How to find this two-variables limit of $f$ as $(x,y)$ approaches to $(0,0)$? I am trying to evaluate the limit $$\lim_{(x,y) \to (0,0)}\frac{2x^6-y^{10}}{x^2-y^3}$$
I have seen that both the reiterated limits are zero. Moreover, I have evaluated the limit by different curves such as $y=x^2$ and the limit equals zero as well.
However, the polar coordinates $$\lim_{r \to 0}\frac{r^4(2\cos^6(\theta)-r^4\sin^{10}(\theta))}{\cos^2(\theta)-r\sin^3(\theta)}$$ make me think that the limit does not exist because it depends on the $\cos^2\theta$ factor, but I don't know how to prove it or which curve to use.

Consider the sequence
$(\frac{1}{ n},0)\to(0,0)$
Then $f(\frac{1}{n},0)=\frac{1}{n^4}\to 0 $
Again the sequence $(\sqrt{\frac{n^2+1}{ n^3}},\frac{1}{ \sqrt[3]n})\to(0,0)$
But $\begin{align}f(\sqrt{\frac{n^2+1}{ n^3}},\frac{1}{ \sqrt[3]n})&=\frac{\frac{2(n^2+1)^3}{n^6}-\frac{1}{n^3\sqrt[3]n{}}}{\frac{1}{n^3}}\\&=2(1+\frac{1}{n^2}) ^3-\frac{1}{\sqrt[3]n}
 \end{align}$
Hence $f(\sqrt{\frac{n^2+1}{ n^3}},\frac{1}{ \sqrt[3]n})\to 2$
Hence $\lim_{(x,y) \to (0,0)}\frac{2x^6-y^{10}}{x^2-y^3}$ doesn't exists.
