The limit of the derivative of an increasing and bounded function is always $0$?

No. Let $h\colon\mathbb R\to\mathbb R$ be a smooth function with $h(x)>0$ for $x\in(0,1)$ and $h(x)=0$ otherwise (a smooth "bump"). Then let $$g(x)=\sum_{n=1}^\infty h(n^2(x-n))=\begin{cases}h(n^2(x-n))&\text{if }n\le x<n+1, n\in\mathbb N\\0&\text{otherwise}\end{cases}$$ And let $f(x)=\int_0^xg(t)\,\mathrm dt$. Then

• $f$ is smooth because in each bounded interval only finitely summands are nonzero
• $\lim_{x\to\infty}f'(x)=\lim_{x\to\infty}g(x)$ does not exist
• $f$ is increasing because $f'\ge 0$
• $f$ is bounded by $\sum_{n=1}^\infty\frac1{n^2}\int_0^1h(t)\,\mathrm dt<\infty$

Hint: One way to look at this is by the Fundamental theorem of Calculus:

$$f(b) - f(a) = \int_a^b f'(x) dx\ .$$

As the function $f$ is bounded we have

$$\int_a^\infty f'(x) dx <\infty$$

Does that imply that $f'$ tends to $0$?