Computing Multivariable Limits Algebraically

TYPO: The point (2,3) in the second example really should be (3,2) throughout.

In this video we're going to show that the limit of a multivariable function DOES exist by using a range of algebraic tricks such as factoring or multiplying by the radical conjugate. The idea is to manipulate the indeterminate form until such point as you can plug in the numbers. We also take a look at algebraic limit laws for multivariable functions.

0:00 Intro & 1st Example
1:50 Limit Laws
4:50 Factoring Example
7:17 Radical Conjugate Example

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This video was created by Dr. Trefor Bazett. I'm an Assistant Teaching Professor at the University of Victoria.

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[huseyn.asadullah]

5:14 I think the numerator and the denominator are not going to zero at that point!🙄

[bebarshossny]

Bro you legit need to make more multivariable calc videos
I have never seen a better educator
You, sal khan and 3blue1brown are legit gems when it comes to giving the intuition behind the math
Thanks a lot man

[connorkokora]

Teachers like you present math in color, and I don't say that just because this is a video in in color. There is energy in the words you say that illuminates the abstract concepts. Thank you.

[sjn]

In your 2nd example, isn’t the bottom 3 (3*3-2*3)?

[tanayc]

It's awesome that you are making such helpful math videos through this pandemic - thank you!!

[ayushpatel]

This series is really helpful! Appriciate it a ton!!

[gabenuss]

5:53 I would point out that the numerator factors by grouping. I think that would clear some of the ambiguity of guessing and checking how the multivariable polynomial could be factored. In general, factoring a multivariable polynomial involves some kind of factoring by grouping, a concept familiar to many from algebra class.

[amansingh-wwqc]

Thanks and love u, no one taught better than u I am far away from my class syllabus because to take classes from u seems to be watching movie in theatre

[balerion_the_dread]

Sir there was three in the denominator. So we could plug the values of x and y directly, i guess. (3y-xy)

[pragalbhawasthi]

Really helpful!! Please keep uploading such informative videos..

[madghostek]

6:30 here's a explaination why its allowed to cancel these out without disturbing the limit (this applies for single variable limits as well):

lim (x, y)->(3, 2) (y^2+2)(x-3)/y(3-x) = (y^2+2)/y * -(x-3)/(x-3)
the first term is continuous at 3, 2 so the limit is just value of function, equals 11/3.
The second term is pretty much a piecewise function of form :
{undefined for x=3, 1 for any other x}.

Now, recall the scary epsilon delta definition of limit, if you can find arbitrarily small region around x0 (not necessarily including x0), where value of function is arbitrarily close to some value L, then the L is a limit at x0. From the piecewise form we clearly see the function is 1 everywhere but at x=3, so clearly the limit will be L=1.

So by cancelling we are removing a discontinuity, I actually don't exactly know why algebraic cancelling works, maybe it just happens to, but it can be justified by this longer proof.

[ryan]

You have great videos but there is a mistake here. You have (x, y) going to (2, 3) meaning x is 2 and y is 3. Substituting those values into the denominator 3(3) - 2(3) = 3, not zero. Likewise, the numerator goes to -11. I think that perhaps you meant (x, y) is going to (3, 2)?

[continnum_radhe-radhe]

Thank you so much sir for this wonderful course 🔥

[marcelukeje]

Great video sir.🎉🎉

Although I thought you were going to teach how to prove that a limit exists when we have tried all those paths and we got the same answers.

Do you have a video for that?

[dantepillon]

Hi, I had a question, but firstly great video and thank you for making series like these :)
So I was wondering in the conjugate example, why do we need that restriction, should we be able to still evaluate the limit in the same way without it, and if not, how do we know when we don't have restrictions, that the limit is undefined? Also another question I had was, if a limit goes to multiple values depending on the path, is there a way we can define the set of all values it approaches? It seems that it is against the spirit of limits to just say it doesn't exist and then job finished, isnt it is better to make a distinction between a limit that really doesn't exist anywhere on the function and a limit that exists for a set of restrictions and can be defined in a set of real numbers?

[Bermatematika]

You missed two more ways. First, by using the squeeze theorem and second by using the epsilon-delta definition :D.

[BC-zvdt]

At 6:13, in order to explain the cancellation would it be more straightforward to factorise a -1 from the (x-3) or (3-x) to cancel the the notation out? It seems more intuitive to explain that way.

A huge fan of the series by the way. Thanks for putting together!

[markbell]

"We don't care what's happening at the point, we care what is happening around the point."

[henriqueserra]

why did you, in 3d example, said x + y =/ 4, but then in the last step you have that x + y = 4 ?

[husamalsayed]

i did the second example in factoring the numerator with (x + y - 4) = (sqrt(x + y) - 2)(sqrt(x + y) + 2)

and then canceling the (sqrt(x + y) - 2) in the bottom

it turns out to be the same thing

just like to share

thanks for the video