# Showing a limit exists using polar coordiantes

So for
$$\lim_{(x,y)\rightarrow (0,0)} f(x,y)=L$$
to exist, the limit as any path towards the origin of $f$ must exist and equal $L$.
Is it enough to show that if I let $x=r\cos \theta$, $y=r\sin\theta$ and show $\lim_{r\rightarrow 0}f(r\cos\theta,r\sin\theta)$ is independent of $\theta$, and goes to $L$, then the limit is independent of theta, does this mean the limit equals $L$?
Basically, does showing that for any angle $\theta$, the limit equals $L$?
I fear not because I feel like I'm approaching rays towards the origin here, and I'm not considering paths such as $y=x^2$ or something like that.

• You can use polar coordinates to show that a limit does NOT exist. If polars do "work", then essentially it is inconclusive. i.e. The limit may or may not exist. You need then to resort to other methods. – imranfat Aug 22 '17 at 3:13
• math.stackexchange.com/questions/753381/… – Will Jagy Aug 22 '17 at 3:14
• Your fear is wise, and correct. – Fimpellizieri Aug 22 '17 at 3:15
• @imranfat: Actually, polar coordinates used correctly are an excellent way of showing that a limit is zero. By this, I mean that one must take into account that $\theta$ may depend on $r$. More precisely, if one can write $f(r \cos\theta,r\sin\theta)=g(r) \, h(r,\theta)$ where $g(r) \to 0$ as $r \to 0$ and $h(r,\theta)$ is bounded (no matter what $\theta$ is doing), then it is safe to conclude that $f(x,y)\to 0$. But assuming $\theta$ to be constant doesn't work, of course (and I assume this is what you are referring to). – Hans Lundmark Aug 22 '17 at 12:42