Counterexample to "$\lim_{t\to\infty}\frac{f(t)}{t}=0$ implies $\lim_{t\to\infty} f'(t)=0$"

Question

From the previous question that I posted, it's true that $\lim_{t\to\infty} f'(t)=0$ implies $\lim_{t\to\infty}(\frac{f(t)}{t})=0$, but does anyone have a counterexample for the following false statement?

$\lim_{t\to\infty}\frac{f(t)}{t}=0$ implies $\lim_{t\to\infty} f'(t)=0$.

score 7 · Accepted Answer · answered Aug 03 '17 at 20:48

7

It's even possible to have $\lim_{t\to\infty}\frac{f(t)}{t}=0$ and $\limsup_{t\to\infty}f^{\prime}(t)=\infty$, as the example $f(t)=\sin(t^2)$ shows.

answered Aug 03 '17 at 20:48

carmichael561

53,688

Clement C. · Answer 2 · 2017-08-04T03:07:40.833

4

Yes, for instance take $f$ defined by $f(x)=\cos(x)$.

edited Aug 04 '17 at 03:07

answered Aug 03 '17 at 20:47

Clement C.

67,323

$f=cos$ is not correct – Stu Aug 03 '17 at 21:27
1

@Stu Why is that? – Clement C. Aug 03 '17 at 21:27
because he wants a counter example – Stu Aug 03 '17 at 21:28
2

@Stu And it is one. $\lim_{t\to\infty}\frac{\cos t}{t} = 0$, but $\cos' t = -\sin t \not\xrightarrow[t\to\infty]{} 0$. – Clement C. Aug 03 '17 at 21:29
oh sorry I thought it was $\frac{f'(t)}{t}$ – Stu Aug 03 '17 at 21:33
1

@Stu Fair enough. I guess the downvote is no longer warranted, then. – Clement C. Aug 03 '17 at 21:34
@ Clement C it's locked – Stu Aug 03 '17 at 21:52
@Stu Oh. I honestly have no idea how these things work. – Clement C. Aug 03 '17 at 21:53
You need to make a minor edit. Just add one character save edit and then again delete the character and save. After this the downvote can be reversed if the downvoter desires. But yes this procedure is a bit irritating. +1 to compensate anyway. – Paramanand Singh Aug 04 '17 at 03:02
@ParamanandSingh Thanks! Just had a go at it – Clement C. Aug 04 '17 at 03:09

Counterexample to "$\lim_{t\to\infty}\frac{f(t)}{t}=0$ implies $\lim_{t\to\infty} f'(t)=0$"

2 Answers2