Limit of a composition function exists, but limit of the outer function does not exists

Question

Limit of a composition function exists, but limit of the outer function does not exists

Given:

1. $f,g$ are real functions
2. $\lim_{x \rightarrow x_0}g(x)=u_0$
3. $\exists \delta > 0, \forall x \in \mathring U(x_0, \delta), g(x) \neq u_0$
4. $\lim_{u \rightarrow u_0} f(u)$ does not exist.

Is it possible for the limit $\lim_{x \rightarrow x_0} f(g(x))$ exists?

This question is motivated by the possibility that, even if the limit of $f$ does not exist in general, there may be some particular way that $g$ "enables" the existence of the limit of $f\circ g$.

I tried some functions like $f(x)=\sin \frac{1}{x}$. We know that $\lim_{x \rightarrow 0} f(x)$ dose not exists. But I can't find the appropriate $g$.

Answer 1

The counterexamples depend on what kind problems you want to find with the limit of $f$. If you are allowing that $\lim_{x\to u_0^+}f(x)$ and $\lim_{x\to u_0^-}f(x)$ exist but are not equal, then the following example should work:

Consider $f(x)= \frac{\vert x\vert}{x}$ and $g(x)=\vert x\vert$ at $x_0=0$, then this holds with $u_0=0$ and $\underset{u\to u_0}{\lim} f(u)=1$.

A somewhat more complicated example would be $f(x)=\chi_{\mathbb{Q}}(x)$ and

$$ g(x)=\begin{cases} 1 &; \vert x\vert\geq 1\\ \frac{1}{n} & ; \frac{1}{n+1} \leq\vert x\vert \leq \frac{1}{n} \\ 0 &;x=0 \end{cases} $$

would work for $x_0=0$, $u_0=0$ and $\underset{u\to u_0}{\lim} f(u)=1$.

Answer 2

It is possible. Take $$f : \mathbb R \to \mathbb R, f(x) = \begin{cases} 1 & x > 0 \\ 0 & x \le 0 \end{cases} $$ $$g : \mathbb R \to \mathbb R, g(x) = - \lvert x \rvert $$ For $x_0 = 0$ and $u_0 = 0$ we obtain an example as requested. The functon $g$ is even continuous on $\mathbb R$.

For $$f(x) = \begin{cases} \sin \frac 1 x & x \ne 0 \\ 0 & x = 0 \\\end{cases}$$ we can take $$g(x) = \begin{cases} \frac{1}{n\pi} & \frac{1}{n+1} \le \lvert x \rvert \le \frac 1 n \\ 0 & \text{else} \\\end{cases}$$