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Evaluate the following limit: $$\lim_{(x,y) \to (0,0)}\frac{-xy}{x^2 + y^2}$$

My try:

Using polar substitution - putting $x =r\cos(\theta)$, $y = r\sin(\theta)$,

$$\implies\lim_{r \to 0} \frac{-r^2\cos(\theta) \sin(\theta)}{r^2(\sin^2\theta+\cos^2\theta)} =\lim_{r \to 0} -\cos(\theta) \sin(\theta) = -\cos(\theta) \sin(\theta)$$

Does this make any sense?

Also, approaching from two different sides, $$\begin{align}\lim_{(x,y) \to (0,0)}\frac{-xy}{x^2 + y^2} & = \lim_{x\to 0}\frac{-x(0)}{x^2 + (0)^2} = 0\\\lim_{(x,y) \to (0,0)}\frac{-xy}{x^2 + y^2} & = \lim_{y\to 0}\frac{-(0)y}{(0)^2 + y^2} = 0 \end{align}$$

Both are equal... but the limit doesn't exist which I verified from Symbolab. What should I do to evaluate this?

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    $\begingroup$ You cancel the $r^2$ so the limit depends essentially on your angle. Wouldn't that indicate that the limit does not exist? Just to be clear: You can use polar coordinates to prove that the limit does NOT exist. You cannot use polars to prove that the limit exist. $\endgroup$
    – imranfat
    May 8, 2022 at 3:58
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    $\begingroup$ What happens if you approach to the origin along the line $y=x$? $\endgroup$
    – azif00
    May 8, 2022 at 3:58
  • $\begingroup$ Well, there can be only one limit, and $\theta$ can change. $\cos\theta\sin\theta$ is a “directional limit.” But $\lim_{(x,y)\to (0,0)}$ doesn’t specify a direction. $\endgroup$
    – Thomas Andrews
    May 8, 2022 at 3:58

3 Answers 3

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Let $y=ax$. Then $$ \lim_{x\to0}f(x,ax)=\frac{-a}{1+a^2} $$ and certainly depends on $a$.

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Consider $(x, y) = (\frac{1}{n}, \frac{1}{n})$ as $n \to \infty$. In that case, your limit evaluates to $-\frac{1}{2}$, which is a contradiction with the other two limits you found so the limit doesn't exist.

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Using polar coordinates you find that $$\lim_{(x,y)\to (0,0)}\frac{-xy}{x^{2}+y^{2}}=\lim_{r\to 0;r>0}\frac{-r^{2}\cos \theta\sin \theta}{r^{2}}=-\cos\theta\sin\theta, \quad (*)$$ So the limit doesn't exists, it depends of values for $\theta$ , just take $\theta=\pi$ and we have $0$ as limit but $\theta=\pi/6$ we have $-\sqrt{3}/4$ as limit, contradiction. I think it's perfectly valid argument. If you need use path, just see in the path $r(t)=(t,mt)$ with $m\in \mathbb{R}^{*}$.

EDIT:

As pointed out Hans Lundmark in the comment, the line in $(*)$ has problems about rigor mathematics. We will try to fix with change of variables to polar coordinates we have for $x^{2}+y^{2}\not=0$, that, $$\frac{-xy}{x^{2}+y^{2}}=-\frac{r^{2}\cos\theta\sin\theta}{r^{2}}=-\cos\theta\sin\theta, \quad r>0$$ and since for different values of $\theta$ we get differents values so the $\displaystyle \lim_{(x,y)\to (0,0)}\frac{-xy}{x^{2}+y^{2}}$ doesn't exists.

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  • $\begingroup$ The argument is valid, but it shouldn't be written that way. The first equals sign that you have written is false, since the left-hand side doesn't exist, while the right-hand side does, if you interpret it as a one-variable limit in $r$ for a fixed value of $\theta$, which is precisely what you do when you write your second equals sign. $\endgroup$
    – Hans Lundmark
    May 8, 2022 at 7:16
  • $\begingroup$ Hi @HansLundmark thank you for the feedback and I understand your remark. Can you show how I should write it, formally? Maybe as like: Change of variables to polar coordinates give $\frac{-xy}{x^{2}+y^{2}}=-\cos\theta\sin \theta$ and then fix $\theta$ for some values? I understand the problem is in left-hand the limit doesn't exists but right-hand the limit exists so not matching. $\endgroup$
    – A. P.
    May 8, 2022 at 8:48
  • $\begingroup$ Yes, that's much better. Write it like that, and then say something like “since we get different values for different $\theta$, the two-variable limit doesn't exist”. $\endgroup$
    – Hans Lundmark
    May 8, 2022 at 9:40
  • $\begingroup$ @HansLundmark I edited, what do you think now? Thanks for you feedback. $\endgroup$
    – A. P.
    May 8, 2022 at 9:45
  • $\begingroup$ It still looks weird to me, since you write that something (which actually doesn't exist, but we don't know that yet) is simultaneously equal to $0$ and to $-\sqrt{3}/4$. I would avoid using the notation $\lim_{(x,y)\to (0,0)}$ at all, until after the limit has been shown to exist (or not). $\endgroup$
    – Hans Lundmark
    May 9, 2022 at 13:10

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