Let $f:(a,b) \rightarrow \mathbb{R}$ be a one to one function. If $x_0$ is a point of the open interval $(a,b)$ such that $\lim_{x \rightarrow x_0} f(x) = l$, is it necessary that $\lim_{x \rightarrow l} f^{-1}(x) = x_0$?
My guess is no. I have found out pretty interesting things for cases where $x_0$ is either $a$ or $b$ or where the domain is not an interval, but i can't arrive to a final conclusion about the interval case. I am asking about general functions, the only nice property I assume is that the domain is an interval, i.e. I don't expect continuity or even boundedness.
So, if this holds, I need a proof and possibly a reference to some textbook. If this doesn't, I want a counterexample. Thanks a lot in advance!
Ok, so this is pretty horrid, but define $f:(-1,1)\rightarrow \mathbb{R}$ in the following way:
$f(x) = x$ for $x \leq 0$.
For $\frac{1}{2}\geq x >0$, $f(x) = x + \frac{1}{2^n}$, where $\frac{1}{2^n}$ is smallest such that $\frac{1}{2^n} \geq x$. This leaves a bunch of "gaps" near zero which are not in the image of $f$ yet.
For $\frac{1}{2} < x < 1$, $f(x) = x -\frac{1}{2}$ if $x - \frac{1}{2}$ is not already in the range of $f$.
There are continuum many remaining values of $x$ (where $f(x)$ is not yet defined), biject them with the rest of $\mathbb{R}$ in any way you like.
Then $\lim_{x\rightarrow 0} f(x) = 0$, but $\lim_{x\rightarrow 0}f^{-1}(x)$ does not exist as, for any $\delta>0$, $f^{-1}(x)$ takes on values arbitrarily close to $0$ and to $\frac{1}{2}$ for $x \in (-\delta, \delta)$.