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
 A: 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.
A: This is in response to an interesting observation made by Rhythmic Fistman in a comment below my (first) answer. We suppose that $\lim _{x \to \infty } f'(x) = L$ for some $L \in \mathbb{R}$. Then, from the definition of the derivative,
$$
L = \mathop {\lim }\limits_{x \to \infty } \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}.
$$
As Rhythmic Fistman observed, naively changing the order of the limits gives rise to the equality
$$
L = \mathop {\lim }\limits_{h \to 0} \mathop {\lim }\limits_{x \to \infty } \frac{{f(x + h) - f(x)}}{h} = 0,
$$
where the last equality follows from the assumption that $\lim _{x \to \infty } f(x)$ exists (finite). ``Hence'' the desired result $\lim _{x \to \infty } f'(x) = 0$. However, as the following counterexample shows, this procedure is not allowed in principle. Define a two-variable function $f$ by $f(x,h)=xe^{-|h|x}$.
Analogously to the case in the original question (where the role of $f(x,h)$ is played by $\frac{{f(x + h) - f(x)}}{h}$),
$$
\mathop {\lim }\limits_{x \to \infty } f(x,h) = \mathop {\lim }\limits_{x \to \infty } xe^{ - |h|x}  = 0,
$$
for any $h \neq 0$. Hence, 
$$
\mathop {\lim }\limits_{h \to 0} \mathop {\lim }\limits_{x \to \infty } f(x,h) = 0.
$$
Also, for any $x \in \mathbb{R}$,
$$
\mathop {\lim }\limits_{h \to 0} f(x,h) = \mathop {\lim }\limits_{h \to 0} xe^{ - |h|x}  = x
$$
(this is analogous to the case in the original question, where $f'$ is assumed continuous on $\mathbb{R}$).
Hence,
$$
\mathop {\lim }\limits_{x \to \infty } \mathop {\lim }\limits_{h \to 0} f(x,h) = \infty  \neq 0 = \mathop {\lim }\limits_{h \to 0} \mathop {\lim }\limits_{x \to \infty } f(x,h).
$$
A: The Mean value theorem,reveals that,  for each $x$ there exists $c_x\in (x,x+1)$ such that 
$$f'(c_x)=f(x+1)-f(x)$$
Hence since, $c_x\to\infty$ as $x\to\infty$ we have
$$\lim_{x\to \infty}f'(x)=\lim_{x\to \infty}f'(c_x) =\lim_{x\to \infty}[f(x+1)-f(x)] =0 $$
A: You know that $\lim_{x \to \infty}f(x)$ and $\lim_{x \to \infty}f^{'}(x)$ exists. Then by Lagrange's theorem there exists $c_n \in (n,n+1)$ such that $f(n+1)-f(n)=f^{'}(c_n)$ Taking the limit as $n \to \infty$ you get that $\lim_{n \to \infty}f'(c_n)=0$. Since the limit exists, and there exists a sequence for which the limit of the function is $0$ it follows that $\lim_{n \to \infty}f^{'}(x)=0$.
A: 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. 
A: 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 $\frac{xf(x)}{x^2}$ ;)
