$\lim_{(x,y)\to 0} \frac{x\sin(y)- y\sin(x)}{x^4 + y^4}$ without polar coordinates? I have the following limit: 
$$\lim_{(x,y)\to 0} \frac{x\sin(y)- y\sin(x)}{x^4 + y^4}$$
And I must evaluate it without polar coordinates. I have tried a lot of stuff but nothing works. Can someone give me a hint?
 A: Let $f(x,y)=\dfrac{x\sin(y)-y\sin(x)}{x^4+y^4}$
We have $f(x,x)=0$.
On the other hand we also have 
$f(x,2x)=\dfrac{x\sin(2x)-2x\sin(x)}{17x^4}=\dfrac{x(2x-\frac 43x^3+o(x^3))-2x(x-\frac 16x^3+o(x^3))}{17x^4}=\dfrac{-x^4+o(x^4)}{17x^4}$
Thus $\begin{cases}\lim\limits_{x\to 0}f(x,x)=0\\\lim\limits_{x\to 0}f(x,2x)=-\frac 1{17}\end{cases}$
Since they are different, we can conclude that the limit $\lim\limits_{(x,y)\to (0,0)}f(x,y)$ doesn't exists. 
A: You can easily disprove that the limit exists by considering the one dimensional family of rays $y=\lambda x, \lambda\in\mathbb{R}$ and taking the limit to the origin along them instead, for a given $\lambda$. If the limit exists then it shouldn't depend on $\lambda$. 
The limit on the rays boils down to evaluating:
$$L=\lim_{x\to 0}\frac{x\sin(\lambda x)-\lambda x\sin x}{(1+\lambda^4)x^4}$$
which one can evaluate straightforwardly using L'Hopital's rule or a Taylor expansion to yield:
$$L=\frac{\lambda-\lambda^3}{6(1+\lambda^4)}$$
which does depend on $\lambda$ and therefore the limit cannot exist.
A: The limit does not exist. Take limit along $y=2x$ for example. You get $\lim \frac {2x\sin \, x(\cos \, x -1)} {(1+2^{4})x^{4}}$.  Use the fact that the $\frac {\sin\, x} x \to 1$ and use L'Hopital's Rule to see that the limit is $-1/17$. Since the limit along $x=y$ is $0$ it follows that the given limit does not exist. 
