Find all points on which a function is discontinuous. $ f(x,y) = \begin{cases} \dfrac{x^3+y^3}{x^2+y^2} &\quad\text{if} [x,y] \neq [0,0]\\[2ex]  0 &\quad\text{if}[x,y] = [0,0]\\ \end{cases} $
The only point it could be discontinuous in is [0,0]. How do I find the limit of the function for $(x,y) \rightarrow (0,0)$? $ \lim_{(x,y) \rightarrow (0,0)} \frac{x^3+y^3}{x^2+y^2} $ seems pretty hard to analyse.

$x = r\cos \theta$, $y = r\sin \theta$
instead of $(x,y) \rightarrow (0,0)$ I can now use $r\rightarrow0$
$$\begin{align}
\lim_{r\to0} \frac{r^3\cos^3\theta + r^3\sin^3\theta}{r^2\cos^2\theta + r^2\sin^2\theta}
&=\, \lim_{r\to0} \frac{r (\cos^3\theta + \sin^3\theta)}{\cos^2\theta + \sin^2\theta} \\
&=\,0 
\end{align}$$
