Existence of a limit of a function Let let $$f(x,y)=\frac{x^3-y^3}{x^2+y^2}$$ and $(x,y)\not=(0,0)$ 
How to prove that the function has a limit as $(x,y)\to (0,0)$ 

Should I look at firstly on the x axis and then on y-axis? 
 A: Hint: Polar coordinates.${}{}{}{}{}$
A: We can deduce the limit using these inequalities:
$$0\leq|f(x,y)|=\frac{|x^3-y^3|}{x^2+y^2}\leq \frac{|x|x^2+|y|y^2}{x^2+y^2}\leq|x|+|y|\to0 $$
A: As Andé Nicolas said, you should use polar coordinates. If you don't want to use polar coordinates :
Let $\Vert \cdot \Vert$ be the usual euclidean norm in $\mathbb{R}^{2}$. $(x,y) \, \rightarrow \, (0,0)$ is equivalent to $\Vert (x,y) \Vert \, \rightarrow \, 0$. We have :
$$ \vert f(x,y) \vert \leq \frac{\vert x^{3} - y^{3} \vert}{x^{2}+y^{2}} \leq \frac{\vert x \vert^{3} + \vert y \vert^{3}}{x^{2}+y^{2}} $$
(we used the triangle inequality). You can also notice that, for every $(x,y) \in \mathbb{R}^{2}$, $\vert x \vert \leq \Vert (x,y) \Vert = \sqrt{x^{2}+y^{2}}$. Using this, we have :
$$ \vert f(x,y) \vert \leq \frac{\Vert (x,y) \Vert^{3} + \Vert (x,y) \Vert^{3}}{\Vert (x,y) \Vert^{2}} $$.
For $(x,y) \neq (0,0)$, we get :
$$ \vert f(x,y) \vert \leq 2 \Vert (x,y) \Vert $$
So, we conclude that $\lim \limits_{(x,y) \, \rightarrow \, (0,0)} f(x,y) = 0$.
