Does finiteness of $\lim\limits_{x\to\infty}f(x)$ and $\lim\limits_{x\to\infty}f'(x)$ imply $\lim\limits_{x\to\infty}f'(x)=0$?

You are assuming that the limit of the derivative exists. Say $$\lim_{x\to\infty}f'(x) = L.$$

Case 1. $L\gt 0$. Then there exists $M\gt 0$ such that for all $x\geq M$, $|f'(x) - L|\lt \frac{L}{2}$. Therefore, for all $x\geq M$, $$\frac{L}{2} \lt f'(x) \lt \frac{3L}{2}.$$ In particular, $f$ is increasing on $[M,\infty)$.

By the Mean Value Theorem, for each natural number $n$ there exists a $c$ (which depends on $n$), $M+n \lt c\lt M+n+1$ such that $$f(M+n+1)-f(M+n) = \frac{f'(c)}{(M+n+1)-(M-n)} = f'(c) \geq \frac{L}{2}.$$ Inductively, we conclude that $$f(M+n)\geq f(M)+\frac{Ln}{2}.$$ But as $n\to\infty$, $f(M)+\frac{Ln}{2}\to \infty$; hence $f(x)\to\infty$ as $x\to\infty$, contradicting our assumption that $\lim\limits_{x\to\infty}f(x)$ is finite.

Case 2. $L\lt 0$; a similar argument shows that $f(x)\to-\infty$ as $x\to\infty$ in this case.

Therefore, the only possibility left is $L=0$, as desired.

The trick to getting geometric information out of the derivative is the mean value theorem. A corollary of the MVT is that, if $m\leq f'(x)\leq M$ for all $x\in [a,b]$, then $m(b-a)\leq f(b)-f(a) \leq M(b-a)$.

So let's proceed to our proof of the the problem.

By subtracting off a constant, we can assume $\lim f(x)=0$. Let $\lim f'(x)=L$. Assume $L\neq 0$. Without loss of generality, $L>0$ Then for $N$ large enough, we have that $x>N$ implies that $|f(x)|<1$ and $L/2<f'(x)<3L/2$.

If $N<a<b$ with $b-a>4/L$, then $f(b)-f(a) > L/2(b-a)> L/2(4/L)=2$, which is impossible.