Question: > 
> Let $f:\mathbb{R} \to \mathbb{R}$ be a strictly increasing bounded function of class $C^1$. Prove that there exists a sequence $\{x\_n\}\_n$ of real numbers such that $x\_n\to\infty$ and $\lim\_{n \to \infty} f'(x\_n)=0$.
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> Then construct a strictly increasing bounded function $f:\mathbb{R} \to \mathbb{R}$ of class $C^1$ such that the $\lim\_{x \to \infty} f' (x)$ does not exist.
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I know that if we assume $a\_n$ increases ($a\_{n+1}\ge a\_n$), then either it is bounded, or not. If yes, then it converges to the $\sup a\_n$, else it goes to $+\infty$. Since it is bounded here it goes to a limit and so eventually it approaches the limit and derivative approaches 0. But I dont really know how to prove it rigorously and construct such a function such that limit DNE.

Answer: 
By the Mean Value Theorem, for any $n\in\mathbb{N}$ there is $x\_n\in (n,n+1)$ such that
$$f'(x\_n)=f(n+1)-f(n).$$
The sequence $(x\_n)\_n$ is strictly increasing and goes to $+\infty$. Then $$\lim\_{n\to \infty }f'(x\_n)=\lim\_{n\to \infty }(f(n+1)-f(n))=\lim\_{n\to \infty}f(n+1)-\lim\_{n\to \infty }f(n)=M-M=0$$
where $M=\sup\{f(x):x \in \Bbb{R}\}$.

Hint for the second part. Consider the continuous piecewise function $f$ defined in $\mathbb{R}$ which is $e^{x}$ for $x\in (-\infty,0]$, it is linear with slope $1$ in each interval $[n,n+\frac{1}{2^n}]$ and it is linear with slope $\frac{1}{2^n}$ in each interval $[n+\frac{1}{2^n},n+1]$ for any $n\in\mathbb{N}$.
The function $f$ is strictly increasing and bounded (why?). Moreover, $f'(x)$ attains the value $1$ and values $<1/2$ infinite times as $x$ goes to infinity, and therefore $\lim\_{x \to \infty} f' (x)$ does not exist.
Note that $f$ is not $C^1$, but it can be made smooth by changing it suitably around the joint points.
An increasing and bounded from above function $f:\Bbb{R} \to \Bbb{R}$ has always a limit as $x \to +\infty$ which is $L=\sup\{f(x):x \in \Bbb{R}\}$