Prove, using first principles, that $\lim\limits_{(x,y)\to(0,0)}{x.y^2=0}$ My try is:
$\left|x.y^2-0\right|<\left|\sqrt{x^2+y^2}.(x^2+y^2)\right|< \delta.\delta^2=\epsilon$
where $0<\sqrt{x^2+y^2}< \delta$ what did I miss here?

Consider $x^2+y^2  <1$, 
then $|y| \le 1$, and $y^2 \le |y|$.
Hence:
$|xy^2| \le |xy|  \le (1/2)(x^2+y^2)$ .
Let $\epsilon >0$ be given.
Choose $\delta = \min (1,2\epsilon)$
Then $|x^2+y^2| \lt \delta$ implies 
$|xy^2| \le$
$(1/2)(x^2+y^2) \lt (1/2)\delta =\epsilon$.
Used: $a^2+b^2 \ge 2|ab|$
