Prove that $ \lim_{x \to \infty} f'(x) = 0$ 
Prove that if $\displaystyle \lim_{x \to \infty} f(x)$ and $\displaystyle \lim_{x \to \infty} f'(x)$ are both real numbers, then $\displaystyle \lim_{x \to \infty} f'(x) = 0$.

Attempt
Intuitively this makes sense to me. Take $y = x$. This slope is constant but it increases arbitrarily, and it seems that we can't make both the slope and value of $f(x)$ to be real numbers without "flattening" out the graph. I tried first saying $\displaystyle \lim_{x \to \infty} f(x) = a$ and $\displaystyle \lim_{x \to \infty} f'(x) = b$. Then we might be able to do something with the L'Hospital's rule.
 A: $\displaystyle f(n+1)-f(n)=\frac{f(n+1)-f(n)}{n+1-n}=f'(c_n)$ where $c_n\in (n,n+1)$
Letting $n\to \infty$, $$0=\lim_\infty f'$$
A: If $\lim_{x\to\infty}f(x)$ exists, then clearly
$$\lim_{x\to\infty}\left(1+{f(x)\over x}\right)=1$$
But ${x+f(x)\over x}$ qualifies for L'Hopitation, which gives
$$\lim_{x\to\infty}\left({x+f(x)\over x}\right)=\lim_{x\to\infty}\left({1+f'(x)\over 1}\right)=1+\lim_{x\to\infty}f'(x)$$
Hence $\lim_{x\to\infty}f'(x)=0$.
A: Hint: $f(x)-f(x_0)=\int_{x_0}^x f'(t)\,dt$, especially for some $x_0$ sufficiently large and $x\ge x_0$.
A: I like Barry Cipra's answer. Here's another proof along the same lines:
$$\lim_{x\to \infty} f(x) = \lim_{x\to \infty} {x f(x) \over x} 
= \lim_{x\to \infty} {x f'(x) + f(x) \over 1} = \lim_{x\to \infty} \big(x f'(x) + f(x)\big)
$$
Now if $\lim_{x\to \infty} f'(x)>0$, $\lim_{x\to \infty} f(x)= +\infty$, 
and if $\lim_{x\to \infty}f'(x)<0$, $\lim_{x\to \infty} f(x) = -\infty$, 
both of which violate an assumption; hence $\lim_{x\to \infty} f'(x)=0$.
A: Employing the Mean value theorem, entails that,  for each $x$ there exists $c_x\in (x,x+1)$ such that 
$$f'(c_x)=f(x+1)-f(x)\to0$$
Hence $\lim_{x\to \infty}f'(x)=\lim_{x\to \infty}f'(c_x) =0$ since, $c_x\to\infty$ as $x\to\infty$ we have

