If limit of $f'(t)$ approaches $0$, then limit of $f(t)/t$ approaches $0$.

Question

Let $f: (0, \infty) \longrightarrow \mathbb{R}$ be differentiable. Show that if $\lim_{t \to \infty}f'(t) = 0$, then $\lim_{t \to \infty}\frac{f(t)}{t} = 0$.

I don't really even know where to start. If we take $\epsilon > 0$, there is $M \in \mathbb{R}$ for which $t > M$ implies that $|f'(t)| < \epsilon$. I'm assuming that we should use the same $M$, but I'm not sure what else.

Hint: use the fundamental theorem of calculus. Can you bound the growth of $f(x)$? — Richard, Jan 05 '18 at 00:48
The context of the problem doesn't involve integration or anti derivatives. — Sprinkle, Jan 05 '18 at 00:52
The limit doesn't approach but rather is. To say that $f $ approaches $L $ is the same as saying that the limit of $f $ is $L $. Mixing both expressions is meaningless. — Andrés E. Caicedo, Jan 05 '18 at 01:18
Special case of this question (and its duplicates): https://math.stackexchange.com/questions/1044668/if-fx-rightarrow-l-as-x-rightarrow-infty-infty-leq-l-leq-infty — Hans Lundmark, Jan 05 '18 at 05:09

score 3 · Answer 1 · answered Jan 05 '18 at 00:51

3

Hint: L'Hopital's rule can be applied to $\lim_{t \to \infty} \frac{f(t)}{g(t)}$ merely on the condition that $g(t) \to \infty$.

answered Jan 05 '18 at 00:51

RRL

90,707

I did not know that, +1 – operatorerror Jan 05 '18 at 00:52
1

@qbert: Often overlooked. Now in Wikipedia. Case 2 of general proof here – RRL Jan 05 '18 at 00:56

Alex Zorn · Answer 2 · 2018-01-05T20:07:35.373

2

Fix $m > 0$ and pick $T$ such that for $t > T$, $|f'(t)| < m$. Then for $t > T$, by MVT there exists $c \in (T,t)$ such that $f(t)= f(T) + f'(c)(t - T)$. Hence $f(t)/t = f(T)/t + f'(c)(1 - T/t)$, so $|f(t)/t| < |f(T)|/t + m(1 - T/t)$. For $t$ large enough we can guarantee that $|f(T)|/t < m$, hence $|f(t)/t| < 2m$, and since $m$ was arbitrary we have $|f(t)/t| \rightarrow 0$.

edited Jan 05 '18 at 20:07

answered Jan 05 '18 at 00:59

Alex Zorn

4,304

You're right! Edited. – Alex Zorn Jan 05 '18 at 20:09

If limit of $f'(t)$ approaches $0$, then limit of $f(t)/t$ approaches $0$.

2 Answers2