$\lim_{x\rightarrow\infty} f(x)=a$ then $\lim_{x\rightarrow\infty} f'(x)=0$ What this means is that if limit $x$ approaches infinity $f(x)$ is a real number other than plus-minus infinity then $f'(x)$ as $x$ approaches infinity will be $0$ or nonexistent. I need mathematical proof or a counterexample.
 A: Let $D$ contain a neighborhood of $+\infty$ and let $f\colon D\rightarrow\mathbb{R}$ be, s.t. $f$ is differentiable in a neighborhood of $+\infty$ and $\lim\limits_{x\rightarrow\infty}f(x)=L\in\mathbb{R}$. 
For sufficiently large $n\in\mathbb{N}$, the MVT guarantees the existence of $\xi_n\in(n,n+1)$, such that
$$f(n+1)-f(n)=\frac{f(n+1)-f(n)}{n+1-n}=f^{\prime}(\xi_n).$$
The LHS goes to $L-L=0$ as $n\rightarrow\infty$, so we also have $f^{\prime}(\xi_n)\rightarrow0$. Now if $\lim_{x\rightarrow\infty}f^{\prime}(x)\in\bar{\mathbb{R}}$ exists, it will be equal to $\lim_{n\rightarrow\infty}f(\xi_n)=0$, because $\xi_n\rightarrow\infty$ as $n\rightarrow\infty$. However, the limit does not need to exist as the counter-example $f\colon(0,\infty)\rightarrow\mathbb{R},\,x\mapsto\frac{\sin(x^3)}{x}$ shows.
A: A counterexample:
Let $a_n$ be a sequence such that $a_n>0$ and $\sum_{n=1}^\infty \frac{1}{\sqrt{a_n}} = g < \infty$. 
Let $f(x) = \int_{-\infty}^x \big(\sum_{n=1}^\infty e^{-a_n(y-n)^2}\big) dy$.
We have:
 $$f'(x) = \sum_{n=1}^\infty e^{-a_n(x-n)^2} > 0 $$
so $f(x)$ is strictly increasing; we have also
$$\lim_{x\rightarrow\infty} f(x) = \int_{-\infty}^\infty \big(\sum_{n=1}^\infty e^{-a_n(y-n)^2}\big) dy = \sum_{n=1}^\infty \sqrt{\frac{\pi}{a_n}} = g \sqrt{\pi} < \infty$$
However
$$ f'(n) = \sum_{m=1}^\infty e^{-a_m(n-m)^2} \ge e^{-a_n(n-n)^2} = 1$$
so $f'(x)$ does not converge to $0$ for $x\rightarrow\infty$.
Basically, the trick is for $f'(x)$ to be very small most of the time, but on very short intervals it can be arbitrarily big; if the interval is short enough, a big derivative doesn't have to increase $f(x)$ by much.
