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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 $|f(g(x))-f(g(a))|<\epsilon$.

dfeuer
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Mr.Fry
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  • 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. – Devin Murray Oct 16 '13 at 01:03
  • 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. – Devin Murray Oct 16 '13 at 01:07

4 Answers4

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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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    Given that $f$ is continuous at $g(a)$, why are you allowed to substitute $g(x)$ for $x$ in the statement $0\lt |x-g(a)|\lt \delta_1 \implies |f(x)-f(g(a))|\lt \epsilon$? Shouldn't you first specify that $x$ is in the domain of $f\circ g$ first? Otherwise, we cannot know if $g(x)$ or $f(g(x))$ are defined. – Display name Feb 19 '24 at 02:14
  • You are right. To be precise, it should be $|y-g(a)|<\delta_1\Rightarrow|f(y)-f(g(a))|<\varepsilon$. Then, the proof is correct. – user408858 Mar 27 '24 at 21:19
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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.
Alex S
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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 > 0, we have, 0<|-a|< g =>|g()-g(a)|< , 0<|x’-a’|< f =>|f(x’)-f(a’)|< . If we choose = g•f, then |(f•g)(x)-(f•g)(a)|=|(())−(())|< |f(x’)-f(a’)|< .///

Dr T
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If $f \circ g$ 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.

obr
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tattwamasi amrutam
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