Does this prove that the limit does not exist? $$\lim_{(x,y)\to(0,0)}\frac{x}{x^{2}-y^{2}}$$
I tried with $y=mx$ and lateral limits. I got that: $$\lim_{x\to 0^{+}}\frac{1}{x(1-m^{2})}=+\infty$$ and $$\lim_{x\to 0^{-}}\frac{1}{x(1-m^{2})}=-\infty$$ Assumming $1-m^{2}>0$. So the limit does not exist.
It's correct? 
 A: Hint.
$$
\frac{x}{x^2-y^2} = \frac 12\left(\frac{1}{x+y}+\frac{1}{x-y}\right)
$$
A: Converting to polar coordinates, 
\begin{align}
\lim\limits_{r\to 0} \frac{r\cos\theta}{r^2 \cos^{2} \theta + r^2 \sin^2 \theta} = \lim\limits_{r\to 0} \frac{\cos \theta}{r} \implies \lim\limits_{r\to0^+}\frac{\cos \theta}{r} \neq \lim\limits_{r\to 0^-}\frac{\cos\theta}{r},\\
\end{align}
Therefore the limit does not exist. Basically same argument as you provided but in polar coordinates, suggesting your result is correct.
A: Ok, what happens if $\ m=1$? Since $\ f(x,y)$ is a surface, you can't ignore any selective path as the point $\ (x,y)$ can move along any path. But you can try in the following way. Let us move $\ (x,y) \to (0,0)$ along $\ y^2=mx$ . Then 
$$\begin{align}\lim_{(x,y)\to(0,0)} f(x,y)&= \lim_{x\to0}\ \frac{x}{x^2-mx}\\
&=\lim_{x\to0} \frac{1}{(x-m)}[x \ne0\, as\, x \to 0]\\
&=-\frac{1}{m}\\
\end{align}$$ which depends on $\ m$. So for different values of $\ m$, we find different limits of $\ f(x,y)$. That leads us to the fact that $\lim_{(x,y)→(0,0)} \ f(x,y)$ does not exist.
