Computation of a 2D limit I came across this limit and I used Mathematica to solve it, (it shows that it is equal to $0$). 
Any thoughts on how to prove this ? The limit is:
$$\lim \limits_{(x,y)\to (4,0)} \frac{y^2(x^2+y^2)}{(x-4)^2+y^2}$$
Thanks.

Let $u=x-4\to 0$ then
$$\lim \limits_{(x,y)\to (4,0)} \frac{y^2(x^2+y^2)}{(x-4)^2+y^2}=\lim \limits_{(u,y)\to (0,0)} \frac{y^2((u+4)^2+y^2)}{u^2+y^2}$$
and for


*

*$y=0$
$$\frac{y^2((u+4)^2+y^2)}{u^2+y^2}=0$$
but for


*

*$u=y$
$$\frac{u^2((u+4)^2+u^2)}{2u^2}=\frac{2u^2+8u+16}{2}=u^2+4u+8\to 8$$
therefore the given limit doesn't exist.
