Let $g$ be a continuous function at a point $a \in \mathbb{R}$, and let
$$\lim_{x\rightarrow g(a)} f(x) = L$$
Show that
$$\lim_{x\rightarrow a} (f\circ g)(x) = \lim_{x\rightarrow g(a)} f(x)$$.
How do I prove this in the case that $f$ is not continuous. My current proof is incorrect since it assumes $f$ is continuous.
Let $\lim_{u\rightarrow g(a)} f(u) = L$. Also, since $g$ is continuous, $\lim_{x\rightarrow a} g(x) = g(a)$. If $\lim_{x\rightarrow a} (f\circ g)(x) = L$, then for all $\epsilon > 0$ there exists a $\delta > 0$ such that $0<|x – a|< \delta$ implies that $|(f\circ g)(x) – L | < \epsilon$.\
Now we will use the continuity of $g$ to prove the desired result. Since $\lim_{x\rightarrow a} g(x) = g(a)$, for all $\delta_1 > 0$ there exists a $\delta > 0$ such that $0<|x – a|< \delta$ implies that $|g(x) – g(a) | < \delta_1$.\
Since $\lim_{u\rightarrow g(a)} f(u) = L$, for all $\epsilon > 0$ there exists a $\delta_1 > 0$ such that $0<|u – g(a)|< \delta_1$ implies that $|f(u) – L | < \epsilon $.\
Now if we let $u = g(x)$, then we achieve the desired result since a $\delta$ and a $\delta_1$ can be chosen such that, $0 < |x – a| < \delta$ implies that $0 < |g(x) – g(a)|< \delta_1$, which finally implies that $|(f\circ g)(x) – L | < \epsilon$.\
Therefore
$$
\lim_{x\rightarrow a} (f\circ g)(x) = \lim_{x \rightarrow g(a)} f(x).
$$
Best Answer
Always use the standard theorem for limit of composite functions:
Theorem: If $\lim\limits_{x \to a}g(x) = b$ and $g(x) \neq b$ in a certain deleted neighborhood of $a$ and $\lim\limits_{x \to b}f(x) = L$ then $$\lim_{x \to a}f(g(x)) = L$$
The result you have mentioned is false (for example when $g$ is a constant function say $g(x) = k$ and $f$ is not defined at $k$). A more complicated example of the failure of your result is when $g(0) = 0, g(x) = x\sin(1/x), x\neq 0$ and $f(x) = \dfrac{\sin x}{x}$ and $a = 0$.
Note that the above theorem is very powerful and is the basis of all substitutions used in evaluation of limits (for example replacing $x \to a$ with $x = a + h$ and $h \to 0$).
Also see this answer and related discussion.