# 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

## 1 Answer

Even in case of Cartesian System You wont be moving across y = x^2. In the Cartesian system you would be moving across the lines parallel to x and y axes. So what you do in the polar system still says the same. But in polar system you need to make sure that the result is valid for all the values of theta