If $\lim\limits_{x\to a^+} f(x)=L$ and if $c$ is a function such that $a < c(x) < x$ for all $x > a$, then $\lim\limits_{x\to a^+} f(c(x))=L$. Note: there has been no discussion about continuity or any discussion about the limits of composition functions at this point. Any help would be appreciated!
Composition of Functions Proof Help
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real-analysis
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0For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latex). – 2012-10-10
1 Answers
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Let $\epsilon>0$. Then there exists a $\delta>0$ such that
$|f(x)-L|<\epsilon$ whenever $0
$|f(c(x))-L|<\epsilon$
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0I was definitel$y$ over-anal$y$zing the problem. Your explanation makes a lot of sense.Thanks – 2012-10-10