# $f:[0,\infty)\rightarrow\mathbb{R}$ is differentiable and $\lim_{x\to\infty}f'(x)=0$, prove $\lim_{x \rightarrow \infty}[f(x+1)-f(x)]=0$. [duplicate]

Assume that $$f: [0, \infty) \rightarrow \mathbb{R}$$ is differentiable for all $$x>0$$ and $$\lim_{x \rightarrow \infty} f'(x) = 0$$. Prove that $$\lim_{x \rightarrow \infty}[f(x+1)-f(x)] = 0$$

I was hinted that I should use the mean value theorem here. My attempt is as follows. Consider the closed interval $$[x, x+1]$$ where $$x>0$$. Clearly, $$f$$ is continuous on $$[x, x+1]$$ and also differentiable on $$(x, x+1)$$ by the assumptions of the question. So we can apply the MVT and conclude that there exists a $$c \in (x, x+1)$$ such that $$f(x+1) - f(x) = f'(c)$$. Now if I take the limit to infinity on the left hand side, I can see the $$\lim_{x \rightarrow \infty} f(x+1) - f(x)$$ come into play, but what is $$\lim_{x \rightarrow \infty} f'(c)$$?

I thought about something like this, but not sure if it is right. Clearly, $$c = x+t$$ for some $$0, so $$f'(c) = f'(x+t)$$, so $$\lim_{x \rightarrow \infty} f'(c) = \lim_{x \rightarrow \infty} f'(x+t)$$. Now I am not sure how to bring $$\lim_{x \rightarrow \infty}f'(x) = 0$$ into the picture.

## marked as duplicate by Nosrati, Lord Shark the Unknown, user91500, choco_addicted, CesareoSep 23 '18 at 12:32

Your proof is almost complete - to nail it down, you may argue as follows: Since $\lim_{x\to\infty}f'(x)=0$, for each $\varepsilon>0$ there exists some $M>0$ such that for all $x>M$ you have $|f'(x)|<\varepsilon$. In particular, if $x>M$, then for $x<c<x+1$ you have: $$|f(x+1)-f(x)|=|f'(c)|<\varepsilon$$ so by definition, $\lim_{x\to\infty}(f(x+1)-f(x))=0$

Let $\epsilon >0$. Then there is $a=a(\epsilon)>0$ such that $|f'(t)|< \epsilon$ for all $t>a$.

Now let $x>a$. Then there is $c \in (x,x+1)$ such that

$|f(x+1)-f(x)| = |f'(c)|$.

Since $c>a$, we have $|f'(c)|< \epsilon$ , hence

$|f(x+1)-f(x)|<\epsilon$ for all $x>a$.

This means: $\lim_{x \rightarrow \infty}(f(x+1) - f(x)) = 0$.