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I’m studying multivariable calculus.

When trying to solve a limit $\lim_{(x,y)\to(x_0, y_0)}f(x,y)$ with $f:\mathbb{R}^2\to\mathbb{R}$, our professor told us that, in some cases, the limit is more easily ”computed” by changing the coordinate system to polar coordinates (using $x=ρ\cosθ$ and $y=ρ\sinθ$) and solving the limit for $\lim_{ρ\to0}f(ρ\cosθ, ρ\sinθ)$. He also tells us to consider $\theta$ as a function of $\rho$, so $\theta=\theta(\rho)$.

  • note that I’ve always dealt just with limits $(x,y)\to(0,0)$, but I wrote it in general form because I don’t know if this changes something.

Now, the idea of changing system to polar coordinates is ok, but what bothers me is that I can’t find any underlying rigorous result/theorem/algebraic process or whatever that lets me say that $\theta$ is a function of $\rho$, which seems wrong to me because the “purpose” of the coordinates should be to be independent (am I wrong? maybe it’s just linearly independent?). Is it a “trick” to help us compute some limits to avoid the underlying theory? Note that, that lets me change a two-variables limit $(x, y)$ in a one-variable limit $\rho$ (which is particularly useful if all the terms with $\theta$ cancel in the limit), so it seems a crucial observation.

EDIT

Thanks to @bb_823 I saw a similar question, but still not exactly what I was asking for: here seems to be the underlying theory of doing multivariable limits in polar coordinates, and while the formulation in the link clarifies the whole question to me (“the value of the limit doesn’t depend from θ” and the uniformity), the formulation of my professor “assume that θ is function of ρ” still seems “wrong” and I don’t see how it’s equivalent to the linked answer.

selenio34
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  • @bb_823 thanks, I already saw that question but didn’t really see an answer. Here instead there’s something similar to what I was asking https://math.stackexchange.com/a/3136487/843088 but, while the formulation in the link seems reasonable to me (“the value of the limit doesn’t depend from $\theta$” and the uniformity), assuming that $\theta$ is function of $\rho$ still seems “wrong” and I don’t see how it’s equivalent to the linked answer. – selenio34 Dec 20 '23 at 07:41

1 Answers1

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Maybe this counterexample can help you:

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

Then you might think that switching to polar coordinates it results in $$\tag{*}\lim_{r\to 0}\frac{r^3\cos(\theta)\sin^2(\theta)}{r^2\cos^2(\theta)+r^4\sin^4(\theta)}=\lim_{r\to 0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}=0$$ As if $\cos(\theta)=0$ is zero and for $r\to 0$ is $0$ as well. Thus the limit seems to be $0$.

Seems: if instead you look at the path $x=y^2$ (along a parabola) and let $y\to 0$, then: $$\lim_{y\to0}\frac{y^2y^2}{(y^2)^2+y^4}=\lim_{y\to0}\frac{y^4}{2y^4}=\frac{1}{2}.$$

So what went wrong here?

The main problem is that by switching to polar coordinates and evalutating the limit (*) you assumed that $\theta$ would be fixed, which is the same as considering only straight paths towards the origin.

As the second limit shows, using non-straight paths results in different limit, so the limit does not exist.

In conclusion $\theta$ must be able to depend on $r$ (or viceversa) in order to allow all paths to the origin.

b00n heT
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  • I noticed just now that we actually can’t conclude anything about the value of $\theta$ from the equations of the change of coordinates $x=\rho \cos\theta$ $y=\rho \sin\theta$, so putting $(\rho, \theta)\to(0,0)$ would be simply wrong. But that means that assuming that $\theta$ is a function of $\rho$ is just something to remind ourselves that it’s not fixed but the limit is still in the only variable $\rho$, correct? – selenio34 Dec 20 '23 at 09:27
  • Also one last question, if you can help me I would be extremely grateful: what can we say about $\theta$ as a function of $\rho$? We can’t use any asymptotic equivalences or any single-variable calculus tool because we can’t assume anything about it, correct? So we can only use these tools on $\rho$-only expressions? – selenio34 Dec 20 '23 at 09:37
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    It's not "assuming that $\theta$ is a function of $\rho$" but $\theta$ is a function of $\rho$.

    There is no constraint on the function $\theta(\rho)$ as long as it is not unbounded as $\rho \to 0$: it can even be discontinuous. This is because the $\rho\to 0$ part in the polar coordinates makes it so that the limit will be zero, and this is the most important part.

    Please let me know if this answers your question or not... I am unsure if I read your question correctly

     – b00n heT Dec 20 '23 at 11:07
  • you perfectly answered my question about the constraints of $\theta$ as a function, and I got why the fact that $\theta$ is a function of $\rho$ helps us solve limits in polar coordinates. My problem is with the fact that $\theta$ must be a function of $\rho$ in order to allow all possible paths towards the origin: why can’t $\theta$ be an indipendent variable, like I think should be with my (little) knowledge of basis/coordinates from linear algebra (like $x$ and $y$ with each other)? If $\theta$ would be an indipendent variable it could assume all possible values for all possible values of… – selenio34 Dec 20 '23 at 11:51
  • …$\rho$, and with this we could achieve all possible paths. I acknowledge that seeing $\theta$ as an indipendent variable would mean that our limit in polar coordinates wouldn’t be in one variable anymore and thus it wouldn’t be useful, but to me $\theta$ is an indipendent variable/coordinate, otherwise I wouldn’t get how $(\rho, \theta)$ could represent a system of coordinates for $\mathbb{R}^2$. So, saying that $\theta$ is a function of $\rho$ still seems, to me, to come out of nowhere. – selenio34 Dec 20 '23 at 11:56
  • so, in order to summarize to avoid confusion, I have two problems: 1) why I must say that $\theta$ is function of $\rho$ to consider all the possible paths towards the origin 2) where does “$\theta$ is function of $\rho$” really come from rigorously speaking, because, to my limited knowledge, they should be indipendent variables in order to be a coordinate system of $\mathbb{R}^2$ – selenio34 Dec 20 '23 at 12:04