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EDIT: this is wrong, in fact the limit does not exist, as indicated in the answers.

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

It is easy to show, without using polar coordinates, that such a limit is $0$. Indeed, $$ 0 \leq\Bigg|\frac{x^2}{x + y^2}\Bigg| = \Bigg|\frac{x}{x + y^2}\Bigg||x| \leq |x|, $$ and the argument follows by the squeeze theorem. Now assume that we do not know this fact and we start using polar coordinates instead. Write $$ x = r\cos(\theta), y = r \sin(\theta). $$ Then, $$ \frac{x^2}{x + y^2} = \frac{\cos^2(\theta)}{\frac{\cos(\theta)}{r} + \sin^{2}(\theta)}. $$ It is obvious that, for a fixed $\theta$, the above expression goes to zero as $r \to 0$, but of course, this is not enough to show what we want. What we have to see is that there exists a one-variable function $h = h(r)$, depending only on $r$, such that $$ 0 \leq \Bigg|\frac{\cos^2(\theta)}{\frac{\cos(\theta)}{r} + \sin^{2}(\theta)}\Bigg| \leq h(r) $$ for every $\theta$ and $r$, and with $\lim_{r \to 0}h(r) = 0$.

The question is: how do we find the function $h = h(r)$ in our case?

EDIT: this is wrong, in fact the limit does not exist, as indicated in the answers.

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  • $\begingroup$ What about $|r\cos\theta|$ ? :-) $\endgroup$ – Yves Daoust Nov 29 '20 at 16:52
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    $\begingroup$ Clearly, $x+y^2=0$ is problematic. $\endgroup$ – Yves Daoust Nov 29 '20 at 17:16
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The limit is NOT zero. It does not exist!

If $x=y^2$, as $y\to 0$, then $x\to 0$ and $$\frac{x^2}{x + y^2} =\frac{y^4}{y^2 + y^2}=y^2\to 0.$$ On the other hand, along the curve $x=-y^2+y^4$, as $y\to 0$, then $x\to 0$ and $$\frac{x^2}{x + y^2} =\frac{(-y^2+y^4)^2}{-y^2+y^4 + y^2}=\frac{y^4-2y^6+y^8}{y^4}= 1-2y^2+y^4\to 1.$$ Indeed, in your first approach, $\left|\frac{x}{x + y^2}\right|$ is NOT bounded in a neighbourhood of the origin.

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  • $\begingroup$ Yes you are right, the limit does not exist! $\endgroup$ – Smm Nov 29 '20 at 16:49
  • $\begingroup$ Yes, it does not exists. $\endgroup$ – Robert Z Nov 29 '20 at 16:51
  • $\begingroup$ Sorry for the mistake and thank you! $\endgroup$ – Smm Nov 29 '20 at 16:52
  • $\begingroup$ @Smm You are welcome! $\endgroup$ – Robert Z Nov 29 '20 at 16:52
  • $\begingroup$ This makes me wonder another thing: in a previous answer, which was deleted, it was suggested to use $x=r^2\cos^2(\theta)$ and $y=r\sin(\theta)$, or something like that. Can we use polar coordinates with $x=r^a\cos^{b}(\theta)$ and $x=r^c\sin^{d}(\theta)$, where $a,b,c,d \in \mathbb{R}$? I guess that the answer is no, otherwise we would get a contradiction with this limit, since it does not exist... $\endgroup$ – Smm Nov 29 '20 at 16:56

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