The problem says determine whether this function is continuous at $(0,0)$:
$$f(x,y)=\begin{cases} \frac{x^3-y^3}{x^2+y^2} & (x,y) \neq (0,0) \\ 0 & (x,y)=(0,0) \end{cases}$$
Well since I couldn't find any paths with different values I think it is continuous. However when using the epsilon delta definition to prove that the limit is 0 when (x,y) approaches (0,0) I couldn't manage to choose a delta.
$0<\sqrt{x^2+y^2}<\delta$ $\to$ $|\frac{x^3-y^3}{x^2+y^2}-0|<\epsilon$
Should I maybe use the fact that $(x^3-y^3)=(x-y)(x^2+xy+y^2)$?
Let $r=\sqrt{x^2+y^2}$. Then both $|x|$ and $|y|$ are smaller than or equal to $\delta$, and therefore $|x^3-y^3|\leqslant2r^3$. On the other hand, $x^2+y^2=r^2$. So,$$\left|\frac{x^3-y^3}{x^2+y^2}\right|\leqslant2r.$$So, for any $\varepsilon>0$, take $\delta=\frac\varepsilon2$.