Real Analysis: Continuity of a Composition Function 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 $|f(g(x))-f(g(a))|<\epsilon$.
 A: 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 $f(g(a))$.


*

*Let $a_n$ be any convergent sequence such that $a_n\to a$.

* Since $g(x)$ is continuous, $g(a_n)\to g(a)$ as $a_n\to a$.

* Since $f(x)$ is continuous, $f(g(a_n))\to f(g(a))$ as $a_n\to a$ as required.

A: 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.
A: 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.
A: 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’)|< .///
