Skip to main content
Loading Explore guides…
Limits and Continuity · Acedra
Home
Explore
Mathematics
Limits and Continuity
Sign in to save
Chat
If YouTube asks you to sign in, open the video on YouTube.
Watch on YouTube
Reload player
Transcript
154 segments
Copy
Hide video
0:00
the idea of a limit is the key foundational concept that distinguishes calculus from algebra this is part one of some problem solving with limits and continuity we start with the problem where we are given the graph of a function f as shown below part a says to find all values of x for which this function f is not continuous okay so we want to think intuitively at the start and then relate this to limits intuitively a function is continuous over its entire domain
0:30
if you can draw its graph without picking up your writing utensil so where a function fails to be continuous is where you have to pick up your writing utensil so we look at this graph and we see that try to imagine drawing this and we see that there are two spots two values of x for which you'd have to pick up your writing utensil at x equals negative one and at x equals positive one so the answer to part a is that x equals negative one plus or minus 1 are the two values of x for
1:00
which f is not continuous that's the answer for part a this is where f is not continuous
1:11
so evidently then that means it's continuous everywhere else and part b says to find the largest open intervals on which f is continuous now these open intervals in this particular picture happen to have the same length however if you had a problem where they had different links i think the author of the textbook probably would want you to write down all open intervals on which f is continuous so for example it's continuous from negative 3 to negative one
1:40
certainly not including the end point negative one there is no open circle here at negative three but they do say make the answer an open interval so we'll go ahead and use open interval notation also from negative 1 to positive 1 and then from positive 1 to positive 3. so in terms of answering the question as it is phrased i'll go ahead and use interval notation to start with for example negative 3 to negative 1 that looks like a point but it's actually an interval
2:07
if you set builder notation you'd say this is the set of all x with the property that x is strictly between negative 3 and negative 1. either kind of notation is fine the next interval again goes from negative one to positive one so you could use open interval notation or set builder notation like this and the third open interval is from one to three we're using open interval notation with parentheses instead of square brackets saying we're not including the endpoints
2:40
again because of how the problem is phrased in this particular graph again there's no open circles here or here so technically you could include those endpoints as part of the domain in which f is continuous but the problem says open intervals okay so we'll go ahead and go with that
2:57
part c find all values for the following limits if they exist if the limit does not exist say so okay so take these one at a time number one the limit as x approaches negative two of f of x does that limit exist and can we find its value okay so this may be a new concept for you what we're after is we are after the output values of f of x what are they getting closer and closer to as x gets closer and closer to negative 2 without actually equaling negative 2 that's where the
3:38
biggest confusion comes into play especially when we're related to continuity because we're ultimately in this particular example going to think about the function value at negative 2 so it is a really confusing concept that way but it is key to fully understanding what calculus is about so x is approaching negative 2. so i imagine x getting closer and closer to negative 2 either from the right or from the left the function outputs which are the second coordinates of
4:07
these points on this graph are getting closer and closer to 0. do they equal 0 not unless x actually equals negative two they don't but they are getting closer and closer to zero closer and closer enough that we can say that that limit does exist the limit okay so this is parts part c part one here the limit as x approaches negative two again either from the right or from the left of this function that limit does exist and does equal zero okay say it again the function
4:43
values as x gets closer and closer to negative two without actually equaling negative negative two are the second coordinates of these points on this graph they are getting closer and closer to zero their y values are getting closer and closer to zero without actually equaling zero unless x actually equals negative two we can make them as close to zero as we like if we make x close to negative two now again the function value at negative two
5:09
actually is zero and that's related to the fact that this function is continuous at negative two
5:15
but the more important thing is the function values get arbitrarily close to zero as x gets close enough to negative two and that's what makes this limit zero it's a very very subtle concept let's go on to the second one of part c the limit as x approaches negative one of f x okay this is where these this subtle concept is really going to come into play because at the function is not continuous at negative 1 so is the limit the second coordinate of that missing point of
5:46
the graph or is it the second coordinate of that point maybe you want to pause the video and think about it the limit as x approaches negative one of the function
5:56
so once again x is getting closer and closer to negative one either from the right or from the left without actually equaling negative one what are the function outputs getting arbitrarily close to think about the second coordinates of these points on the graph here and here they're getting closer and closer to one not two this limit does exist and equals 1 and it doesn't matter that the
6:22
function value at negative 1 is actually 2 that is completely irrelevant for the value of the limit
6:30
this limit over here happened to equal the function value at negative 2 and this limit right here does not equal the function value at negative 1. what's the difference why did it work there and not there because here at negative 2 the function is continuous but at negative 1 the function is not continuous so the fact that these two things are not equal is related to the fact that the function is not continuous at x equals negative 1. making
7:03
these key points here all right on to the next part part 3 the limit as x approaches 0 of f x
7:14
limit as x approaches zero of f of x so now we're letting x get closer and closer to zero without actually equaling zero the function values the second coordinates of these points their y values are getting closer and closer to negative one and it doesn't matter that this graph has what you might call a corner in it it's not smooth when x equals zero that's completely irrelevant for the value of the limit does look like the function value does equal negative one there's
7:44
no open circle there this is equal to negative one and that does equal the function value when x equals zero the function is continuous at x equals zero and it's again completely irrelevant for continuity the fact that there's a corner there it's not completely irrelevant for derivatives for something called differentiability but that's a later topic on to the fourth part here of part c
8:11
fourth part looking over here the limit as x approaches one what's this minus sign there is that is that should that be in front is that a typo no it's not a typo we are letting our x approach one from the left this negative sign does not mean negative number it means from the left this is weird notation but it is standard notation we're letting x approach 1 from the left not from the right let x be 0.5 then 0.9 then 0.99 then 0.999 what are the function values approach the
8:44
second coordinates of these points are approaching 1 which is the second coordinate of that point the limit as x approaches 1 from the left second chord into the missing point i should say
8:58
does exist and does oops sorry does equal the second coordinate of the missing point one the function never equals one okay over here the function output is never equal to one that just gets arbitrarily close to one really really close as close as i like by making x sufficiently close to one on the left that means this one-sided limit
9:25
does exist and does equal one but that's not the same as the function value when x equals one the function is not continuous at one there's a jump in the graph
9:38
okay so we're seeing again the relationship between continuity and limits here part five
9:45
we're letting x approach one there's a plus sign there that means from the right not necessarily through positive values though of course numbers that are bigger than 1 are positive but from the right we're letting x approach 1 from the right the function values are up here now they're getting closer and closer to 2.
10:02
for that one the limit as x approaches one from the right does exist and equals two not one
10:13
this is also a one-sided limit
10:20
there's other names for these things this one where we're letting one x approach one from the left is also called a left-hand limit and this one where we're letting x approach one from the right is called a right-hand limit i'll just leave it as is the labeling that does happen to equal the function value when x equals one but that's not enough to say f is continuous at one because this one-sided limit did not equal the function value and these two things in part four and five
10:47
giving different answers one and two is enough to say that for part six the limit does not exist the two-sided limit where you don't see a minus sign or plus sign to the right of the number does not exist because the one-sided limits even though they exist they're not equal to each other
11:05
for the two-sided limit to exist both one-sided limits must exist and equal each other so
11:16
this limit here does not exist dne is a standard shorthand for does not exist okay and the reason is because the two the one-sided limits are not equal to each other
11:29
and this limit not existing means it also can't equal f of one so that's another reason that the function is not continuous at one you could say this is a little funny you could say the function is a right continuous at one because the right hand limit does equal the function value there would also be a corresponding concept of left continuous but we're not going to go there that's it's just not continuous and the last part the limit as x approaches 2 of the function
11:58
part 7 here now we're letting x approach 2 from the left or right it's a two-sided limit
12:07
x approaches 2 what are the function outputs get close to they get close to the second coordinate of the point here which would be 3 and that does equal the function value at 2 this function is continuous at 2. number 2 is easier than number one let f of x be one divided by the square root of three x minus nine the goal is to find the largest interval over which f is continuous it's implicit in a problem like this that you're allowing
12:37
yourself to plug in any value of x for which this function makes sense it's also implicit in a calculus class that we're not going to take square roots of negative numbers so we don't want to get complex non-real number outputs okay that's also implicit in a calculus class which i'm assuming you are taking when you watch this so what do we do it turns out that for functions like this where
13:00
are either rational functions which are ratios of polynomials or things involving square roots where we want to avoid taking square roots of negative numbers and we also want to avoid dividing by zero and that's the key insight this function will be continuous in fact wherever it's defined and not equal to some non-real number okay you can also use your graph to figure it out but we can figure it out without it without a graphing calculator um you want
13:29
to again avoid dividing by zero so you don't want 3x minus 9 to equal you also don't want it to be negative you want it to be positive so we need to solve this inequality effectively it's equivalent to saying by adding 3 9 to both sides that 3x is bigger than 9 and that's equivalent to saying by dividing both sides by 3 that x is bigger than 3. so the largest open interval the largest interval on which this function is continuous
13:59
could be written in interval notation as the open interval from 3 to infinity again it's an open interval i'm not using the square bracket there i'm not including that endpoint and in set builder notation that would be the set of all real numbers x such that x is bigger than three either kind of notation is fine if you wanted to draw it on a number line you'd mark the number three here and you'd make an open circle there and then shade everything to the right of three
14:28
that's the domain of the function it's also going to be the the set where the function is continuous
14:36
if you graph this with your technology here what will we see i hope you can kind of predict what you're going to see to some degree
14:45
you are dividing by something that gets close to 0 as x gets close to 3 in fact that means the output of this is going to get very very large go off to infinity you might say as x gets close to three from the right and then as x gets big it's going to go down towards zero so you should see some kind of decreasing graph with a vertical asymptote and also a horizontal asymptote three x minus nine
15:15
if i on my window screen if i just keep it say this window um i don't think we'll get an error i think it'll graph it but it will only start the graph once we get to x equals three there it is over there it's just this little piece it doesn't look like it's a vertical asymptote um but it should be it's just maybe not plotting enough points here let's start it at 3
15:35
see if we see any more yeah it's hard to see i mean if we if we plotted it really close
15:42
to say 3.5 here or something let's see if we see anything any more detail
15:49
yeah you can see it is going up fairly rapidly as your x approaches 3 from the right
15:54
yes it actually is a vertical asymptote even though it doesn't quite look like it it's a case where your technology could be misleading you as x approaches 3 from the right the value of 3x minus 9 is approaching 0 the square root of something approaches 0 is also approaches 0 meaning this fraction is getting very large it's 1 over something very very close to 0. it's going off to infinity it's not a proof but that is what's going on in this problem problem
16:22
number three is about continuity in an applied problem a 0.8 milliliter dose of a drug is injected into a patient steadily for half a second so whether it's like they're getting injected with a needle here at the end of this time that half second the quantity q of the drug in the body starts to decay exponentially at what's called a continuous rate of 0.1 percent per second okay you might wonder is this something that could actually happen don't worry about it if this is
17:02
mathematical modeling it's not going to be perfect we just want a decent model using formulas the goal is to express q this quantity as a continuous function of time t in seconds
17:19
seems pretty hard it's probably best to try to draw a reasonable graph first
17:26
for q as a function of t so the horizontal axis is t time in seconds and the vertical axis is q the quantity in milliliters other kinds of units could have been like milligrams or something like that
17:45
all right how do you deal with this how do you even draw a graph going in steadily means the rate at which it's going into your body is constant it's like walking at a constant rate
17:57
for that first half second from time 0 to time 0.5 the graph has got to be a straight line going from 0 you don't have any drug right away 2 0.8
18:12
like this
18:15
okay that's the fact that it's going in your body steadily for half a second
18:20
what would the slope of this be rise over run 0.8 divided by 0.5 would be 1.6 1.6 milliliters per second would be the slope of that line then right away the instant they stop injecting it it starts to decay exponentially at a continuous rate of 0.1 percent per second exponential decay graphs look like that they're decreasing but still concave up
18:49
and you want this to be a continuous function which makes good intuitive sense for real life you don't want there to be some sort of jump discontinuity in the amount of drug that wouldn't make any intuitive sense so you're going to get a graph going down about like this it's got a whole the t axis is a horizontal asymptote there is an exponential decay kind of graph all right so that's helpful i think to writing the formula now the directions say using formulas
19:18
the plural there is really referring to the fact that this is going to have to be a piecewise formula we need two formulas for the two pieces of the graph one when t is between zero and zero point five and one when t is bigger than 0.5 and again slope of this is
19:37
1.6 so and it goes through zero zero so the formula for that first piece is going to be 1.6 t when t is between zero i'm including zero and 0.5 should i include 0.5 in this piece of the formula or not the answer is is that it doesn't matter at least if you design your formula to be continuous like this i could put a less than there i could also put a less than or equal to i think i'll put a less than or equal to okay but it could be either way either answer is okay
20:11
the exponential decay part that's a little trickier um it's also made a little trickier by the fact that we're starting at time 0.5 here instead of time 0 for that part of the formula
20:25
um the initial value at that time and when i say initial i mean time 0.5 is the 0.8 so i want a 0.8 in front the fact that it's a continuous rate instead of say an actual rate per second means we need e the number e as the base of our exponential function if if it were an actual rate of 0.1 percent per second it's a little trickier you'd need like a base of in that case like 0.999 but here we need e because it's said to be a continuous rate
21:04
and it's going to be exponential decay so i need a negative sign here and 0.1 percent per second means the coefficient of t needs to be negative 0.00 one then times t uh careful though not just times t times the quantity involving t that is zero when time is 0.5 and it's not going to be a t that's going to be a t minus 0.5 there's a linear function of t that is 0 when t equals 0.5 up here in the exponent of the e it's going to get multiplied by negative .001 and this
21:43
is for t greater than 0.5 though i could put a strict less than there and it'll greater than or equal to there and it'll be the same function and this is a continuous graph as t approaches
21:57
we can write down left hand and right hand limits here the limit as t approaches 0.5 from the left of this function i can use the formula in that case 1.6 times t that's the same as the limit as t approaches 0.5 from the left of 1.6 times t that's a continuous expression of t i can just plug in 0.5 and get 0.8 so that's the left-hand limit the right-hand limit
22:25
as t approaches 0.5 from the right put a plus sign here
22:29
we need the other formula because that's the formula that is relevant when t is bigger than 0.5
22:40
all right and even though it's complicated looking it's a fairly simple limit as t gets closer and closer to 0.5 this expression up here is getting closer and closer to zero e to something getting closer and closer to zero gets closer and closer to one point eight times something closer and closer to one gets closer and closer to 0.8 so these two limits are the same these one-sided limits therefore the two-sided limits also exists and equals that same number
23:12
0.8 and that is the function value as 0.5 this is certainly a continuous function
23:19
at 0.5 and in fact everywhere in problem 4 we are given a piecewise function and we're trying to estimate a limit numerically based on using our technology the function f of x equals e to the 2x minus 1 all over x if x is non-zero and is equal to c if x equals 0. now there might be some confusion here going on in your mind you might be looking at that formula and saying how can the limit as x approaches 0 if this thing exists wouldn't this have a vertical
23:48
asymptote and amazingly it doesn't that's a pretty important thing actually for calculus part a says to make a table of values of f x for values of x that are close to 0 negative 0.1 negative 0.01 positive 0.01 positive 0.1 actually we'll use the table feature on our calculator to look at the values of this for more values other than just these but those are the buttons so that we'll write down we're ultimately going to use the table to estimate the value of the limit
24:22
which evidently exists based on the way this problem is phrased and then we're going to ask in part c what value of the constant c will make the function f continuous at x equals 0. i won't say the answer right now but i hope you realize what to do for part c based on the previous problems that we've done so in part a we'll go ahead and write down a little table on the paper here
24:47
for these values of x negative 0.1 negative 0.01 positive 0.01 and positive 0.1
24:59
i'll go ahead and do say this one with a calculator by plugging numbers in um but then we'll use a the graphing feature where we can actually make a table as well to do it more quickly so we can just do it step by step when we plug in negative 0.1 here we first find e to the negative 2 times 0.1 or 2 times negative 0.1 which would be e to the negative 0.2 subtract 1 and then divide by negative 0.1 divide by negative 0.1
25:39
we get about 1.8 okay let's let's go 1.81 is what that is okay and i could do that for the other ones as well i could also do it all at once let's just double check it that way i could do e to the negative 0.2 and then minus 1 divided by negative 0.1 yup about 1.81 okay let's use the the table feature in the graphing calculator portion i'll plug the function in here so again it's e to the looking at the formula up here 2x 2 times x and then minus 1 all divided by x
26:23
and let's now make sure our table settings are good i'd want to start at negative 0.1 and do the delta table to be 0.01 and let's see what we see when we make the table now so this first value of y that's the one i wrote down going down the table there's the next one i want 1.98
26:49
i get an error when x equals zero because the calculator is trying to divide by zero that's an error how about positive point zero point zero one i get two point zero two
27:02
and positive point one
27:06
looks like 2.21
27:12
so as x gets closer and closer to 0 as you move toward the center of the table so to speak what are these outputs getting closer and closer to well they're pretty close to 2
27:21
let's go ahead and assume that that's just the limit now is that a good assumption it does turn out to be the right answer here for the true value of the limit as x approaches 0 of this function does in fact equal to two but it's no proof okay it's just numerical evidence i suppose we could if we wanted extra confirmation go back to our table settings start at negative point zero one and make this be .001 and see how it looked then and we get really close to 0
27:54
without equaling 0 yes the outputs are getting really close to 2 okay and in fact they get closer and closer to 2 as x gets closer and closer to 0 we can make it as close as we wanted to by making x officially close to 0 without actually equaling 0. okay at least in this part of the formula
28:14
that expression that right there gets arbitrarily close to two without actually equaling two but the limit is two this is something i've been emphasizing already in this video
28:25
and that is the answer to part c part c you'd want f of 0 well ok for the value of c i guess that is part c still what value of the constant c will make the function continuous at 0 f of 0 which is c does need to equal 2 to make the function continuous there let's go ahead and make a graph of the function as well
28:53
check my window here negative let's just do like negative five to five
29:01
and then maybe um negative five to five here as well so there's the function as far as the calculator is concerned it looks like the function value when x equals 0 is 2. the calculator technically is just skipping over plugging in 0 to get it's not giving us any kind of error it's just plotting points and connecting the dots it doesn't know that the function if i did not include the c here is undefined when x equals zero
29:29
all right so it looks like the limit is too graphically as well that is the answer problem five is about trying to evaluate certain limits using algebra and it's kind of a hard problem because you can't just plug in what x is approaching because you'd be dividing by zero and so again the intuition is you might think that these functions that you see here function of x two functions of h might have vertical asymptotes at what x or h is approaching but in this case
30:02
this particular problem these limits actually do exist in equal particular numbers that's not obvious at all if these functions happen to be continuous at what x or h was approaching then you could plug in those values of x or h into the formulas and get the answer those would be easy problems these are harder problems part a evaluate this limit as x approaches negative four
30:28
of this expression now it's also possible in situations like this that the limit does not exist okay that is possible but in these particular problems that i picked the limit does exist how do you know in this case with part a factor the top we can factor the top as x plus 4 times x minus five
30:52
and we can simplify this expression cancel those x plus fours and hmm is this valid or not can we say the limit as x approaches negative four of this expression is the same as the limit as x approaches negative four of just plain old x minus five is that okay question mark to do that it turns out it is okay though you can't just do it without thinking about it
31:23
this these two expressions here and here are equal and it's these are both rational functions of x and they're undefined when x is negative four i'm not doing piecewise formulas here
31:38
this thing this expression is defined when x is negative four
31:44
so how can you do this these two expressions here's the key these two expressions are equal as long as x is not equal to negative four
31:56
and remember with limits in this case as x approaches negative 4 either from the left or from the right it does not actually equal negative 4.
32:05
we don't care about what happens to these functions technically speaking at negative 4 when x equals negative 4. it doesn't matter that that's undefined the limit could still exist
32:16
and this function is a nice simple function it's a linear function it's graph is a straight line
32:23
it's defined everywhere and continuous everywhere and i could evaluate the limit of this one by plugging in negative 4 and get negative 4 minus 5 is negative 9.
32:35
that's no problem that's continuity continuity of the expression of the function x minus 5 everywhere it's certainly continuous everywhere it's also continuous at negative 4.
32:53
this thing is okay the reason is it's irrelevant what happens to this function at negative four
33:02
this function equals this function when x is not negative four and so if this function has a limit this one does too the graph of this function in part a in either form is actually a straight line except with a hole in it it's a line with a slope of 1 and a y intercept to negative 5.
33:25
when x is negative 4 technically speaking the graph the original function
33:31
has a hole in it a point missing right you when i draw a circle like that an open circle means a point is missing that's a little bit misleading because points don't really have any area
33:44
that's what's going on that's the graph of the original function it's undefined at negative four no vertical asymptote the limit does exist and equals negative 9. this function x minus 5 would fill in that hole so to speak this is what's called a removable discontinuity similar things happen with b and c let's go ahead and just do the algebra without talking about graphs
34:05
these kind of limits that you see in b and c are going to come up a lot in calculus with derivatives so they are worth getting a little practice with
34:16
so here's the one for part b
34:21
what you want to do is you want to do some algebra first
34:26
expand the top here you got to do 5 plus h quantity squared meaning use foil method 5 plus h times 5 plus h first times first is 5 times 5 is 25 outside times outside and inside times inside are both 5h add those and get 10h and then the last times last is h times h is h squared simplify the 25s cancel on the top we'd like to be able to plug in zero we can't yet because we'd be dividing by zero but you can factor out an h out of the top like this
35:01
and then yes you can divide out the h
35:07
and say the limit is the limit of this thing
35:10
as h approaches 0 this is a continuous function of h you can just plug in h equals 0 to get 10 plus 0
35:17
is 10. once again there is the trickiest step to justify as a function of h this is undefined at h equals 0 but this is defined at h equals 0 but these are equal as long as h is not zero so if the limit of this one exists so we'll have a limit to this one
35:33
the limit of that one is definitely ten therefore the limit of this one must be ten if you graph this expression as a function of h it will look like a straight line again with a point missing at h equals zero just like what happened in part a part c
35:53
yikes there is part c there what in the world letting h approach zero can't plug in h equals zero fractions within fractions what in the world am i going to do with this let's write it down
36:13
there's two you want to use algebra to simplify it there's two things you could do here you could either subtract these fractions right away by getting a common denominator of four times four plus h and then divide all that by h same as multiplying by 1 over h or what i find a little quicker is to multiply the top and the bottom of this big fraction by the common denominator four times the quantity four plus h if you do it to the top
36:46
you gotta do it to the bottom to compensate um so i'm also ah that's a little bit messy so i let me redo it over here
36:57
just to be clear about what i'm doing i'm multiplying both the top
37:02
and the bottom of the original fraction by the common denominator of the smaller fraction so to speak in the numerator of the bigger fraction because then when i do this when i distribute i'll get cancellation you know i'm writing that down but if you know how to do this in your head you don't even have to bother writing that down which is why i think it's a little quicker but
37:26
so i get cancellation of the 4 plus h there and just get a 4 get cancellation of the 4 there and you get a 4 plus h careful with parentheses there and subtracting that whole thing and the bottom becomes more complicated but that's okay the top the 4 is canceled distribute the minus sign through i get the limit as h approaches 0 of negative h divided by 4 h times 4 plus h the negative h and the h can divide out to leave a negative 1 up top
37:57
let me go ahead and write it negative 1 over that h divided out so i have 4 times 4 plus h
38:05
and now you can plug in h equals 0 because you're not dividing by 0. this is a continuous function of h at h equals 0. of course it's also discontinuous at h equals negative 4 but that's not where we're letting h approach
38:16
so i get negative 1 over 4 times 4 negative 1 16 is the limiting value and by the way you can check parts b and c with your calculator if you graph these functions of h you'll see that these things are right our last problem for this video is about what you might call a wild function sine of 1 over x we want to intuitively explain why this limit as x approaches 0 of sine of 1 over x does not exist certainly this function is undefined when x equals zero one over zero is undefined
38:51
can't take the sine of an undefined number but that doesn't prove that this limit does not exist the limit still could exist as we've seen in some examples in this video
39:02
it's not clear what's going on here why this wouldn't necessarily exist
39:09
let's go ahead and use the calculator to see what seems to be happening here so i'll type sine of 1 over x you would assume your calculator is in radian mode which mine is though in degree mode you'd see something somewhat similar
39:24
it's assumed everything's in radians here almost said otherwise sine of one over x let's pick a window oh let's go ahead and keep it like this to start with although these are well okay let's make it smaller in the y direction the outputs values are outputs of sines they're between negative one and one negative one to one and make this maybe point one let's just see what we see here
39:49
okay whoa what in the world's going on there
39:53
something a little strange does seem to be going on near zero it's almost like it had a vertical asymptote but then it went back up again and some it's sort of oscillated kind of rapidly there let's let's zoom in closer to zero and see what we see
40:11
oh something definitely weird going on you might wonder is this graph accurate and i would say to you actually not necessarily as accurate as could be at least let's try negative 0.1
40:29
to positive 0.1
40:36
whoa whoa
40:42
okay first of all word of warning sometimes when you see graphs like this in problems like this for different examples excuse me it's actually calculator error okay and there is some calculator error here going on especially like in there for example that
41:01
is making this graph not as good as it could be you might say however this function does still behave in a wild way near zero and that's a graphical reason why the limit as x approaches zero does not exist technically speaking on any tiny interval near zero no matter how small the graph actually oscillates in reality infinitely often whoa does that blow your mind infinitely often but why why would that happen
41:38
as x approaches 0 let's just say from the right just to keep it simple to keep it simple
41:52
one over x
41:56
is getting very very large right if you do one over a really small number it's really big if x is 0.1 then one over x is 10. if x is 0.01 and that one over x is a hundred if x is .001 then one over x is a thousand this is getting bigger and bigger and if i'm letting x approach the positive values you could say it's approaching plus infinity
42:20
i'm not saying infinity is a number i'm saying 1 over x is getting as large as i like positively speaking so the input for the sine function is getting larger and larger without bound going off to infinity the ordinary sine function has a graph that looks like this as its input along the x-axis grows larger and larger it oscillates between negative one and one in a periodic way period two pi and as x goes off to infinity you could say there are infinitely many oscillations
42:59
the sign of this quantity that's going off to infinity has to go through infinitely many oscillations between negative one to one
43:08
so you could say if you're just keeping yourself to an intuitive explanation you could say that so
43:14
sine of 1 over x is oscillating
43:21
infinitely often and for extra emphasis between negative 1 and 1
43:32
as x approaches 0 from the right and the same kind of thing would be true as x approaches 0 from the left if you want a little bit more rigor
43:43
you could for example ask what values of x will lead to the sine of one over x equaling say one or negative one or zero for example sine of 1 over x equaling
43:55
1 means what means the input where is the sine equal to 1 if you think about the unit circle definition it's a pi over 2 and go around another 2 pi to what would it be threw it in my my head here 5 pi over 2 and then 9 pi over 2 and then 13 pi over 2.
44:19
if if x is positive here this means 1 over x would have to be pi over 2 plus some integer multiple of 2 pi 2 n pi for n equals 0
44:34
1 2 etc i could also think about negative values of x and then i could allow n to be negative
44:42
and that would be if i solve that for x by getting first of all say a common denominator here of 2 pi plus 4 n pi over 2. i could solve for x by taking the reciprocal of that 2 over pi i'll factor out the pi in the bottom 1 plus 4n
45:03
n equals 0 1 2 3 etc and those numbers
45:10
are all 2 2 over pi or less as n gets bigger and bigger these numbers get smaller and smaller when n is 0 it equals 2 over pi on a number line if 0 is here 2 over pi you know is a little less than 1 if one's right there 2 over pi might be right there
45:28
and then when n is 1 it would be 2 over 5 pi i guess be right there 2 over 5 pi
45:37
then it'd be 2 over 9 pi then 2 over 13 pi et cetera these numbers get closer and closer to 0 and at all those numbers the function output equals 1. and for it equaling negative 1
45:51
that means 1 over x is going to be
45:55
let's see down at negative 1 would be 3 pi over 2 plus 2 n pi n equals 0 1 2 3 et cetera
46:06
get a common denominator of 2 this is 3 pi plus 4 n pi all over 2. and we could solve that for x
46:16
2 over again i'll factor out the pi pi times in parentheses 2 plus 4 n n equals 0 1 2 3 etc and these are all numbers that are not equal to these numbers but are still behaving in a similar way going towards zero and there's infinitely many of them so you know there is some algebra that you want to add that kind of explanation to to help you finish the problem if you if your teacher is looking for a little bit more explanation than just what i had up here
46:46
to intuitively explain why this limit does not exist the function is oscillating infinitely often with a constant amplitude of one between negative one and one as you approach zero on any tiny little interval around zero the function oscillates instantly often between negative one and one there's no way for this limit to exist thanks for watching
Here to help you learn
Sign in to keep a private conversation with your course tutor.
Ask me anything about the material…
Sign in to chat