Real Analysis: Continuity of a Composition Function

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.

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.