<s>[INST] Does the limit $\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4}$ exists? I was trying to find if the the following limit exists:
$$
\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4}
$$
From this topic I tried to switch to polar coordinates:
$$
\frac{x^3y^2}{x^6+y^4}=\frac{r^3\cos^3\theta\cdot r^2\sin^2\theta}{r^6\cos^6\theta+r^4\sin^4\theta}=\frac{r\cos^3\theta\sin^2\theta}{r^2\cos^6\theta+\sin^4\theta}\to 0
$$
But apparently this limit does not exists (proved it in another way). Why this method worked for the limit in that topic and here it didn't work?
 [/INST] The conversion to polar coordinates has not failed.  Note what happens when $r=\tan^2(\theta)\sec(\theta)$, $|\theta| > \pi/2$.  Then,
$$\frac{r\cos^3(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}=\frac{\sin^4(\theta)}{2\sin^4(\theta)}=\frac12$$
And of course, if $\theta =0$ or $\theta =\pi/2$, then
$$\frac{r\cos^3(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}=0$$

EXPLANATION:  So what is happening here?
Note that $x^6+y^4\ge 2\sqrt{x^6y^4}=2|x|^3y^2$, with equality for $y^2=|x|^3$. Therefore, we have
$$\left|\frac{x^3y^2}{x^6+y^4}\right|\le \frac12$$
with equality when $y^2=|x|^3$.
If we transform to polar coordinates, this analysis is identical to
$$\left|\frac{r\cos^3(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}\right|\le \left|\frac{r\cos^3(\theta)\sin^2(\theta)}{2\sqrt{r^2\cos^6(\theta)\sin^4(\theta)}}\right|=\frac12$$
with equality when $r=\frac{\sin^2(\theta)}{|\cos^3(\theta)|}=\tan^2(\theta)|\sec(\theta)|$.
</s>