If $\lim_{x\to a}(f\circ g)(x)=L$, $\lim_{x\to a}g(x)=b$ and $g$ is 1-1 near $a$ then $\lim_{x\to b}f(x)=L$?

Question

The question is clear (every composition etc. is defined). I personally think this is not true. If this were true then the proof of $f$ is continuous $\Rightarrow$ $f^{-1}$ is continuous would be much simpler by noting that $f^{-1}(f(x))=x$. If it is not true then provide a counterexample. If it is true then provide a hint to the $\epsilon-\delta$ proof.

Answer 1

Take $X = (-\infty, 0] \cup \mathbb N \subset \mathbb R$ and $a = 0$. Now $$g(x) = \begin{cases} x &\text{ if } x\leq 0 \\ x^{-1} &\text{otherwise}\end{cases}$$ and $$f(x) = \begin{cases} x &\text{ if } x \leq 0 \\ 1 &\text{otherwise}\end{cases}$$ give a counterexample.

Note however if we assume that there is some $\epsilon > 0$ such that $A = (a- \epsilon, a+\epsilon) \subset X$ and $g\vert_A$ is continuous and injective, then the statement becomes true.