Why does evaluation of a two-variable limit fail when using polar coordinates?

Question

The definition of the limit of a two-variable function:

$\lim\limits_{(x,y)\to (a,b)}f(x,y)=L\,$ if and only if for all $\epsilon>0$ there exists a $\delta >0$ such that $$0<\sqrt{(x-a)^2+(y-b)^2}<\delta \implies |f(x,y)-L|<\epsilon$$

Consider the following proposition (I do realize that it is not true):

Let $f^*(r,\theta) := f(a+r\cos\theta,b+r\sin\theta)$. Then $$\lim\limits_{(x,y)\to(a,b)} f(x,y) = L \iff \lim\limits_{r\to0^+} f^*(r,\theta) = L$$

Proof.

1. Suppose that $\lim\limits_{(x,y)\to(a,b)} f(x,y) = L$. This means that for all $\epsilon>0$ there exists $\delta>0$ such that $$0<\sqrt{(x-a)^2+(y-b)^2}<\delta \implies |f(x,y)-L|<\epsilon$$ If we let $x=a+r\cos\theta$, $y=b+r\sin\theta$, then we get that $$0<\sqrt{r^2\cos^2\theta+r^2\sin^2\theta}=r<\delta \implies |f^*(r,\theta)-L|<\epsilon$$ Thus, by definition, $\lim\limits_{r\to0^+}f^*(r,\theta) = L$.

2. Now suppose that $\lim\limits_{r\to0^+}f^*(r,\theta) = L$. This means that for all $\epsilon>0$ there exists $\delta>0$ such that $$0<r<\delta \implies |f^*(r,\theta)-L|<\epsilon$$ Again, letting $x=a+r\cos\theta$, $y=b+r\sin\theta$, we get $$0<r=\sqrt{r^2\cos^2\theta+r^2\sin^2\theta}=\sqrt{(x-a)^2+(y-b)^2}<\delta \implies |f(x,y)-L|<\epsilon$$ Again, by definition, $\lim\limits_{(x,y)\to(a,b)} f(x,y) = L$. Q.E.D.

I am aware that the above proof can be done by directly proving the equivalence, but I didn't want to risk making it less clear that way.

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The problem

The proposition is incorrect, or so I am inclined to believe. Consider the following function:

$$f(x,y) = \frac{x^2y}{x^4+y^2}$$

and the following limit:

$$\lim\limits_{(x,y) \to (0,0)} f(x,y)$$

The function is taken from, and my question heavily relies on, this post.

Changing to polar coordinates, and after some arrangements, the limit of $f(x,y)$ as $(x,y) \to (0,0)$ is

$$\lim_{r \to 0^+} \frac{r (\cos^2\theta\sin\theta)}{r^2\cos^4\theta + \sin^2\theta}$$ Some answers to that question say that you have to be careful with that limit. My understanding is that this refers to "substituting $r$ with 0". I know that this is the wrong approach. But, I get that this limit is always 0, regardless of $\theta$. And I didn't find it very difficult to arrive at this conclusion.

I break it down into two cases.

1. First, I assume that $\sin\theta\neq0$. The numerator tends to 0 and the denominator tends to $\sin^2\theta$. Therefore, the limit is just $0/\sin^2\theta$, i.e. zero.

2. The second case is $\sin\theta = 0$. But, now the under-limit function is identically zero for all $r\neq0$, yielding a limit that is zero.

This proves that the limit is zero. Have I done something wrong?

One answer of the above mentioned post says that, when looking for this limit, you have to analyze the case where $\theta$ is a function of $r$. Why?

With the limit being zero regardless of $\theta$, my proposition would imply that the limit of $f(x,y)$ is 0. Yet, I know that this is not the case, because if I let $y=x^2$, the "limit" evaluates to $\frac{1}{2}$. I have always been told that for a limit to exist, it needs to be the same for every path you approach the limit point on. I've always taken that for granted, and it made intuitive sense to me. But now, thinking more deeply about it, I don't really know why that is. This also has to do with the fact that this comes up nowhere in the proof of my proposition.

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My thoughts

My proof relies on (or so I think) the fact that every point $(x,y) \in \mathbb R$ is representable in polar form as the pair $(r,\theta)$ and that this representation is unique if we restrict $\theta \in [0, 2\pi)$. Is this correct?

As far as I can see $r$ and $\theta$ can be independent variables. I cannot figure out why one would need to allow for $\theta$ to be a function of $r$.

I also rely on the fact that $\sqrt{(x-a)^2+(x-b)^2}$ and $r$ are always equal. Am I missing something?

I will sum up my questions:

1. How does the definition of the limit imply that all paths of approach yield the same result? I am looking for an intuitive explanation.

2. What is wrong with my proposition/proof? How does the fact that it fails for the function $f(x,y)$ and the path $(x,x^2)$, relate to the proof of my proposition. In other words, can you pinpoint exactly where the proof fails?

3. Where does the notion to let $\theta = \theta(r)$ (or even $r=r(\theta)$?) come from? I suppose that this is closely related to question 1.

4. When can/should I use polar coordinates to prove that a limit exists?

Thank you for your patience.

Answer 1

What you showed was that for each fixed $\theta\in [0,2\pi),$ the limit

$$\tag 1 \lim_{r\to 0^+} f(r\cos \theta, r\sin \theta) = 0.$$

A more informal way to say this: "$f$ has limit $0$ at the origin along every ray emanating from $(0,0).$"

THAT DOES NOT IMPLY $\,\lim_{(x,y)\to (0,0)} f(x,y) = 0.$ And it's crucial for you to understand this; otherwise limits in higher dimensions will be this weird vague mystery that forever makes you uneasy.

Let's look at $(1)$ more closely. In that process you are fixing $\theta$ and then letting $r\to 0^+.$ Yes, you get $0$ for the limit for every fixed $\theta,$ but that doesn't cut it. The definition of a limit as $(x,y)\to (0,0)$ doesn't involve looking only at rays, right? What about letting $(x,y)\to (0,0)$ when $(x,y)$ is on the parabola $y=x^2?$ Why would that be covered by the special case of rays? For the existence of a limit in this setting, the main idea is that when $(x,y)$ gets close to $(0,0),$ in any way whatsoever, $f(x,y)$ gets close to a limit $L.$

In fact in your problem we get

$$f(x,x^2) = \frac{x^2\cdot x^2}{x^4 + (x^2)^2} = \frac{1}{2}.$$

So $f$ is simply equal to $1/2$ at every point of the parabola $y=x^2$ where $x\ne 0.$ Since points on this parabola can be made as close to $(0,0)$ as we like, $f$ cannot have limit $0$ as $(x,y)\to (0,0).$ In fact $\lim_{(x,y)\to (0,0)} f(x,y)$ fails to exist (we get $1/2$ along the parabola and $0$ along each ray).

I think this specific problem is easier in rectangular coordinates, but if you want to think about it in polar coordinates, consider traveling to $(0,0)$ along the curve $\sin \theta = r$ as $\theta \to 0^+.$ Along this curve, which looks a lot like the parabola discussed above, we have

$$f(r\cos t, r\sin t) = \frac{r^2\cos^2 \theta\cdot r^2}{r^4\cos^4 +r^4} = \frac{\cos^2 \theta}{\cos^4\theta +1}.$$

As $\theta \to 0^+,$ the right side $\to 1/2,$ so again we see the limit of $f$ at the origin fails to exist.