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 I've been trying to solve the following problem:
Suppose that $f$ and $f'$ are continuous functions on $\mathbb{R}$, and that $\displaystyle\lim_{x\to\infty}f(x)$ and $\displaystyle\lim_{x\to\infty}f'(x)$ exist. Show that $\displaystyle\lim_{x\to\infty}f'(x) = 0$.
I'm not entirely sure what to do. Since there's not a lot of information given, I guess there isn't very much one can do. I tried using the definition of the derivative and showing that it went to $0$ as $x$ went to $\infty$ but that didn't really work out. Now I'm thinking I should assume $\displaystyle\lim_{x\to\infty}f'(x) = L \neq 0$ and try to get a contradiction, but I'm not sure where the contradiction would come from. 
Could somebody point me in the right direction (e.g. a certain theorem or property I have to use?) Thanks

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 $\frac{xf(x)}{x^2}$ ;)
