Prove that $[f(x+1)-f(x)] = 0$ I know this is a very elementary level question. But, I still need t0 understand this in term of mean value theorem. So here it goes:
Suppose $f$ is differentiable on $(0,\infty)$ and lim$_{x \to \infty} f'(x ) = 0$. Prove that lim$_{x \to \infty}[f(x+1)-f(x)] = 0$. 
Here, can we assume $f'(x ) = \frac{f(x+1) - f(x)}{(x+1) -x}$ and proceed to conclusion? 

For each $x>0$ there is a $c_x\in(x,x+1)$ with
$$ f'(c_x) = \frac{f(x+1) - f(x)}{(x+1) - x} = f(x+1) - f(x). $$
Now, $c_x > x$ tends to $\infty$ for $x\to\infty$, thus
$$ \lim_{x\to\infty} f(x+1) - f(x) = \lim_{x\to\infty} f'(c_x) = 0. $$
