Verify the following limit using epsilon-delta definition: $ \lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}=0$ Show that $$ \lim\limits_{(x,y)\to(0,0)}\dfrac{x^2y^2}{x^2+y^2}=0$$
My try:
We know that, $$ x^2\leq x^2+y^2 \implies x^2y^2\leq (x^2+y^2)y^2 \implies x^2y^2\leq (x^2+y^2)^2$$
Then, $$\dfrac{x^2y^2}{x^2+y^2}\leq x^2+y^2 $$
So we chose  $\delta=\sqrt{\epsilon}$

Tips
$\lim\limits_{\left(x,y\right)\rightarrow\left(0,0\right)} \dfrac{x^2 y^2}{x^2+y^2} = \lim\limits_{\left(x,y\right)\rightarrow\left(0,0\right)} \dfrac{1}{\frac{1}{x^2}+\frac{1}{y^2}}$
