Heya guys. I have to study the continuity of the following function at (0,0):

$\displaystyle f(x,y) =\begin{cases}

\frac{x^3y}{x^4+y^2} & (x,y)\neq(0,0) \\

0 & (x,y)=(0,0)

\end{cases}$

Thus I need to check whether

$\displaystyle \lim_{(x,y)\to (0,0)} \frac{x^3y}{x^4+y^2}$

exists or not.

I converted to polar coordinates:

$\displaystyle \lim_{r\to 0} \frac{r^4 cos(\theta)^3 sin(\theta)}{(r cos(\theta))^4+(r sin(\theta))^2} = \lim_{r\to 0} \frac{r^2 cos(\theta)^3 sin(\theta)}{r^2 cos(\theta)^4+sin(\theta)^2} = \frac{0}{sin(\theta)^2}$

Thus the limit seems to be 0 except where

$\displaystyle sin(\theta)^2=0 \implies sin(\theta)=0 \implies \theta = 0 \vee \theta = \pi$

If $\displaystyle \theta = 0 \vee \theta = \pi$, then

$\displaystyle \lim_{r\to 0} \frac{r^4 cos(\theta)^3 sin(\theta)}{(r cos(\theta))^4+(r sin(\theta))^2} = \frac{0}{0}$

I then used L'Hopital's rule:

$\displaystyle \lim_{r\to 0} \frac{r^2 cos(\theta)^3 sin(\theta)}{r^2 cos(\theta)^4+sin(\theta)^2} = \lim_{r\to 0} \frac{cos(\theta)^3 sin(\theta)}{cos(\theta)^4} = 0$

since $\displaystyle sin(\theta)=0$.

However, by using such paths as $\displaystyle y=x$ and $\displaystyle y=2x$, the limit equals 1 and 2, respectively. What's going on that I don't understand here?

Thanks!