Limit of composite function where the outer function is not continuous

Question

$\lim_{x\to a}g(f(x))=g(\lim_{x\to a}f(x)) $

My question is what happens when $g$ is not continuous say $\lim_{x\to a}f(x)= 3$ and

$g(x) =\begin{cases} x\space, x\neq 3 \\ 5\space, x=3\end{cases}$

will $g$ be $3$ or $5$?

Answer 1

That very much depends on $f$ and $g$. If for example $f(x)=x$, then $$g(f(x))=\left\{\begin{array}{c} x,\ x\neq 3\\ 5,\ x=3\end{array}\right.$$ Hence you limit may behave like this: $$\lim_{x\rightarrow 3, x\neq 3}g(f(x))=3.$$ On the other hand if $f(x)=3$ for all $x$, you get $g(f(x))=5$ for all $x$. Hence $$\lim_{x\rightarrow 3}g(f(x))=5.$$

It is even possible to construct $g$, such that no limit even exists. For example $$g(x)=\left\{\begin{array}{c} \sin(\frac{1}{x}),\ x\neq 0\\ 0,\ x=0\end{array}\right.$$ This $g$ is not continuous at $x=0$ and $$\lim_{x\rightarrow 0} g(x)$$ does not even exist. If you now set $f(x)=x$ your composite function does not have a limit for $x\rightarrow 0$ as well.

To see that this $g$ is not continuous, just examine the following two sequences: $$a_k=\frac{1}{2\pi k}$$ $$b_k=\frac{1}{2\pi k + \frac{\pi}{2}}$$ Then $g(a_k)=0$ and $g(b_k)=1$, but $a_k,b_k\rightarrow 0$ for $k\rightarrow \infty$