If $f(u(x))$ is continuous, then so is $u(x)$?

Question

I have read some of the other claims regarding the fact,

If $f(u(x))$ is continuous, then $$\lim_{x \rightarrow x_0} f(u(x)) = f(\lim_{x \rightarrow x_0} u(x))$$

I am not familiar with real analysis and the proofs do not make sense to me so I would like a more simple explanation.

So far I know that if $f(u(x))$ is continuous, by definition

$$\lim_{x \rightarrow x_0} f(u(x)) = f(u(x_0))$$

So in order for the first equation above to be true I would like to say

$$f(\lim_{x \rightarrow x_0} u(x))=f(u(x_0))$$

but this would only hold when $u(x)$ is continuous.

Therefore, I am thinking that somehow the continuity of $f(u(x))$ implies that $u(x)$ will also be continuous... but I cannot prove that.

May I have some help, please?

$$\space$$ Added:

Thank you for your response, everyone.

What I really am trying to do is to know the condition where

$$\lim_{x \rightarrow x_0} f(u(x)) = f(\lim_{x \rightarrow x_0} u(x))$$

Particularly, why

$$\lim_{x \rightarrow x_0} \ln{u(x)} = \ln{ \lim_{x \rightarrow x_0} u(x)}$$

I did not think that my claim would be incorrect because when I looked it up on google it was the first thing that popped up as if it were true.

It sucks that there are false facts on the internet D:

Answer 1

Continuity of $f(u(x))$ does not imply continuity of $u(x)$. For example, if $u(x)=1$ for $x$ rational and $-1$ for $x$ irrational and $f(x)=x^{2}$ then $f(u(x))$ is continuous but $u$ is 'badly discontinuous'.

Answer 2

It is definitely not true.

Let $f$ be a constant function.

Then if composition $f\circ u$ exists then it will be constant hence continuous for every function $u$.

So it would lead to the (false) conclusion that every function is continuous.

Answer 3

First of all, continuity of $f(u(x))$ only means that $\lim_{x\to x_0}f(u(x))=f(u(x_0))$, while your statement is instead about the continuity of $f$. If $f\circ g$ is continuous, we cannot say anything about $f,g$, as this example shows: let $\chi_{\mathbb{Q}}$ be the indicator function of the rationals, then $\chi_{\mathbb{Q}}\circ \chi_{\mathbb{Q}}=1$ is continuous, but $\chi_{\mathbb{Q}}$ isn't. On the other and, if you assume that both $f\circ g$ and $f$ are continuous, you can obtain more interesting results

Answer 4

In context:

Let $f$ be continuos in $\mathbb{R}.$

1)$ \lim_{x \rightarrow x_0} u(x)= L,$ i.e. the limit exists and $L \in \mathbb{R}$.

Then

$\lim_{x \rightarrow x_0} f(u(x))=f(\lim_{x \rightarrow x_0}u(x))= f(L)$.

2) If $u(x)$ is continuos we have

$\lim_{x \rightarrow x_0} u(x)=L=u(x_0)$, and

$\lim_{x_ \rightarrow x_0} f(u(x))=f(u(x_0))$.