3
$\begingroup$

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 $0<t<1$, 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.

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

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, $\lim_{x\to\infty}(f(x+1)-f(x))=0$

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.