Prove that this function is continuous at $(0,0)$ I need to prove that the function is continuous at $(0,0)$.
$$f(x,y)=\frac{2x^2y^2}{x^2+y^2}, \text{ if } (x,y)\neq(0,0)$$
and $0 \text{ if } (x,y)=(0,0)$ 

I'm trying to prove that $$\lim_{(x,y)\to (0,0)}\frac{2x^2y^2}{x^2+y^2}=0$$
but I don't know how to relate it to $$0\leq \sqrt{x^2+y^2}\leq  \delta.$$ Finally I have $$2(x^2+y^2)\leq \epsilon.$$ Any ideas that can help me?

Since $x^2,y^2\leqslant x^2+y^2$,$$0\leqslant\frac{2x^2y^2}{x^2+y^2}\leqslant2\frac{(x^2+y^2)^2}{x^2+y^2}=2(x^2+y^2).\tag1$$So, by the squeeze theorem,$$\lim_{(x,y)\to(0,0)}\frac{2x^2y^2}{x^2+y^2}=0.$$
Or, given $\varepsilon>0$, you can take $\delta=\sqrt{\frac\varepsilon2}$. If $\sqrt{x^2+y^2}<\delta$, then, by $(1)$,$$\frac{2x^2y^2}{x^2+y^2}<2\sqrt{\frac\varepsilon2}^2=\varepsilon.$$
