# 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 $\lim_{x\to b}f(x)=L$?

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.

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But $g(x) = \lvert x\lvert$ is not 1-1 near $0$. – Thomas Nov 13 '12 at 19:08
You are right... – noone1 Nov 13 '12 at 19:09
What is the (co-)domain of $f$ and $g$? – Alexander Thumm Nov 13 '12 at 19:19
@AlexanderThumm $g:X\to \mathbb{R}$, $f:g(X)\to \mathbb{R}$ and $X\subseteq \mathbb{R}$ – noone1 Nov 13 '12 at 19:21
Is $f$ or $g$ assumed to be continuous? Is $X \subset \mathbb R$ arbitrary? – Alexander Thumm Nov 13 '12 at 19:23

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.
$\mathbb{N}$ has no accumulation points (save for $+\infty$)... – noone1 Nov 13 '12 at 19:37
But isn't then $\lim_{x\to 0} f(x) = 0$ ? – Thomas Nov 13 '12 at 19:39
No, because the sequence $1/n$ converges to $0$ and lies in $g(X)$, but $f(1/n) = 1$. – Alexander Thumm Nov 13 '12 at 19:41