# Calculating Limits Using Polar Coordinates-Phylosophical question!

I'm having difficult times understanding the rules regarding polar coordinates, in the context of calculating limits.

On the one hand, I understand that when we take $\theta$ constant, then the path $f(r\cos\theta, r\sin\theta )$ is a straight line, which means that in order to encounter all possible paths, we need also to check what if $\theta$ depends on $r$.

But if so, let's look at the following: $\lim_{r\to 0^+} r\sin(\theta)$ which is obviously zero. But, I can take $\theta$ to depend on $r$ by : $\theta = \arcsin (\frac{1}{r^2} )$ , and then get that $\lim_{r\to 0^+} r\sin(\theta)=\infty$. What am I getting wrong ? $\theta$ can obviously depend on $r$ , so where is my mistake in the calculation above ?

I don't understand what this means: "I can take $\theta$ to depend on $r$". –  wj32 Apr 25 '13 at 7:14
If $|r|\lt 1$, there is no angle whose sine is $\dfrac{1}{r^2}$.
OOPS. sorry about that. But in the general framework, $\theta$ can be a non-constant right? like $\theta = kr$ or something... ? –  czash Apr 25 '13 at 8:41
Certainly it can, picking $\theta$ constant is calculating limits only along lines. There are plenty of cases in $2$-variable calculus where a function $f(x,y)$ approaches $0$ as $(x,y)\to (0,0)$ along any line, but the limit as $(x,y) \to (0,0)$ of $f(x,y)$ does not exist because, for example, there is trouble if we approach $(0,0)$ along $y=x^2$. –  André Nicolas Apr 25 '13 at 8:54