus one is less than one open it we get x minus one it's in between of one here and then put negative one here and then just add one to everybody so we see that's zero and then that's x and then oh i'm trying to get to there now but let's do the things that we did earlier uh two okay okay fine i need a 2x so let's multiply everybody by two so multiply by two multiply by
6:28:27
two multiply by two so 2x is in between of zero and then four and then let's add one to everybody
6:28:37
ah so yeah i have one one and one things like this pretty cool i would say so we have two x plus one is in between of one and five
6:28:49
what did we do earlier we take the reciprocal first right so we take the reciprocal this is going to be one over one which is one and flip 1 over 2x plus 1 and then flip and then also reciprocal that which one has a bigger absolute value this does so this means
6:29:11
1 over two x plus one in the absolute value is less than one really interesting because we saw the one earlier too yeah and we want the street right there huh so let's just say that
6:29:29
to having a space one over three absolute value two x plus one is less than one third
6:29:38
okay all right this thing we have the
6:29:45
less than two steel two this is less than delta i forgot to say that earlier i was too excited to work out that all right and this thing let me just do like purple there
6:30:01
it's less than one over three so multiply this and we get one over three
6:30:07
now ladies and gentlemen all the way at the end we want to end up with epsilon so what shall we pick for delta we better have a delta we better have epsilon here right we divided by 3 so we better multiply by 3 we multiply by 2 here so we show divided by 2. and when we have that
6:30:33
we can see 2 and 2 cancels 3 as you can so and we'll just end up with epsilon and delta is equal to that so we will come here and put down 3 over 2 epsilon and then hurry up let's close this and then put down and then
6:30:58
yes yes
6:31:03
you think we're done no we have 10 more pretty hard questions why are they
6:31:11
riman are they any easier depends they definitely look scarier because they are longer you see this right here they look like the question looks like this but like the works a lot
6:31:26
for the remain sum for the in definition of integral pretty much um the the the
6:31:35
question is like that long and then the work is actually not bad
6:31:44
i'm going to simplify for you guys right i'm not going to over complicate the riemann sum because once you know what to look for then yeah you know that you're already i do have a very detailed explanation for the muslim as well so yeah ladies and gentlemen number 91 will come to a very very long limit limit as n goes to infinity ready first one 1 over n times 1 over n squared next one 1 over n times 2 over n square next one 1 over n
6:32:28
times 2 over n squared i'm not going to write that down i'm just going to put on plus data 1 over n n over n squared all right so there are a few ways that you can do it if you recognize what the region is then you can convert that into a
6:32:48
integral question right away so that's that but for this case right i want to showcase at this one time because this is everybody's first definite integral so i'm showing you guys how to do it a very classic way all right so i'll show you firstly
6:33:08
you keep the limit as n goes to infinity and notice everybody has the one over n so i'm going to factor that out in the meantime though n square and square everybody also has n square on the bottom as well right so i'm also going to factor that out right here and you will notice that this right here will just have one square left next we will have two square left and there's three square left and so on and so on up to the very last part which is n squared like so okay
6:33:44
then you are going to see this is actually just very nice
6:33:48
to be approaching infinity this is 1 over n to the third power and all this we have extremely nice formula we have extremely nice formula
6:34:13
what is it ready n times m plus one times two n plus one over six check my other video to see how we can prove this formula all right
6:34:31
yeah in fact this is really cool because this formula actually contains 1 plus 2 plus 3 plus up to n because this right here is n times m plus 1 over 2. so we have the n times n plus 1. break this down as 2 times
6:34:52
well that's a discussion for another video anyway
6:34:57
does this look familiar you should the limit as something to the approach infinity right
6:35:06
so this is the limit as n goes to infinity on the bottom we have six and cube on the top just for five i'm going to multiply out for you guys here we have an squared plus n and then we multiply 2 n plus 1. so this right here is 2n to the third power and then n squared and then 2n squared yes together is 3 n square and lastly we have n
6:35:39
right you really don't have to multiply everything but number question number 21 my soul is loading
6:35:50
leaving my body
6:35:56
limit is unapproaching infinity this is everybody's favorite calculus question on the top we care about 2 to the so we care about 2n to the third power on the bottom we also have the same thing and to the third power they cancel out reduce 2 and 6 we get 1 3 hey the answer is just 1 3 very very nice very very nice all right now it's about time to tell you guess what this represents this as i said
6:36:25
is everybody's first integral equation and it's just going to be the area under x square
6:36:36
from 0 to 1. how well i'm going to show you don't worry when we have 0 to 1 the idea is that we are going to cut this into n equal with rectangles that's why we have all the 1 over n so just imagine kind two unequal piece rectangles so this is the first piece and i'm going to use the right end point to construct a rectangle meaning that use this x value go up hit the curve and then we are going to just construct our rectangle right here and because we
6:37:13
have n rectangles just imagine a 10. then in this case would be like 1 over 10 right but against n so this right here is 1 over n the height of this rectangle it's just a y value we use this x value 1 over n we put it here and we get 1 over n squared so this height is 1 over n squared times the base is one over n so this right here is the area of the first rectangle next
6:37:44
where we have this rectangle here but you go up by another one over n so this x value is two over n so go up hit the curve and then construct the rectangle the width is still 1 over n though but the height we put 2 over n into the x so we get 2 over n squared so this right here represents the area of the second rectangle so on so on so on and the very last one is one which is just the same as n over n we go up hit the curve and then we construct our rectangle
6:38:18
yeah which is the one over and the highest and over n square the area of the last rectangle
6:38:26
as we have n approaches infinity each rectangle is going to be so so so so thin like infinite has more small yeah so you can actually get the area okay all right so that's cool but i want to show you guys what's cooler check this out this right here is just a way to say we have the limit as n goes to infinity and then you see that we have all the addition which is the summation which we can use sigma for it we can write it as sigma like this and then
6:39:02
you see we have the one two three up to n that changes right they change right so that's where we need to use the running index we'll say i just run index i goes from 1 to n inside here we have i
6:39:20
over n yeah and then we square that and we multiply by 1 over i just put one over at the end
6:39:29
this right here is the width aka delta x delta x another way to look at it is
6:39:37
from where to where and we write this b minus a over n so the range right it's only one unit away if we start with zero we end up with one if we start with 17 we end up at 18. so that's the structure and guess what if you know the structure of this we can change this to a super version that is the circle integral what's the function you see the input is i over n that's what we talked about we enter all this into the x so that's the function so the function in this case is x squared
6:40:15
yeah and then you can actually look at it from here one over n right and based on this we know b minus a is one so just go ahead and start with zero you can always start with zero you will see another question later on so we can just go from zero to one and then the delta x
6:40:38
that's so i don't know it's 20 22 already don't write delta x write dx
6:40:46
so anyway this right here is the explanation of the integral notation right so you can ignore that i want you guys to be able to go from here to here but of course it's hard sometimes so definitely draw the picture and convince you guys that all the setup is correct it's correct though because you see we just want the area going from zero to 1 under x squared and the dx is just the width of the rectangle so it does make sense
6:41:15
when we have the dx in the integral notation right so how do we do integrals well we are going to use the reverse power rule so if you haven't learned about the integrals by now you can watch my 100 integrals video or you can pause for a little bit for now if you are in calc one then this is for try to focus on and maybe come back after a week or two then this is what you will be doing anyway we use the reverse power rule meaning that we find the entire properties l1 to the power
6:41:52
which is three and then divided by the new power so we have one third so an anti-derivative for that is just one over three x to the third power and then we're going to plug in plug in plugging one and subtract plugging zero this process is called the fundamental zero calculus part two anyway i'll show you one over three one to the third power this is the first part minus plugging zero we have one over three zero third power this is zero this is yes you know it
6:42:27
one third so yeah this weird curve shape has the area of a very nice number one third
6:42:42
all right so keep in mind of how we go from here to here but i want to warn you guys though a lot of times
6:42:50
the questions will put one over to here so you will see we have the limit as n goes to infinity and then one over n and then the sigma and then i goes from one to infinity and whatever this is
6:43:06
all right but just worry about this right here will give you the function part this right here will tell you from where to where and change that to the integral and that's it all right all right the next example is going to illustrate why we can always just downwards zero
6:43:29
the soda is not cold anymore so i'm
6:43:34
i don't know if i shall finish it anybody wants it probably not don't
6:43:49
okay number 92 here it's very similar to earlier you'll see this is the limit as x of x n goes to infinity and then we have 1 over n and then we are going to have 2 plus 1 over n squared and the next 1 over n times 2 plus 2 over n square and so on so on so on the next one will be here 2 plus 2 over n and all that stuff okay you can see on the handle but i'm just not writing it down and then one over n 2 plus and over n square
6:44:30
all right
6:44:33
so here we go let's combine this into a sigma notation it will be the limit as n goes to infinity everybody has one over we can factor it and then you see i'm going to put it right here okay usually nobody really put it at the end but that's the dx point you just have to realize that and then we have the sum i goes from 1 to n and i is this 1 2 3 4 and so on up to n that's the changing part and this time you see we have two plus i over n and then square
6:45:16
okay now based on this what region are we talking about
6:45:24
well again let's focus on this right here first we have one over n huh so this right here tells us b minus a will give us one
6:45:36
if we start with zero we end with one if we start with 25 we end up with 26 etc etc
6:45:46
all right does that help not really okay just start with zero that helps just start with zero that helps okay just start with zero so i'm going to claim that this right here it's actually just the integral going from 0 to 1 just our zero and then the function is actually 2
6:46:09
plus this right here i over n it's just going to be the x and then square and then dx so you can change that to integral this integral of course represents the following
6:46:27
have a look if you graph 2 plus x it's just like so if you have 2 plus x squared it's just like you have a parabola by put to the left 2 units right so when x is negative 2 here we have a parabola
6:46:45
something like this okay and i'm saying that the region that we want is from here to here is root 1
6:46:53
and then have a look this is y equals 2 plus x of uh to the second power
6:47:00
does this satisfy our original it does have a look we go from here to here so the whole thing is one cut into n pieces so each piece is 1 over n so the first x value that we can use is 1 over n
6:47:18
put it here the height is what 1 over n in here so we have 2 plus 1 over n square the next x value 2 over n
6:47:31
put it here we have 2 plus 2 over n squared that's the height and then everything here the width is 1 over n so yeah that's your confidence you guys know
6:47:44
okay what if you don't want to start with zero
6:47:49
ah i don't know if you want to start with two it's okay too show you if you start with two then you go up to three per hour discussion earlier right and in that case the function is no longer the same right here but you put it as just x squared dx
6:48:14
why let me show you this right here will give you that as well check this out
6:48:22
so we have our x square horrible picture i know i'm sorry but if we go from two to three so this is the region that we are talking about now okay
6:48:35
why we start with 2 because this right here is a 2.
6:48:40
check this out from here to here the whole thing is still 1 cut into n pieces so this right here the width is still 1 over n so we still have one over and there good but what's the x value
6:48:55
from here to here is two we have to add one over n so this x value is two plus one over n
6:49:05
and then you see our function is x squared we are going to put this whole thing 2 plus 1 over n 2 here and then go up you see that's how we can get the height which is that same thing next this right here is what from here to here is 2 1 over n another one over x all together is 2 plus 2 over n whoa look at that this is the x value put it here look at that same thing so that's why i said i said just always start with zero for this
6:49:40
kind of questions everybody will be so much happier if you don't want to start with um
6:49:47
if you don't want to start with zero that's totally okay it's just that you kind of move the region around this is the microphone i moved to here let's do the microphone same thing so it doesn't matter how you move it horizontally you will still have the same area under the curve this right here is definitely easier to compute so that's the advantage compared to that all right
6:50:10
so i'm going to compute this for you guys do the same thing anti-derivative so you get one third x to the third power and then we're plugging two three this right here one third three to the third power that's the first part and then minus one third two to the third power yeah okay that's 27
6:50:40
27 minus nine
6:50:46
eighteen
6:50:51
no that's eight oh my goodness twenty-seven minus eight nineteen right seven minus eight is eight this is this nineteen over three there we go almost right here i was like oh how can this be so nice to be like three not possible
6:51:05
if you do this you get the same answer all right let me see if the okay yes my phone is recording almost seven hours oh my god i thought i was able to finish this thing six but no all right
6:51:22
so with that being said i'm just going to show you guys how we can quickly set all this up and then
6:51:30
if necessary i'll show the pictures to convince you guys that it's actually correct and then um we'll do the integrals and then that's it that's it number 93 oh my god number 93
6:51:50
yeah i want to thank everybody though whoever is still watching right now thank you guys so much i want to thank all my subscribers all my channel members all my patreons thank you guys so so so much almost 1 million subscribers too thank you guys so much
6:52:11
93
6:52:14
limit s and approaching infinity it's just so much fun due to math questions um and then you guys are watching and participating and giving me your thoughts and all that stuff it's also so much fun love it so much and it's a lot of fun to do 100 questions in one go
6:52:35
why well i draw my resonance so yeah same idea but i'm hungry right now also let me just finish
6:52:46
all right six over and square the next one will be knight over n okay inside and then
6:53:00
cool now put this in the sigmund formation form here we have the limit as n goes to infinity everybody has u over n so let's write it right here sigma i goes from 1 to n
6:53:18
i want to see 1 2 3 4 all the way to n so you do this this is the same as 3 times 1 this is the same as 3 times 2 and here we have 3 times n right so that's the part of changing so we have 3 i over n and then square so the question can be presented this way or this way so be careful with both ways and here's an important point you see the three as you match it's crucial you'll see but anyway though let's do what we did earlier three over n this three tells us b minus a
6:53:58
is equal to three so if we start with five we go to eight if we start with zero which you should always start with zero you go to three so here's the deal integral zero to three perfect now
6:54:13
what's the function we have a 3 right here that's bothering us is the function 3x squared
6:54:21
or are we having just like 0 to 3 x squared dx which one
6:54:32
the answer is this right here not that this right here why because the reason that we have a studio right here is because the whole thing is three and we kind to end pieces so that's why and uh i'm going to perhaps just
6:54:52
erase this and i'm going to just give you guys a picture here real quick so you see when we have uh x to the some power but it's actually just x to the third power no no no no x to the second power
6:55:09
because we are going from 0 to 3 right
6:55:14
so each width is 3 over n the next one is 6 over n so you see that's why i said students you match so you see you put this to the x value we get that for the height you put this to b x value and here you get the height so the function is actually just x squared it's not three x squared be really careful with that so it's x squared dx
6:55:44
work that out one third x to the third power
6:55:49
0 to 3 plus 3 in here 3 to the third power is 27 divided by 3 is 9.
6:56:01
and then subtract plugging zero just zero so yeah we're good we're good so that's number 20 that's number 193 so be really careful huh this can not match what if we don't have a 3 right here though well you can put this 3 on the very outside that's that works too but the reason that i
6:56:21
have these three questions so similar is just because i want to go over this more details with you guys so that way you guys can fully understand what's going on
6:56:31
maybe you guys have heard my stomach already i'm so hungry
6:56:37
anyway
6:56:40
next i'm going to illustrate what i mean by if the numbers they don't match anyway number 94 here we have limit and goes to infinity one over it's meant to be a one okay oh the way i wrote it is this i don't know why i wrote it as 1 over n and then here we have sine pi over n
6:57:08
plus sine of 2 pi over n and then sine of three pi over n and then so on so on and then sine of n pi over n all right let's put this in the sigma summation first which is just the limit as n goes to infinity 1 over n all this can be put it as the summation i goes from 1 to n
6:57:36
1 pi 2 pi 3 pi so on so up to m pi so we are talking about psi of let's put this i pi pi
6:57:52
i pi over n okay now this number is just one that's all means b minus a is equal to one
6:58:08
so this right here is equal to what integral going from 0 to 1 only now the hard part is what's the
6:58:18
function is this sine of x or sine of pi x
6:58:26
the answer is sine of pi x if this is a one this pi will just be the coefficient and look at that i over n which is that'll give you the x how confused you guys are this actually the jet sign of pi over x let's just say it looks like this somewhat like that
6:58:49
hopefully this is the chat if not i will have to correct it i'm tired so we'll see
6:58:57
so you see we have sine of pi over we have sine of pi x so this is the curve and the deal is that we are going from zero to one when x is one sine of pi is zero so we do have this part right here okay we want to have
6:59:16
oh the picture is wrong let me let me try to draw a slightly more realistic because this is zero let's say this is one and when x is one over two you get one so
6:59:32
e when x is one over two you get sine of pi over two which is one doesn't matter trust the computation all that stuff anyway going from zero to one this right here is one over n
6:59:47
2 over n 2 over n and so on put this into this x pi over n put this into here 2 pi over n
7:00:01
and so on so and so on okay okay yeah so i want to see how many of you guys are still um
7:00:17
i want to see how many of you guys are still watching this
7:00:20
question for you guys what if today we have the limit as n goes to infinity what if we have a pi here and over n and the summation i goes from 1 to n and then sine of i pi over n
7:00:39
so tell me what the function is and tell me from where to where
7:00:47
i'll give you guys the dx there right do not just put a pi at the front just say hey that's the answer multiplied by pi don't do that don't do that do the things i said that it's better that way so the range will be going from 0 to pi okay the range will be going from 0 to pi let me tell me what the function is but anyway though what's the answer for this
7:01:13
all right i'll show you guys how to do this anti-derivative of sine is negative cosine because the derivative negative cosine is positive sine and then the input is pi x but we will have to divide by this constant here how do we do all the integral of delta i have 100 integrals already
7:01:36
by the way trust me this right here is denser for that yes now put the 1 to the 0 so here be really careful though 1 over negative 1 over pi cosine of pi times 1
7:01:53
that's the first part minus put a zero so we have negative one over pi cosine of pi times zero okay
7:02:10
cosine of pi is negative one cosine of zero that's pi times zero zero is one so here this right here gives us positive one over pi and then negative negative becomes a positive one over pi wow this was crazy the answer for thai is two over pi how pretty is this
7:02:42
okay
7:03:02
all right now i'm just going to
7:03:06
uh put the question in the summation form is better that way so we have the limit
7:03:15
let me just see i don't know why i just want to check wow seven hours oh my goodness
7:03:23
seriously
7:03:27
yeah
7:03:30
limit and we have
7:03:38
n goes to infinity yeah of course and costing what am i talking about over n
7:03:48
and then times the summation
7:03:53
i goes from one to n
7:03:59
square root of 1 plus 3i over n whoa
7:04:09
so how to do this first thing first three right b minus a is three just start with zero yeah zero to three and this right here will give you the function what's the function though
7:04:25
three and three match so you pick the whole thing to be dx so just square root of one plus x
7:04:32
just like that work this out real quick yeah uh okay to do this right here a
7:04:51
okay this right here might not be the easiest to integrate the moment what we can do is right here inside i'm just going to minus one so i'm just going to move it and then here i'm just going to add one this is the same idea like shifting okay but anyway
7:05:12
i'll give you guys the area argument i think that's much better okay so here yeah because i don't want to get into the use substitution thing in this video that's why anyway i'm saying so many times anyway this picture is like this okay and we go from zero to three
7:05:34
okay but the truth is we can move it to the right one unit and that will give us the integral going from one to four and then you can see that it's just
7:05:54
like this it's the same area okay and we can just look at the square root of x much easier to integrate because now we can look at this as integral x to the one-half power dx one to four and then just to the anti-derivative that i showed you i went to the power which is three half divided by the new power which is two over three and this right here is two o plus three x to these three over two and then plugging plugging here
7:06:27
putting the four and i'm putting a one
7:06:35
putting the one
7:06:42
okay
7:06:45
okay four to these three over two is you take the square root of four first and then you do the third power which is two to the third power which is eight
7:06:54
so it's eight times that which is 16 over three this is one so it's minus two over three so i know fourteen over three it's almost like pi but the other way around huh wait this is actually four point no it's not no sorry yes two one four let's do the upside down by the way that will be easier
7:07:20
okay so it does have the appendage if you don't start with zero sometimes but like that's uh if you just want to get the region and start with zero much easier and then especially if you are familiar with like how to do the integral that you can do the so-called u substitution or some kind of magic and then you can figure the answer one way or the other
7:07:43
i wonder what if i don't do all 100 questions is there anybody who's going to notice
7:07:51
if you have watched the whole thing definitely leave a comment down below but don't lie though be honest i'm i did the whole thing
7:08:02
okay 97 no 96 limit and goes to infinity and then we have one over n sum i goes from one to infinity and then we have no no no i always want to write like sigma dancing to the infinity because that's why we're doing k2 all the time but no
7:08:32
e to the
7:08:35
2i over n power okay
7:08:45
okay ready cool it's one over n so just integral going from zero to one here this is one this is 2 so the function is e to the 2x okay integrate this e to the 2x stays but divided by 2 because that's the derivative of 2x and then just plug in numbers 0 and be really careful with when you're plugging zeros sometimes it does matter so don't ignore it so here we have one half e to the two times one
7:09:20
and then minus plugging zero one half e to the two times zero
7:09:27
okay this is just one over two e squared and this is just one so just minus one half
7:09:39
97
7:09:45
okay limit as n goes to infinity and we have pi over 4 and
7:09:56
sum i goes from 1 to n and then secant squared pi i over 4n
7:10:10
pi over 4 pi over 4 match this means we go from zero to pi over 4 and the input is just what secant squared x dx okay yeah very different very different okay and now let's just work this out the derivative of function will give us secant squared the answer is tangent
7:10:37
so tangent x zero pi over four plugging pi over four tangent pi over four is one plugging zero tangent of zero is zero so on all the answer is just one so nice
7:10:52
yeah so i'm just going to write it right here because i want to fit another question
7:10:59
98 number 98 okay here's the limit as n goes to infinity and then we have 1 over n sigma i goes from 1 to n and then we have cosine this is crazy it's just i over n wow
7:11:24
how do we do this well same idea integral zero to one because it's just one yeah one over and one over a match so this is just cosine x dx
7:11:38
okay yeah and then let's just go ahead integrate this derivative for the derivative of function will give us cosine the answer for that is just sine x positive though plugging one sine of one is a number it's legit just leave it it's not like a special value you cannot really simplify it such as sine of one plugging zero sine of zero is zero so the answer is just sine of one
7:12:08
yeah so once you know what exactly you are looking for this right here are so
7:12:14
beautiful
7:12:30
let me see if i have a better marker here
7:12:35
maybe
7:12:42
ladies and gentlemen number 99 all right limit this is another expanded version you guys will see why this is not well this is number 19 already okay focus
7:13:00
n over n squared plus 1 plus n over n squared plus four plus n over n squared plus nine so up to wow and squared over n squared plus n squared oh my god
7:13:29
where is the open end part huh where is it
7:13:34
pause the video give this a try first
7:13:45
done by the way let's do this huh when you see this question do not just do the following and say put infinity here and here and say hey the degree right here is bigger than that so the answer is 0 for the first one and the next one is zero likewise zero and then just zero so we're just adding a bunch of zeros huh so the answer should just be zero no because the limit is zero think about this as 0 plus and another thing is that as n goes to infinity we have how many terms
7:14:21
you see the plus data and this as n we actually have infinitely many of these kind of things
7:14:26
so this is almost as saying 0 times infinity determinant 4 so cannot say any answer unless we do more work it's a different kind of indeterminate form
7:14:40
okay so how exactly do we do it though again we want to see the sound number of m parts
7:14:53
let's see
7:15:07
i know the answer but like i just have to think about how to explain it so that i don't keep you away right away
7:15:15
number 19 is tough it's not a it's not easy all right one four nine height they look suspicious only just a one square two square three square and so on and remember we want to have like i over n and then because it's a 1 square 2 square 3 squared so maybe you want to produce i over and square somehow
7:15:47
so let's do the following right here let's divide this by n square and of course do the same on the top divide this by n square and do the same right here and divide this by n square
7:16:05
do the same and then divide this by a square and then do the same what good does this do let's see
7:16:15
okay now i have to look for a marker that works
7:16:25
this is the one that i let it rest earlier
7:16:31
all right here we have the limit as n goes to infinity check this out if we have n over n squared on the top everybody has a 1 over n perfect let's factor that out put it here and then sum i goes from 1 to
7:16:52
n
7:16:56
all right and then let's see now because i put that over there already so we have a one on the top and over this over there is just 1 and 1 squared over n squared as we said earlier that's exactly what we want huh that's plus parenthesis i over and n squared this is 2 squared over n squared this is 3 squared over n squared so on so on and so on
7:17:24
good all right 1 over n and 1 i over n right so this is good this becomes the integral of going from zero to one the function is one over one plus x squared dx
7:17:39
integrating this we get inverse tangent because the derivative inverse tangent is one over one plus x squared and then we plug in plug in plugging 1 here inverse tension of 1 is pi over 4. plugging 0 is 0. so this right here is it that's number 99
7:18:01
so cool huh and you know what let me let me actually write it down right here this is 0 to 1 1 over 1 plus x squared and this is just inverse tangent
7:18:13
of x is going from zero to one final answer pi over four yeah because this way i can
7:18:21
use the rest right here to do number one hundred number one
7:18:32
right limit as n goes to infinity here we have that this marker is about to try out too it's just two warming side here two over m plus two and then two over n plus four and then two over m plus six
7:18:52
two over and plus 2n so quite similar huh this time we have to recognize that 2 is the same as 2 times 1 um and then this one here is the same as
7:19:07
2 times 2 and 2 times 3 and so on so
7:19:11
maybe we can just divide the type and bottom by n just about we did earlier yes or you can just factor the n from the bottom so it depends on how you like to look at it perhaps let's divide everybody by n it's more consistent than what we did earlier so let's do that
7:19:31
perfect so
7:19:45
so this right here is the limit as n goes to infinity and then we have our two over n let's put that right here
7:19:57
sum i goes from one to n and then now on the times just one and then we have one right and then i'm going to write it as plus and we have this is like 2i yeah 2i over and yes so it's like this yeah
7:20:22
is it
7:20:27
yes yes yes yes okay so yeah here we have the
7:20:37
integral going from 0 to 2 and then the function is 1 over 1 plus is x what works x match right so x okay dx integrating this is ln of 1 plus x we don't need the absolute value because everything is positive ladies and gentlemen
7:21:00
plugging 2 we wiki ln3 plugging 0 ln1 is 0 final answer l13
7:21:11
100 limits there you have it
7:21:17
you know it i don't just want your limits let's see if we can fit in number 101 here as well yes of course all right question number 101 what's the question limit as n goes to infinity and then we have the nth root of n factorial over n to the nth power
7:21:47
why is this one number 101 yeah because this right here has the most amount of view right i have a video just on this over 1 million views right my brilliant limit yeah thanks to brilliant for sponsoring me for throughout this year i really appreciate everybody's help and everybody's watching video and then all the sponsors that i have i'm very grateful
7:22:14
i'm debating to see if i can show you guys two solutions here
7:22:24
you guys want two solutions i i think just to honor the
7:22:30
the brilliant limit i am going to do the same thing uh kind of consistent of what we did let me see if i can fit in everything firstly
7:22:43
actually no i cannot fit in everything so if you want to see the original way that i solved this go ahead and check out my other video right here i would like to show you guys
7:22:55
another way to solve it this is true i don't have a place to prove so i'll just tell you note if we have the limit as n goes to infinity of an n root of some
7:23:12
secrets now a n like this this right here is actually the same as the limit as n goes to infinity of a n plus one over a and yeah
7:23:26
where did i saw this where did i see this i saw this on wikipedia
7:23:32
so yeah this fits this condition and then i can finish this nicely
7:23:39
remember the answer that we got from the boolean limit is one over e so keep that in mind i'll show you that this is actually one over e as well okay this is one over e right so
7:23:56
cool now this right here is our inner limit so it's just a n yeah so let's just go ahead and check i don't know this here okay of course so this right here is the limit as n goes to infinity it's like whenever we have the answer we just do get inside and then just do am plus one meaning that we do n plus 1 and then we factorio that
7:24:26
sorry n plus 1 so n plus 1 factorial that and then n plus 1 raised to the n plus 1 power
7:24:40
and then we want to divide it by a n which is the same as just multiplied by its reciprocal so i'm just going to put n to the nth power over n factorial so if you have done some calc 2 stuff the ratio test or whatsoever yeah pretty much that now break this apart as n plus 1 times n factorial and you see that n factorial cancel very nicely what's this though we can also break it apart write this as m plus 1 to the nth power first times n plus 1 to the first power and we see that
7:25:21
it happens to be that n plus 1 and plus 1 cancel very nicely and this is so wonderful because
7:25:30
this has an s power likewise does so we can put them together as the ones and power like this so this is the limit as n goes to infinity
7:25:39
and over that so n over m plus one power raised to the nth power now how do we do this
7:25:47
we have done some similar like this right yes the secret weapon flip this negate that here this is the limit as n goes to infinity and plus 1 over n raised to the negative nth power
7:26:06
and you know it this right here is just the limit as n goes to infinity 1 plus 1 over n raised to the negative n power
7:26:18
and you know it ladies and gentlemen a is one b is equal to negative one so all the answer is e to the one times negative one which is negative one
7:26:47
this is how dirty my hand is yeah because this thing is so dirty uh i should wash it yeah
7:26:56
i do wash it occasionally like every other week what's up yeah i have a couple of those by the way though 101 limits hopefully in this video i can help you with your calculus 1 calculus 2 class if you are my student hopefully you are watching it uh maybe i'll pick some questions from here to be to put on a test yeah and you know it i told you guys that run my results so i'll show you guys
7:27:30
yes this is the marathon method that i have yeah the very first one definitely means the most to me
7:27:42
i'm gonna put it up ah let me
7:27:51
boy
7:28:08
so finally you have some limits
7:28:12
i i don't know what else to say thank you guys so much thank you for all my subscribers all my viewers and all the sponsors all my patreons and all my channel members everybody
7:28:27
almost 1 million subscribers can you guys believe that insane
7:28:32
fun fact uh i remember i reached i reached 10 000 subscribers on october 20th 2016. so it would be kind of interesting or funny that i reached exactly 100 exactly 1 million subscribers on the same day but uh six years later that's kind of cool i don't know though
7:29:04
anyway you guys know it as always that's it
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