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I am currently teaching a multivariable calculus class and trying to come up with problems for the final exam, in particular a problem about multivariable limits. The following question popped into my head:

Let $f(x,y) = \frac{P(x,y)}{Q(x,y)}$ be a rational function ($P$ and $Q$ are polynomials) with $P(0,0) = Q(0,0) = 0$. For different curves $C: [0,1] \to \mathbb{R}^2$ with $C(0) = (0,0)$, the limits $\lim\limits_{t \to 0} f(C(t))$ may depend on $C$, or might not even exist.

What can we say about the set of (extended) real numbers $L$ which are limits of $f$ along some curve $C$? My gut instinct is that this set should be a connected subset of the extended real line, but I haven't been able to prove it.

Comments:

  1. If it makes the problem easier, feel free to assume that $Q \neq 0$ in a neighborhood of the origin. I don't necessarily care about the most general result, I'm just interested in any kind of result.
  2. I don't think it should matter, but let's let our curves be any continuous curves. Would the answer change if we forced our curves to be $C^1$ or $C^\infty$? Or algebraic?
  3. I only asked that $f$ is a rational function because those are the kinds of functions we tend to give students in multivariable calculus classes -- also, my first attempt at this was to rewrite $\frac{P}{Q} = k$ as $P - kQ = 0$ and try to think about the geometry of that as a polynomial equation in three variables. But I suspect that whatever we could say about rational functions we could also say about functions which are just continuous away from the origin.

My two attempts:

  1. Rewrite $\frac{P}{Q} = k$ as $P - kQ = 0$ which is now just a polynomial equation. I thought at first that maybe something along the lines of "for most values of $L$ which are the limit along some curve, there is in fact a curve to $(0,0)$ along which the function is a constant $L$" but I'm not sure if that's actually true.
  2. Suppose that $A$ and $B$ are curves along which $f$ has limits $a$ and $b$ respectively, and let $c$ be between $a$ and $b$. Reparametrize them so that $|A(t)| = |B(t)|$ for all $t$. For each $t$, look at the circular arc between $A(t)$ and $B(t)$. I want to use the intermediate value theorem to pick points, for each $t$, along this arc which are close to $c$. (The function is continuous and there are points near the origin along $A$ which are close to $a$ and along $B$ which are close to $b$ so there should be points in between them which are close to $c$.) But it's not clear to me that we can choose such points $C(t)$ for each $t$ in a way that makes the resulting function $t \mapsto C(t)$ continuous.
user263190
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  • It is meaningless to say that $P(0,0)=Q(0,0)=0$ since if this is the case, the function is simply undefined at the origin. You want just the denominator to be zero. Consider some function like $\overbar{z}/z$ from complex analysis, which has a nonexistent limit from different directions. Also, if a limit has multiple values, then it does not exist (because limits are unique) so the phrasing of your question should be edited. – user404974 May 07 '21 at 04:04
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    I think you have severely misunderstood my question. I'm considering functions like $f(x,y) = \frac{x^2y}{x^4 + y^2}$. The overall limit doesn't exist as $(x,y) \to (0,0)$; that's the whole point. But along different curves, the limit will exist. If only the denominator is zero, then the limit along every curve will be $\pm \infty$. We need the numerator to be zero as well in order to get something interesting. – user263190 May 07 '21 at 04:11
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    @Exponent I think you misread OP's question, it makes sense to me. Given the conditions $P(0)=Q(0)=0$, is the set of all possible limits from all possible paths connected? And we should assume this is an isolated singularity as well. – Ninad Munshi May 07 '21 at 04:13
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    Possibly relevant: Discontinuities and smooth curves in n-space by Peter J. Nyikos AND google search for "curvilinear cluster set" AND my answer to Name for multi-valued analogue of a limit AND my 4 June 2005 sci.math post Continuity in each variable vs. joint continuity. – Dave L. Renfro May 07 '21 at 06:24
  • This problem might be more suitable for your students, unless their ability is well beyond any I've ever taught at this level (3rd semester of a US style elementary calculus sequence). – Dave L. Renfro May 07 '21 at 06:26
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    @DaveL.Renfro Thank you for the references! I will look into them this afternoon. And don't worry -- this problem was not going to make it onto my exam! It just popped into my head as I was walking home and trying to come up with some more traditional examples. – user263190 May 07 '21 at 15:17

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