Calculating lim using polar coordinations?

Question

How to calculate the following limit:

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

I replaced $x$ with $r\cos(\theta)$ and $y$ with $r\sin(\theta)$ and got $-2\sin(\theta)\cos^3(\theta)$

and got stuck there... any help?

Answer 1

There is no need to use polar coordinates here. Just note that $f(x,x)=-\frac12$ if $x\ne0$, which implies that $\lim_{x\to0}f(x,x)=-\frac12$. But $\lim_{x\to0}f(x,0)=0$.

If you want to do it with polar coordinates, your computations are correct: in fact, $f(r\cos\theta,r\sin\theta)=-2\sin(\theta)\cos^3(\theta)$. Now, note that this is equal to $0$ if $\theta=0$ and that it is equal to $-\frac12$ if $\theta=\frac\pi4$. Therefore, there is no limit at $(0,0)$.

Answer 2

the limit does not exist, since if you approach along the path $y=x$ the limit is $-\frac 12$ and if you approach along $y=-x$ you get $+\frac 12$