Given $f$ is strictly increasing, is $f$ bounded?

Question

Let $f \in C^1(\Bbb{R})$ such that $f'(x)>0$ always with $\lim_{x\to \infty} f'(x)=0$, then is $f$ bounded above?

Tried many functions but couldn't find any counter, all seems to be satisfying, not sure if the statement is true or not.

Answer 1

No, consider,

$$f(x) = \begin{cases}x-1 & x\leq 1,\\ \log x & x>1 \end{cases}$$

Answer 2

Try $\tanh(x)$, it fullfills all of your assumptions.

\edit This is a copunterexample: $f(x)=\sqrt{x}$ for $x\ge1$ and $f(x)=\frac{x}{2}+\frac{1}{2}$ for $x\le 1$

Answer 3

Here's a counterexample with no need to split into cases: $$ f(x) = \ln(\ln(1+e^x)) . $$