How to prove that $\lim_{(x,y) \to (0,0)} \frac{x^3y}{x^4+y^2} = 0?$ 
How to prove that $\lim_{(x,y) \to (0,0)} \dfrac{x^3y}{x^4+y^2} = 0?$

First I tried to contradict by using $y = mx$ , but I found that the limit exists.
Secondly I tried to use polar coordinates, $x =  \cos\theta $ and $y = \sin\theta$,
And failed .. How would you prove this limit equals $0$?

Let $y=kx^2$, then the expression is $\frac{kx}{1+k^2}$.
$|\frac{k}{1+k^2}|\leq 1/2$ so the expression is bounded by $|x/2|$.  The path along $x=0$ can be done separately.
