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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.

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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.

Thorgott
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  • It's a huge assumption to assume $\lim\limits_{x\to\infty} f'(x)$ exists. That certainly does not follow from the hypotheses. – Ted Shifrin May 30 '19 at 01:57
  • @TedShifrin I agree it is a huge assumption. It was my interpretation of "does not oscillate", but I was misinterpreting the question. I've edited the answer in a way that complies with OPs latest edit. – Thorgott May 30 '19 at 02:36
  • @TedShifrin That huge assumption has been meanwhie edited into the question ("... will be $0$ or nonexistent") – Hagen von Eitzen May 30 '19 at 09:56
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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.

Adam Latosiński
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  • How do you explain the answer by @Thorgott? There seems to be a contradiction, since the derivative $f'(n)$ converges to a value for all finite n. Maybe then your function is non-analytic at infinity? How does one reconcile that with the fact that the mean value theorem only requires the existence of the first derivative? – DinosaurEgg May 29 '19 at 22:54
  • Thorgott assumes that the limit $\lim_{x\rightarrow\infty} f'(x)$ exists. In my counterexample it does not exist. It's true that $\lim_{n\rightarrow\infty} f'(n) = 1$, but $\lim_{n\rightarrow\infty} f'(n+\frac12) = 0$, and altogether in my example $\lim_{x\rightarrow\infty} f'(x) $ does not exist. – Adam Latosiński May 29 '19 at 23:15
  • thank you, however, your answer is not what I was looking for, I made clarifications to make this clearer. –  May 30 '19 at 01:42
  • In that case, Thorgott's answer is ok. If $f'(x)$ exists for all $x$ and $\lim_{x\rightarrow\infty} f'(x)$ exists, then indeed $\lim_{x\rightarrow\infty} f'(x)=0$ as Thorgott proves. – Adam Latosiński May 30 '19 at 06:49