Two Variables Walked Into A Bar

Calculus Level 3

$\large\displaystyle\lim_{(x,y) \to (0,0)} \dfrac{xy^{2}}{x^{2} + y^{4}} =\, ?$

0

\dfrac{1}{2}

1

\dfrac{3}{4}

\infty

The limit does not exist

1 solution

Brian Charlesworth
Apr 23, 2016

Along the horizontal path $y = 0$ excluding the origin we have that $\dfrac{xy^{2}}{x^{2} + y^{4}} = 0$ , so the limit along this path is $0$ . However, along the path $x = y^{2}$ to the origin the limit becomes

$\displaystyle\lim_{(x,y) \to (0,0)} \dfrac{y^{4}}{2y^{4}} = \dfrac{1}{2} \ne 0$ .

Thus since the limit is path-dependent it does not exist as $(x,y) \to (0,0)$ .

I haven't reached this in my studies yet, so alas I got the problem wrong because I wasn't sure how to approach it.

But nevertheless, I do have a question! Since x and y are both approaching the same number, would it be incorrect to write y as x and evaluate the limit just as x approaches 0?

Andrew Tawfeek - 5 years, 1 month ago

For a two-variable limit to exist it must be independent of the path taken to the origin. (With one-variable limits we only need consider approaching from the left or the right, but with two variables the possibilities are infinite.) If we take the path $y = x$ as you suggest the limit would be

$\displaystyle\lim_{x \to 0} \dfrac{x^{3}}{x^{2} + x^{4}} = \lim_{x \to 0} \dfrac{x}{1 + x^{2}} = 0$ ,

which is a different value than we found along the path $x = y^{2}$ .

Brian Charlesworth - 5 years, 1 month ago

This is really interesting, thank you for your response!

Andrew Tawfeek - 5 years, 1 month ago

Thanks @Brian Charlesworth You solved my problem too.

Madhukar Thalore - 5 years, 1 month ago

I guess a method prove that a two variable limit doesn't exist is to obtain two different values for two different paths, but how would we prove that a two variable limit exists and is independent of the path ?

A Former Brilliant Member - 5 years, 1 month ago

@A Former Brilliant Member – You simply use the fundamental $\epsilon-\delta$ definition of limit, which works for any number of variables, i.e. show that for every $\epsilon >0$ , there exists a $\delta >0$ , such that $|f(x)-f(\alpha)|< \epsilon, \hspace{10pt} \forall ||x-\alpha||<\delta$

Abhishek Sinha - 5 years, 1 month ago

I had done the same way. I put y=mx and then put the limit x approaching to 0? Is it incorrect ?

Madhukar Thalore - 5 years, 1 month ago

@MadhukarThalore Since the limit is in two dimensions, you need to make sure all possible paths converge, not just y=mx.

Toby M - 2 years, 4 months ago

Brian, nice solution. Letting x = my^2, the limit becomes m/(m^2 + 1), so we have an infinity of limits. I would think that unless x is not proportional to y^2, the limit will be 0. Ed Gray

Edwin Gray - 2 years, 4 months ago

It seems that we share the same opinion.

Alstream Yang - 2 years, 3 months ago

@Brian Charlesworth can we replace x and y with polar coordinates and then say that limit doesn't exist by the fact that theta can take any value ?

Mohit Kumar Basak - 2 years, 3 months ago

Substituting in polar coordinates does work well for some two-variable limits, but in this case with $x = r\cos(\theta), y = r\sin(\theta)$ the limit would become

$\displaystyle \lim_{r \to 0} \dfrac{r^{3} \cos(\theta) \sin^{2}(\theta)}{r^{2}\cos^{2}(\theta) + r^{4}\sin^{4}(\theta)} = \lim_{r \to 0} \dfrac{r\cos(\theta)\sin^{2}(\theta)}{\cos^{2}(\theta) + r^{2}\sin^{2}(\theta)}$ .

which would go to $0$ for any fixed value of $\theta$ . Even if you varied $\theta$ as $r \to 0$ it's not obvious that you would get different results, so for this particular limit it's better to use the method I have outlined in my solution. Your idea would would work well though with limits such as

$\displaystyle \lim_{x,y \to 0,0} \dfrac{xy}{\sqrt{x^{2} + y^{2}}}$ or $\displaystyle \lim_{x,y \to 0,0} \dfrac{x^{2} - y^{2}}{\sqrt{x^{2} + y^{2}}}$ ,

both of which are equal to $0$ .

Brian Charlesworth - 2 years, 3 months ago

Two Variables Walked Into A Bar

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