Given $f$ is strictly increasing, is $f$ bounded? Let $f \in C^1(\Bbb{R})$ such that $f'(x)&gt;0$ 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.
No, consider,

$$f(x) = \begin{cases}x-1 &amp; x\leq 1,\\ \log x &amp; x&gt;1 \end{cases}$$
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 $x\le 1$