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Multivariable Limit Techniques · Acedra
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Multivariable Limit Techniques
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in this video I will show you six ways for evaluating a limit of a multip variable function and this right here will help you for your Calculus three class multiv variable calculus starting with the first one just plug in as long as you don't have in determined forms as long as you end up with a nice number congrats that is the nice answer such as this one X comma Y is approaching 1 comma Pi so let's just pluging one for
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x Pi for y and see what happens so we get 1 * cine of pi over 14x squ Dot and then + 2 pi instead of cosine gives us -1 * 1 is -1 over 1 s is 1 + 2 is three and we end up with 1 over3 and we are done and now for the second way we will just do Algebra just like the good old c one days so for example the limit as X Y approaching four comma 0 < TK of x - < TK y + 4 over x - y - 4 if we just plug in for 4X on the top we get square root of 4 and then minus square root of 0 for y
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so 0 + 4 over 4 - - 0 and then minus 4 you will see that the top will give us zero the bottom also gives us zero which is in determinant form so we must do more work in order to figure this out and to do so just do the algebra just like the Google days here we have a square root case let's just go ahead and multiply the top and bottom by its conjugate and hope for the best so we will have square otk of X Plus Square < TK of y + 4 and do the same thing on the
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bottom okay so this right here is the limit as XY approaching 4 comma 0 just multiply on the top and wait for the good thing to happen when we multiply the top this is a difference of two squares formula we just have to square the first term which is just going to give us uh let me just put down square root X and then we square that and then we minus the second term square y + 4 and then we square that on the bottom just leave it so x - y - 4 * < TK x + S < TK of y + 4
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now check this out this and that cancel so we just have X this and that cancel so we just have y + 4 but of course distribute the negative so we have x - y - 4 this is exactly the same as this so we can actually just cancel cancel so we have a one on the top and now it's the time to plug in 4 comma 0 into here again one 1 over Square < TK of 4 forx + square < TK of 0 for y and then + 44 the
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original from here finally this is 2 + 2 we end up with 1 over four for the answer that's it okay method number three substitution just like in many different every math we use substitution for this one he also works out if we have the limit as X Y approaching 0 0 S of x² y over x² y if we're just plugging Z and Zer into here and here we get Z over zero so we will have to do more work but
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notice the input is the same right here so it's a good idea to just put this to be another variable and see what happens I'm going to let t equal to the input and here instead of saying the limit as X Y approaching 0 0 when X and Y are is equal to Z T is equal to Z right so that means T will be approaching zero and then we just have to look at s of T over T now we are talking about the single
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variable case and this is a famous limit whenever we have t approaching zero of sin t/ T the result is just equal to one and technically you should do a geometric proof for this I do have a video on that so you guys can check that out try not to say by laito rule right mathematician will appreciate that or math professors will appreciate that yeah quot it as like famous result or by one of the
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famous limits I think people will be happier now number four I call this separable but the actual thing that we're doing is just we are using one of the limit laws have a look I will tell you though if we put 0 0 into X and Y we will get 0 over zero so you know the deal we will have to do more work and have a look this is the same as that so we can factor that out so this equals limit as XY approaching z0 I'm going to factor this out and put it at the end so I will first write
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e to d2x minus 1 and then here we have the Ln of 2 y + 1 and then for the denominator we have x * Ln of 3 y + 1 now notice we have this time St on the bottom we have this times that look at this part they just in terms of X this this part is just in terms of Y so now we can separate it so we can look at this as for the first part just say the limit as X approaching zero and then focus on e
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to the 2x -1 /x and then we just have to multiply by the limit as y approaching Z of this Ln of 2 y + 1 over l n of 3 y + 1 this is why I call this separable imagine earlier we have a plus right here unfortunately we cannot break it down like this anymore and this is actually just one of the limit laws which says the limit of a product is the same as the product of the limits provided that they both exist and they do both exist let me show you this one right here real quick so for
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the limit as X approaching zero of e to 2x - 1 overx some people just use laal rule but I will tell you though this right here represents F Prime of zero it's a derivative of some function zero because it satisfy the definition of derivative where the function is just e to the 2x yeah this right here tells you the function X approaching zero that tells you the input is zero so now you can just
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differentiate that frime of X by using the chain rule you get two in the front and e to the 2x now plugging zero so F Prime of 0al 2 * e to the 2 * 0 this is 0 e to 0 is 1 and this is just equal to two so the first one we get two and chose to be told if you want to secretly do Lal R somewhere and go from here to here that's fine too all right for this one right here we will legitimately use the Lal rule so for the limit as y approaching zero
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here we have Ln of 2 y + 1 over Ln of 3 y + 1 we will get Z over zero so laal rule d y ddy so this right here is because 0 ver 0 we're using litos rule we will just have to look at the limit as y approaching zero take the derivative of the top Ln of something will give give us 1 / 2 y + 1 and remember to use the chain rule multiply by the derivative side which we need a two similarly for the bottom we will have 3 y + 1 the bottom and then multiply by three now
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we can just put zero into all the Y not so bad so if you do that you get 2 over 2 * 0 + 1 over 3 over 3 * 0 + 1 it's just 2 over 1 which is 2 is just 3 over one is just three so in fact this is just 2 over three and you can come here now times 2 over three multiply this out we get 4 over 3 and know we are done now number five the polar method meaning that we are going to take take this from the rectangular system to the polar coordinate system and when can we do this though
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the big thing is when we have XY approaching the origin and the reason why we should do now number five the polar method and now number five the polar method meaning we are going to take this from the rectangular coordinate system to the polar system and when can we do that well when we have X Y approaching the Wen it's a good idea try it okay let's see why if you want to get to the waren in the rectangular system in fact we have infinitely many path for example
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we can go along the diagonal line like this or maybe Y is equal to 2X or maybe Parabola like this or maybe just when X is equal to Z when Y is equal to Zer and whever this curve is we have infinitely many path so this is not easy at all because we have to make sure to get the limit being the same along all the path well if you take this to the polar coordinate system in order to get to the origin there's only one condition that is R is equal to zero and of course
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in the polar world not only we have to know R but also Theta in fact here though in the origin Theta can be anything and usually we keep the Theta from 0 to 2 pi quick note some people will want the r to be non- negative so they will say R is going to be approaching Z plus that's fine yeah and some people will say it's okay to have negative R I'm to work it's okay to have negative R so it's kind
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of up to your instructor or up to the textbook depending on how you are going to follow the definition of the Polar coordinates in fact this right here is similar to the substitution but it's a specific substitution here we go we will have to remember that X is equal to R * cine Theta and Y is equal to R * sin Theta and lastly X2 + Y2 = R 2 once you know three you should be good to go why
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are these the formulas well I have done a video on that so you guys can go check out the link will be in the description for your convenience okay here we go this is the limit X Y approaching the w we will say R is approaching zero and once again it's okay to just consider zero plus but keep in mind though Theta is anything right can be anything now on the top x to the thir power so here we
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have R to the thir power also cosine to the third power on the bottom we just have r squ ah so nice because now you can see this and that cancel and we have one more of them so we just have to focus on the limit as R approaching zero R time cosine to the thir power Theta now can we draw conclusion right here yes the reason is because if you plug in R is equal to Z right here we have 0 * cosine
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to the power Theta well this right here let's indicate that what's the biggest value that this can possibly have just one right because the range of cosine is one when you reach that to the third power you still get one and then the smallest value for this is1 so I can tell you that the absolute value of cosine to the 3 power Theta is less than or equal to one so no matter what Theta is we will never have to worry about 0er * Infinity otherwise that will be indeterminant form
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just real quick if you happen to get the limit as R approaching zero and then I say we have R * 1 / cosine to the thir power Theta well unfortunately in this case we get zero times if Theta is equal to Pi / 2 then cosine to third power this right here will give us zero 1 over zero Ah that's no good right also if you just think about as limit then you end up with infinity that will give us in
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determinant form so if you use polar coordinates and if you end up with right here unfortunately we cannot draw any conclusion but for this right here we have zero times something with absolute value less than or equal to one so for sure we will end up with just zero so this is how we can use the PO method to calculate this limit now the last method for you guys is the squee the and in fact earlier
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we were kind of using the squeeing room right here some people might want you to write this limit to be between of zero and zero then argue that this is also approaching zero but I think this right here it's pretty clear because I show you guys that we have something that's multiplied with something that's bounded by one 0 * fin has to be zero now for this though all I did was I change
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the Y to the second power to Y to the force Power so let's try to do this with the PO method first and see what what happens if we do this though we get the limit as R approaching zero X is R cosine Theta but to the third power so put it here put it here this time for the bottom unfortunately there's no nice formula for this we just had to write it out x² gives us R 2 * cine s Theta and then we add Y is R sin Theta and then raise that to the fourth power okay in this case on the
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bottom we could factor out R to the second power that's like the most we can do plus this is r² and we have sin to the 4 power Theta and here we can cancel r s with this and we end up with the limit as R approaching zero I still have R * cosine to the 3 power Theta very similar to this right here but here we have cosine Square theta plus here R 2 * sin 4 power Theta now if you plugging Zer
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into R and 0 into r s right here it looks like we have Z 0 over that isn't the answer just equal to zero because Z divided by right we get roided by cosine Square Theta well remember as I told you guys earlier Theta could be anything what if Theta is equal to Pi / 2 well cosine P two is zero and that will make this bad in fact we cannot draw any conclusion right here unfortunately the difference
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between this and that is we have cosine Square Theta on the bottom if we have Theta is equal to Pi / 2 that might go to Infinity so cannot draw conclusion so how do we do it though of course the squeeze theem first I would like to just point out for you guys that x² + y 4 power this is always positive on the top we have X power I cannot say this is always positive because X could be approaching Zero from the negative Direction so this is what I can do I'm going to put an absolute
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value first around just this expression X power over X2 + y to the 4 power now with this absolute value I can say this has to be non- negative yeah indicate that then we are going to do some algebra work with the absolute value we know the absolute value of a quotient is the same as the quotient of the absolute values but the bottom is never negative so the absolute value on the bottom doesn't matter we can just say this it's equal to the absolute value of x to the third power
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and fact you can also put the third power right here because the absolute value of the product is the same as the product of the absolute value and then over x to the 2 power + y to the 4th power now keep this in mind when we are trying to do the squeeze theorem we will try to show some inequality work we are trying to show that this is in between of zero and also zero because if this is in between of Z and Zer then we conclude that this limit is also equal to zero so now here
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is the idea we have x and x X it would be nice if they cancel right I don't want to have the Y and it's good right here because we know this is plus y to the fourth power which is non negative so I'm just going to take that away and compare this with just absolute value of x to the 3 power over x² now let's thing about it if we take away something from the denominator and this is positive how do
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they compare well in fact this right here will be smaller than or equal to this let me give you a quick example notice the tops are the same right let's say we have 10 and 10 on the bottom here that's say we have two + three on the bottom that say we just have two right here you see how I just take away this three which fraction is bigger this fraction is bigger because this gives us five this
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gives us two so inequality goes like this there are a lot more different cases if you are not familiar with the inequality work that this you will have to for sure watch the other video on how to show a limit does exist but for now we have this inequality next x² it's always non negative you can attach the absolute value to it as well and the reason we do this it's because now we can cancel cancel so this right here is equal to just the value of x and this is great because we know
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X is approaching zero so this must be approaching zero as X goes to zero and as we can see we have this that's in between of zero and zero so we can come here and conclude that this is equal to zero and the idea is that when we have absolute value of something in of 0 and Z take the limit because absolute for is a continuous function so we can just focus on the limit of the inside must be also equal to zero as well and I think if you write this out it should be
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pretty clear and now you might be wondering why did we have to do all this work to show this right here is equal to zero because earlier when we use the polar method we got to the point that we have the limit as R approaching zero of R * cosine to the 3 power Theta over cine 2 Theta + r² sin to 4 power Theta and I told you guys if you put zero in here times this and this is also zero but if Theta it's pi/ 2 it would be bad so I told you guys this was a sad
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face but you can argue zero * anything ided by anything should be just zero and that's all we got right here as well well not all the time you really have to wor about what the angle Theta is here let me give you guys a quick preview in the following video if I change this limit a little bit let's say if we have the limit as XY approaching the origin of x to thir power over x to the 4th power plus y to the second power let me tell you this limit actually does not
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exist even though you still end up with a similar situation in the poal world but you really really really cannot say this is equal to zero because this right here does not exist and you will see that the sque zero wouldn't even work for this one either so that will be in the next video but now hopefully this video helps best of luck in your Calculus 3 class
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