Proving that $\lim\limits_{x\to\infty}f'(x) = 0$ when $\lim\limits_{x\to\infty}f(x)$ and $\lim\limits_{x\to\infty}f'(x)$ exist

Apply a L'Hospital slick trick: $\, $ if $\rm\ f + f\,'\!\to L\ $ as $\rm\ x\to\infty\ $ then $\rm\ f\to L,\ f\,'\!\to 0,\, $ since

$$\rm \lim_{x\to\infty}\ f(x)\ =\ \lim_{x\to\infty}\frac{e^x\ f(x)}{e^x}\ =\ \lim_{x\to\infty}\frac{e^x\ (f(x)+f\,'(x))}{e^x}\ =\ \lim_{x\to\infty}\, (f(x)+f'(x))\qquad $$

This application of L'Hôpital's rule achieved some notoriety because the problem appeared in Hardy's classic calculus texbook A Course of Pure Mathematics, but with a less elegant solution. For example, see Landau; Jones: A Hardy Old Problem, Math. Magazine 56 (1983) 230-232.

Hint: If you assume $\lim _{x \to \infty } f'(x) = L \ne 0$, the contradiction would come from the mean value theorem (consider $f(x)-f(M)$ for a fixed but arbitrary large $M$, and let $x \to \infty$).

Explained: If the limit of $f(x)$ exist, there is a horizontal asymptote. Therefore as the function approaches infinity it becomes more linear and thus the derivative approaches zero.

To expand a little on my comment, since $\lim_{x \to \infty} f(x) = L$, we get

$$\lim_{x \to \infty} \frac{f(x)}{x} =0 \,.$$

But also, since $\lim_{x \to \infty} f'(x)$ exists, by L'Hospital we have

$$\lim_{x \to \infty} \frac{f(x)}{x}= \lim_{x \to \infty} f'(x) \,.$$

Note that using the MTV is basically the same proof, since that's how one proves the L'H in this case....

P.S. I know that if $L \neq 0$ one cannot apply L'H to $\frac{f(x)}{x}$, but one can cheat in this case: apply L'H to $\frac{xf(x)}{x^2}$ ;)