TITLE: Is polar coordinates enough to prove that a limit exists
QUESTION [9 upvotes]: Somewhat of a basic question but I failed to find an answer or come up with a formal one myself.
Suppose I want to find the limit $\lim_{{(x,y)} \to {(0,0)}}f(x,y)$ using spherical coordinates $x:=r\cos \theta$, $y:= r\sin\theta$. Suppose I found that $\lim_{r \to 0} f(r,\theta)$ exists and is equal to $\alpha$ regardless of $\theta$. Did I really cover every possible path? Can we say for sure that the limit is $\alpha$? maybe some other limit exists in another path that we didn't cover.
For example, take $\lim_{(x,y) \to (0,0)}\frac{x^2y}{x^2+y^2} = \lim_{r\to 0}\frac{r^3\cos^2\theta \sin \theta}{r^2} = \lim_{r \to 0}r\cos^2 \theta \sin \theta = 0$.
I agree that IF the limit exists, it has to be zero. But maybe there is some path we didn't cover and from that path the limit is something else?

REPLY [6 votes]: The fine point is in the phrase "regardless of $\theta$". Let $f$ be the characteristic function of the set
$$A:=\left\{{1\over k}\biggl(\cos{1\over k},\sin{1\over k}\biggr)\>\biggm|\>k\in{\mathbb N}_{\geq1}\right\}\ .$$
Then $\lim_{r\to0}\tilde f(r,\theta)=0$, regardless of $\theta$, but the $\lim_{(x,y)\to(0,0)} f(x,y)$ does not exist.
If, however, you can prove that for some $(x,y)\mapsto g(x,y)$ defined in a punctured neighborhood of $(0,0)$ you have
$$|g(r\cos\theta,r\sin\theta)|\leq q(r)\quad \wedge\quad \lim_{r\to0}q(r)=0$$
then you may conclude that in fact $\lim_{(x,y)\to(0,0)} g(x,y)=0$.