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?
 A: Note that for any $\epsilon>0$ we have (using the AM-GM inequality)
$$\begin{align}
0<\left|\frac{2x^2y^2}{x^2+y^2}\right|&\le \left|\frac{2x^2y^2}{2|x||y|}\right|\\\\
&=|x||y|\\\\
&\le \frac12(x^2+y^2)\\\\
&<\epsilon
\end{align}$$
whenever $0<\sqrt{x^2+y^2}<\delta=\sqrt{2\epsilon}$
A: You have
$$\begin{equation}\begin{aligned}
\lim_{(x,y)\to (0,0)}\frac{2x^2y^2}{x^2+y^2} & = \lim_{(x,y)\to (0,0)}\frac{2}{\frac{1}{y^2}+\frac{1}{x^2}} \\
& = 0
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
This is because the numerator is fixed at $2$ but the denominator goes to infinity as $x$ and $y$ go to $0$.
Alternatively, you could also use the squeeze theorem by using
$$\begin{equation}\begin{aligned}
0 \le \frac{2x^2y^2}{x^2+y^2} \le \frac{2x^2y^2}{y^2} = 2x^2
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
A: 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 $(1)$,$$\frac{2x^2y^2}{x^2+y^2}<2\sqrt{\frac\varepsilon2}^2=\varepsilon.$$
A: Use polar coordinates: let $r$ be the polar radius ($r=\sqrt{x^2+y^2}$), $\theta$ the polar angle. If $(x,y)\ne (0,0)$,
$$f(x,y)=\frac{2x^2y^2}{x^2+y^2}=\frac{2r^4\cos^2\theta\sin^2\theta}{r^2}
=\tfrac12r^2\sin^22\theta.$$
Can you take it from there?
