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.
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 $\lim_{x\to b}f(x)=L$?
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limits
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0But $g(x) = \lvert x\lvert$ is not 1-1 near $0$. – 2012-11-13
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0You are right... – 2012-11-13
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0What is the (co-)domain of $f$ and $g$? – 2012-11-13
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0@AlexanderThumm $g:X\to \mathbb{R}$, $f:g(X)\to \mathbb{R}$ and $X\subseteq \mathbb{R}$ – 2012-11-13
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0Is $f$ or $g$ assumed to be continuous? Is $X \subset \mathbb R$ arbitrary? – 2012-11-13
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0$X$ is arbitrary and $f,g$ can possibly have discontinuities – 2012-11-13
1 Answers
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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.
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0Yes, it does. Trivially. – 2012-11-13
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0$\mathbb{N}$ has no accumulation points (save for $+\infty$)... – 2012-11-13
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0Yes but every eventually constant sequence has a limit. – 2012-11-13
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0But isn't then $\lim_{x\to 0} f(x) = 0$ ? – 2012-11-13
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0No, because the sequence $1/n$ converges to $0$ and lies in $g(X)$, but $f(1/n) = 1$. – 2012-11-13
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0Could you state your definition of limit with $\epsilon$ and $\delta$? – 2012-11-13
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0Why does the limit of $f\circ g$ exist?... – 2012-11-13
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0I edited my answer as to not only include the trivial limits. – 2012-11-13
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0@Thomas: you can find this in any textbook. Note that any map from a discrete space is continuous. – 2012-11-13
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0Seems good. Thanks! – 2012-11-13
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0@AlexanderThumm: I still don't see how $\lim_{x \to 0} g(x) = 0$ but $\lim_{x\to 0}f(x) \neq 0$. – 2012-11-13
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0@Thomas $f$ is defined on $\mathbb{R}$ while $g$ is defined in $X$ which includes $\mathbb{N}$ Taking $\delta<1$ exludes all psotive integers and allow as to take $\lim_{x\to 0}g(x)=\lim_{x\to 0^-}g(x)$ – 2012-11-13
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0Almost correct. You said $f$ is defined on $g(X)$, hence the yoga in the definition of $g$. – 2012-11-13
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0@AlexanderThumm Right. $f$ is defined in $(-\infty,0]\cup \left\{\frac{1}{n}:n\in\mathbb{N}^*\right\}$ One has to wonder if non trivial counter examples like yours exist – 2012-11-13
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0Yes they do. Let $g$ be the identity on $(-\infty, 0]$, send the intervals of the form $(1/(n+1),1/n]$ (with $n$ a natural number) to $(1/(2n+1),1/(2n)]$ by translation and scaling and use the rest of $\mathbb R$ to "fill the holes". Now $g(X) = X = \mathbb R$ and $g$ is injective in a neighborhood of $0$ (we can even construct $g$ to be injective everywhere). By construction $g$ is even continuous at $0$ but we have plenty of room to "mess around" with $f$ at $0$ such that $g$ does not notice. – 2012-11-13
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0Well at least $g$ restricted to some neighborhood of $0$ does not notice :D – 2012-11-13