Simple Limit Question

I'm just getting used to limits and formal proofs.

Let g be a real-valued 1-1 function with domain (c − 1, c + 1). Let lim x→c g(x) = A. Prove or disprove: lim y→A g−1(y) = c.

I can't really articulate mathematically what I think, but I'm thinking yes it does have to be. I'm just thinking the first limit wouldn't exist at a discontinuity, and as it's one-to-one the inverse exists.

Any guidance is appreciated.

I'm just getting used to limits and formal proofs.

Let g be a real-valued 1-1 function with domain (c − 1, c + 1). Let lim x→c g(x) = A. Prove or disprove: lim y→A g−1(y) = c.

I can't really articulate mathematically what I think, but I'm thinking yes it does have to be. I'm just thinking the first limit wouldn't exist at a discontinuity, and as it's one-to-one the inverse exists.

Any guidance is appreciated.

Possibly I'm missing something, but the previous posts seem to imply the limit exists, but I think otherwise. (That is, I have a counterexample).

So the question is:
Let $g: (c-1, c+1) \to \mathbb{R}$ be injective, and denote $\displaystyle \lim_{x \to c} g(x) = A$. Is it true that $\displaystyle \lim_{y \to A} g^{-1}(y) = c$?
It looks like we have to assume that the first limit exists. Have you written out the epsilon-delta definitions and compared them? And what does injectivity allow you to do with respect to the inverse? (that is, does it allow you to give a left-inverse or right-inverse?)

I'm not sure exactly what you are allowed to assume, but there are two approaches that immediately present themselves.

One is to simply work through the epsilon-delta definitions to make the deduction, which is grueling but pretty simple.

you can improve it by applying some theorems, like the limit existing implies continuity, and then continuity implies etc. (

I was assuming the limit A exists, because otherwise the question doesn't make sense.

Although thinking more about the initial counter example I had in mind, there's a second problem: the question implicitly assumes that $\lim_{y \to A} g^{-1}(y)$ makes sense. That is, that there's a interval around A s.t. $g^{-1}$ is defined on that interval (possibly excluding the point A itself).

For my counterexample, that's not true. (You can fix it to get a counterexample where the second limit does make sense, but it's a little more work).

..

I think that question has to be read as saying that the inverse function *is* defined, so I don't think that this is a counterexample.

OK, I agree. That's quite nice. And I guess ghostwalker's comment earlier is rather relevant here.

I'm just getting used to limits and formal proofs.

Let g be a real-valued 1-1 function with domain (c − 1, c + 1). Let lim x→c g(x) = A. Prove or disprove: lim y→A g−1(y) = c.

I can't really articulate mathematically what I think, but I'm thinking yes it does have to be. I'm just thinking the first limit wouldn't exist at a discontinuity, and as it's one-to-one the inverse exists.

Any guidance is appreciated.