TITLE: If the limit of a derivative is zero as $x \to \infty$, what can we say about $f(x+1)-f(x)$?
QUESTION [8 upvotes]: Given a differentiable function $f$ such that 
$$
\lim_{x \to +\infty}f'(x) = 0
$$
what can we say about
$$
\lim_{x\to\infty}(f(x+1)-f(x)) \text{ ?}
$$
My first thought was to use mean value theorem on $[x,x+1]$, and I will get that the limit is $0$, is this true?
Thank you for your help.

REPLY [15 votes]: Given that $f$ is differentiable
consider the interval $[x,x+1]$ and apply the mean value theorem 
we get 
$f(x+1)-f(x) = f'(c_x)$ where $c_x\in (x,x+1)$.
Now taking limits both sides 
$$\lim_{x\to\infty}(f(x+1)-f(x)) = \lim_{x\to\infty}(f'(c_x))$$ where $c\in (x,x+1)$ , so as $x$ tends to infinity , $f'(c_x)$ tends to zero as given in the hypothesis . 
Therefore $$\lim_{x\to\infty}(f(x+1)-f(x)) =0$$ .