How to get the following limit:
$$\lim_{(x,y)\to (0,0)}\frac{x^4y}{x^8+y^2}=?$$
If I let $x=r\cos \theta$ and $y=r\sin \theta$ where $\theta\in (0, \pi/2)$, then $$\lim_{(x,y)\to (0,0)}\frac{x^4y}{x^8+y^2}=\frac{r^5\cos^4\theta\sin\theta}{r^8\cos^8\theta+r^2\sin^2\theta}$$
It seems the limit does not exist.