Find the limit or prove that it does not exist $\lim_{(x, \space y) \to (0, \space 0)} f(x, y) $ where $f(x, y) = \frac{x^5-y^5}{x^4-2x^2y^2+y^4}$ Find the limit or prove that it does not exist
$$\lim_{(x, \space y) \to (0, \space 0)}  f(x, y) $$ where $$f(x, y) = \frac{x^5-y^5}{x^4-2x^2y^2+y^4}$$

Iterated limit $\displaystyle{\lim_{y \to 0}} \space \displaystyle{\lim_{x \to 0}} \space f(x, y) = \displaystyle{\lim_{x \to 0}} \space \displaystyle{\lim_{y \to 0}} \space f(x, y) = 0$, but it doesn't mean that $\displaystyle{\lim_{(x, \space y) \to (0, \space 0)}}  f(x, y) = 0 $. 
I've also tried substitution $x = r \cdot \cos \phi, \space y = r \cdot \sin \phi$, which gave me $ \frac{r(\cos^5\phi - \sin^5\phi)}{(\cos^2\phi-\sin^2\phi)^2} $. If $x \to 0$ and $y \to 0$, then $r \to 0$. So, $\displaystyle{\lim_{r \to 0}} \space \frac{r(\cos^5\phi - \sin^5\phi)}{(\cos^2\phi-\sin^2\phi)^2} = 0$. 
But I have a feeling that original limit doesn't exist and WolframAlpha says that too and I'm stuck here. If my assumption is correct, how to prove it properly?
 A: The limit doesn't exist. If $n\in\mathbb N$, then$$f\left(\sqrt{\frac1{n^2}+\frac1{n^4}},\frac1n\right)=\left(\left(\frac1{n^2}+\frac1{n^4}\right)^{5/2}-\frac1{n^5}\right)n^8$$and$$\lim_{n\to\infty}\left(\left(\frac1{n^2}+\frac1{n^4}\right)^{5/2}-\frac1{n^5}\right)n^8=\infty.$$
A: I think your substitution gives you a hint. The angles $\tan\phi =\pm 1$ are the problematic ones, i.e., the limit along the lines $\pm x=y$. You can check that along these lines there's a divergence. 
A: In the posting, $$ f(r,\phi)=\frac{r(\cos^5\phi -\sin^5\phi
)}{(\cos^2\phi -\sin^2\phi)^2} = \underbrace{\frac{r}{\cos\
\phi-\sin\ \phi}}_{=g}\underbrace{\frac{\sum_{i=1}^4\
\sin^i\phi\cos^{4-i}\phi }{(\cos\ \phi+\sin\ \phi)^2} }_{=h}$$
When $f(r,0)=r$, so $\lim_r\ f=0$. 
Further, $\lim_{\phi \rightarrow
\pi/4}\ h $ exists. We have a sequence $\phi_n<\pi/4,\
\phi_n\rightarrow \pi/4$ s.t. $ \cos\ \phi_n > \sin\ \phi_n$. We let
$$r_n =\sqrt{\cos\ \phi_n - \sin\ \phi_n}.$$ Hence $$ g(r_n,\phi_n) =
\frac{1}{\sqrt{ \cos\ \phi_n - \sin\ \phi_n }}\rightarrow \infty$$
so that $f(r_n,\phi_n)$ goes to $\infty$. 
