Prove:
If $f$ is a real-valued differentiable function such that $\lim_{x\to \infty}f'(x) = 0$ then $\lim_{x\to \infty}(f(x+1)-f(x))=0$.
Proof: We know $\lim_{x\to \infty}f'(x) = 0$. (1)
By the mean value theorem there exists $c\in [x, x+1]$ such that $f'(c) = \dfrac{f(x+1)-f(x)}{1}$
$\iff \lim_{x\to \infty} f'(c) = \lim_{x\to \infty} f(x+1)-f(x) = 0$ by (1).
End of Proof.