<s>[INST] 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  $\delta=\sqrt{\epsilon}$
 [/INST] In two variables the epsilon-delta definition for $\lim_{\substack{x\to a\\ y\to b}}f(x,y)=L$ means that for every $\epsilon >0$ there exists a $\delta>0$ such that $\big|f(x,y)-L\big|<\epsilon$ whenever $0<\sqrt{(x-a)^2+(y-b)^2}<\delta$. 
In your case, you want to show that $\big|f(x,y)-0\big|<\epsilon$ whenever $0<\sqrt{x^2+y^2}<\delta$. You did this by showing that
\begin{align}x^2y^2\leq (x^2+y^2)^2\implies\bigg|\frac{x^2y^2}{x^2+y^2}-0\bigg|\le\bigg|\frac{(x^2+y^2)(x^2+y^2)}{x^2+y^2}\bigg|=\bigg|x^2+y^2\bigg|=x^2+y^2\end{align}
so that you could choose $\delta=\sqrt{\epsilon}$ and then get the required form of $\big|f(x,y)-0\big|<\epsilon$.
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