If $f:\mathbb{R}\rightarrow \mathbb{R}$ is differentiable and $\lim_{x\to \infty } f^\prime(x)=0$ show that $\lim _{x\to \infty } (f(x+1)-f(x))=0$.
Solution 1:
Use the mean value theorem. Since $f$ is differentiable on the interval $[x,x+1]$, there exists $c_x \in [x,x+1]$ such that $f'(c_x) = f(x+1)-f(x)$. Thus $$ \lim_{x \to \infty} f(x+1) - f(x) = \lim_{x \to \infty} f'(c_x) = \lim_{c_x \to \infty} f'(c_x) = 0. $$ The last limit substitution is justified because $c_x \to \infty \quad \Longleftrightarrow \quad x \to \infty$ since $x \le c_x \le x+1$.
Hope that helps,