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Suppose $f$ and $g$ are functions such that $g$ is continuous at $a$, and $f$ is continuous at $g(a)$. Show the composition $f(g(x))$ is continuous at $a$.

My idea: Can I go straight from definition and take $\delta=\min\{\delta_1,\delta_2\}$, where $\delta_1$ is used for the continuity of $g$ at $a$ and $\delta_2$ is used for f being continuous at $g(a)$. In my proof I just treat $g(a)$ as a point when referring to the composition. So it goes like this:

Proof: Given $\epsilon>0$, take $\delta=\min\{\delta_1,\delta_2\}$. Then $0<|x-g(a)|<\delta$ which implies $|f(g(x))-f(g(a))|<\epsilon$.

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  • $\begingroup$ What you just wrote isn't quite correct. From your $\delta$ you get that $|f(x) - f(g(a))| < \epsilon$ which isn't what you want. $\endgroup$ – Devin Murray Oct 16 '13 at 1:03
  • $\begingroup$ It's also unclear what your $\delta_1$ and $\delta_2$ are referring to. Once you clear up that confusion to yourself it may be a bit more obvious to you how to complete your proof. $\endgroup$ – Devin Murray Oct 16 '13 at 1:07
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Since $f$ is continuous at $g(a)$, our definition of continuity tells us that for all $\varepsilon > 0$ there is some $\delta_1$ such that $$|g(x) - g(a)| < \delta_1\implies|f(g(x))-f(g(a))|<\varepsilon.$$ Also, since $g$ is continuous at $a$, there is some $\delta$ such that $$|x-a|<\delta \implies |g(x)-g(a)|<\delta_1.$$ I've taken $\varepsilon =\delta_1$ here. Now this tells us that for all $\varepsilon > 0$ there is some $\delta > 0$ (and a $\delta_1 > 0$) such that $$|x-a| < \delta\implies|g(x)-g(a)|<\delta_1\implies|f(g(x)) - f(g(a))|<\varepsilon,$$ which is what we wanted to show.

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Proof

It will be shown that the limit of $f(g(x))$ at any arbitrary point $x=a$ in the domain of $f(g(x))$ is equal to $f(g(a))$.

  1. Let $a_n$ be any convergent sequence such that $a_n\to a$.
  2. Since $g(x)$ is continuous, $g(a_n)\to g(a)$ as $a_n\to a$.
  3. Since $f(x)$ is continuous, $f(g(a_n))\to f(g(a))$ as $a_n\to a$ as required.
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If $fog$ is well defined then without any loss of generality let's assume that $f:A\to R $, and $g:B \to R$ such that $f(A) \subset B$. Also $A$ and $B$ are $\subset R$.

Now since $g$ is continuous at $a$, given $\epsilon >0$,there exists a $δ >0$, such that whenever $0<|x−a|<δ$, we have $|f(x)−f(a)|<ϵ$. Now choose $\delta_1 >0$ such that whenever $0<|y−a|<δ_1$ $\implies |g(y)−g(a)|< \delta$.

Then we are done.

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Continuous Compositions: Thm: The composite denoted by, “•”, of two continuous functions is continuous.

Pf/Let f:A—>B and g:C—>D be continuous functions : g(A) C A. Then for every 𝜖 > 0, we have, 0<|𝑥-a|< 𝛿g =>|g(𝑥)-g(a)|< 𝜖, 0<|x’-a’|< 𝛿f =>|f(x’)-f(a’)|< 𝜖. If we choose 𝛿 = 𝛿g•𝛿f, then |(f•g)(x)-(f•g)(a)|=|𝑓(𝑔(𝑥))−𝑓(𝑔(𝑎))|< |f(x’)-f(a’)|< 𝜖.///

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    $\begingroup$ Welcome to Math.SE! Regarding your answer, please use MathJax so that your answer is easier to read. $\endgroup$ – Ertxiem - reinstate Monica Mar 31 '19 at 18:38

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