4
$\begingroup$

I'm currently facing the following problem, my math teacher told us that the following statement is true:

$\lim_{x\to(a,b)} f(x)=L \iff \forall \theta \lim_{r\to0} f(a+r\cos\theta, b+r\sin\theta)=L$

And in an attempt to better understand why this is true, I researched here for related questions and attempted to prove this myself. But I failed in finding any proof and my research led me to believe that the statement is actually false. Here is a link to one of the questions:

Limit $\frac{x^2y}{x^4+y^2}$ is found using polar coordinates but it is not supposed to exist.

In this question, he shows what I think is a counterexample to the statement above:

$$\lim_{(x,y)\to(0,0)}\frac{x^2y}{x^4+y^2}$$

in which using polar coordinates to find the limit returns $0$ regardless of the value of $\theta$. But by making $y=x^2$ the limit is equal to $1/2$. Therefore, the limit doesn't exists.

The intuitive explanation for this, given in the accepted answer is that using polar coordinates only checks the limit in the directions of straight lines. But that explanation doesn't satisfy me at all, I still can't explain why checking the limit in every straight direction isn't sufficient. I feel that if I could come up with counterexample on my own that would mean I truly understand.

So here are my first questions:

Could you build your own counterexamples? And how would/did you go about doing this?

Are there any counterexamples you know?

Am I really correct in assuming that the statement is actually false?


Since the polar limit checks in every straight direction, the following statement is true?

$\forall y\in\mathbb{R}^2\lim_{t\to0} f((a,b)+ty)=L \iff \forall \theta \lim_{r\to0} f(a+r\cos\theta, b+r\sin\theta)=L$

Don't know if that is particularly useful.

$\endgroup$
  • $\begingroup$ The statement is obviously false in general. You can approach the $(a,b)$ point along any curve, not just a straight line. As you show with your example $\endgroup$ – Yuriy S Oct 27 '18 at 23:33
  • $\begingroup$ Yeah, but why the curve in which you approach matters, can you help me in buildind a picture? Or could you, based on what you said build an example? $\endgroup$ – chris Oct 27 '18 at 23:43
  • $\begingroup$ The curve matters when the function takes different values in the neighborhood of $(a,b)$, no matter how small it is. Then we know the limit doesn't exist. In your polar form, fix some small $\rho$ and consider the dependence on $\theta$ $\endgroup$ – Yuriy S Oct 27 '18 at 23:48
  • $\begingroup$ If the limit exists, then it is the same along any curve, and so we can use the method your teacher suggested. In that case, the dependence on $\theta$ disappears for small $\rho$ $\endgroup$ – Yuriy S Oct 28 '18 at 0:06
  • $\begingroup$ Oh, now it seems trivial to picture what is going on, and to build a counterexample. Here is one, that have the same "flaw" as the given in the cited question $f(x,y) = \begin{cases} 1, & \text{if $y=x^2$} \\ 0, & \text{if $y\ne x^2$} \end{cases} $ But in this case is not so simple to prove that using polar the limit is queal to $0$ $\endgroup$ – chris Oct 28 '18 at 0:21
3
$\begingroup$

The correct statement is the following: $$ \lim_{(x, y) \to (a, b)} f(x, y) = L$$ if and only if there exists a function $g$ of one variable, such that $$ \left\lvert f(a + r \cos \theta, b + r \sin \theta) - L \right\rvert \leq g(r)$$ for all $\theta$ and $r$, and $\lim_{r \to 0} g(r) = 0$.

This is basically a squeeze law, and it is easiest to understand if we pick a particular value of $r$. Suppose we pick $r = 1/3$, then the inequality above is saying that within the disc of radius $1/3$ from $(a, b)$, the function $f$ is at most $g(1/3)$ different from the limit $L$. Since $g(r) \to 0$ as $r \to 0$, each disc gives progressively better bounds on $f$.

An application of this is the function $f(x, y) = \frac{x^2 - y^2}{\sqrt{x^2 + y^2}}$ for example. After changing to polar coordinates, we can see that $$\lvert f(r \cos \theta, r \sin \theta)\rvert = \lvert r(\cos^2 \theta - \sin^2 \theta) \rvert \leq 2 r$$ and hence we could take $g(r) = 2r$, $L = 0$, and the statement above shows that $f(x, y) \to 0$ as $(x, y) \to 0$. Note the really important thing here is that we "got rid" of the $\theta$ dependence, and we could do this here by simply bounding it. It's also good to visualise what is going on: we have shown that when plotted, $f$ has to live within a cone which has vertex at the origin. (Compare to the single-variable squeeze law).

$\endgroup$
  • $\begingroup$ Very nice, I was already thinking there were no correct statements. I wish my teacher showed us this one. $\endgroup$ – chris Oct 29 '18 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.