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 $0$?

With polar coordinates we're cool, too:
$$x=r\cos\theta\;\;,\;\;y=r\sin\theta$$ 
$$\frac{x^3y}{x^4+y^2}=\frac{r^4\cos^3\theta\sin\theta}{r^4\cos^4\theta+r^2\sin^2\theta}=\frac{r^2\cos^3\theta\sin\theta}{r^2\cos^4\theta+\sin^2\theta}\xrightarrow[r\to 0]{}\frac0{0+\sin^2\theta}=0$$
But what if $\;\sin^2\theta=0\;$ ? Well, then also $\;\sin\theta=0\;$ and the first expression above's already zero from the beginning...
