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A lot of questions say "use polar coordinates" to calculate limits when they approach $0$.

But is using polar coordinates the best way to evaluate limits, moreover, prove that they exist?

Do they account for every single possible direction to approach a limit, for example, along a parabola.

Specifically, if I were to show that $$\lim_{(x,y)\rightarrow (0,0)} f(x,y)=L$$ using polar coordinates, is that enough to asser that the limit is indeed, $L$. ?

Hugh Entwistle
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Usually the “use polar coordinates” technique for evaluating limits of two variables works like this: Write $f(x,y) = g(r,\theta)$, and let $r\to 0$. If the limit still depends on $\theta$, the two-variable limit $\lim_{(x,y) \to (0,0)} f(x,y)$ does not exist.

But if $\lim_{r\to 0} g(r,\theta) =L$, it is not sufficient to say that $\lim_{(x,y) \to (0,0)} f(x,y)=L$. For instance, let $$ f(x,y) = \frac{x^2y}{x^4+y^2} $$ Then substituting $x=r \cos\theta$, $y=r\sin\theta$ gives $$ f(x,y) = \frac{r^3\cos^2\theta\sin\theta}{r^4\cos^4\theta + r^2\sin^2\theta} = \frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta + \sin^2\theta} $$ as $r\to0$, the expression on the right tends to zero. But $\lim_{(x,y) \to (0,0)} f(x,y)\neq 0$. If we approach $(0,0)$ along the line $y=x^2$, we get $$ \lim_{x\to 0,y=x^2} f(x,y) = \frac{x^2(x^2)}{x^4 + (x^2)^2} = \frac{1}{2} $$

Matthew Leingang
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    The statement “the expression on the right tends to zero” isn't true if $\theta$ is allowed to vary along with $r$ (which it should be allowed to do). – Hans Lundmark Mar 24 '17 at 09:17
  • @HansLundmark: You're absolutely right. I am assuming that the technique the OP is asking about is just letting $r\to0$ and keeping $\theta$ fixed. This is the more commonly taught technique. – Matthew Leingang Mar 24 '17 at 15:06
  • Really? What's the point of teaching it that way? If all you know is that the limit is $L$ as $r \to 0$ with $\theta$ fixed, then you can't conclude anything about the actual limit as $(x,y)\to (0,0)$, so what has been gained by the investigation? Finding examples of two different approaches giving different limits (in the case that the limit doesn't exist) is usually easier in the original $(x,y)$ coordinates. The point of polar coordinates (as I see it) is to have a tool for proving that the limit is what you think it is (in the case when the limit exists). – Hans Lundmark Mar 24 '17 at 15:23
  • @HansLundmark: Letting $r\to 0$ with $\theta$ fixed is equivalent to approaching $(0,0)$ along a line $y=mx$, in that they give the same information. If you find two values of $\theta$ or $m$ for which this limit is zero, you know the two-variable limit does not exist. – Matthew Leingang Mar 24 '17 at 16:20
  • I'm distracted by other work at the moment, but those are the only situations I'm familiar with, where polar coordinates helps to answer the question in the negative. – Matthew Leingang Mar 24 '17 at 16:23
  • I have been taught that theta is always fixed -- so in a case where I find a limit to be $L$ I would have to then proceed to prove it formally by the epsilon delta definition? – Hugh Entwistle Mar 24 '17 at 22:01
  • @HughEntwistle epsilon/delta or a theorem that uses two-variable limits. For instance, the Squeeze Theorem is true for two-variable limits. This is useful when the expression can be factored into something that tends to zero, times something that is bounded. – Matthew Leingang Mar 25 '17 at 01:23