If $f$ is bounded and $\lim\limits_{x\to+\infty} F(x)=0$ then $\lim\limits_{x\to+\infty} f(x)=0$

Asked Mar 12 '19 at 13:30

Active Mar 12 '19 at 18:57

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Let $f:(0, + \infty) \rightarrow \mathbb{R}$ be a derivative function and $F$ one of its primitives. Prove that if $f$ is bounded and $$\lim_{x\to+\infty} F(x)=0$$ then $$\lim_{x\to+\infty} f(x)=0$$

What I have tried:

I observed that $$(f(x)F(x))'=f'(x)F(x)+f^2(x)$$ It is obvious that the left term is $0$ when $x\rightarrow+\infty$. The problem is that we don't know that $f'$ is also bounded. If that would be the case, then $f'(x)F(x)\rightarrow 0$, so $f\rightarrow 0$.

Can you please help me solve this problem? Is it even correct or should $f'$ be bounded(instead of $f$)?

edited Mar 12 '19 at 18:57

rtybase

17,398

asked Mar 12 '19 at 13:30

razvanelda

1,765

2

I think I'm missing something. If $f = \cos(x^2)$, and $F(x) = \int_0^x f(t)dt - \sqrt{\frac\pi8}$, then $f$ is differentiable and bounded, and $F\to 0$, but $f\not\to 0$. – Arthur Mar 12 '19 at 13:35
@Arthur Indeed, the statement is false. – Lorenzo Q Mar 12 '19 at 13:40
Do you mean "Prove that if $f'$ is bounded and ..."? Because that seems more likely to be true. Besides, what else would be the point of requiring $f$ to be differentiable? – Arthur Mar 12 '19 at 13:43
@Arthur That's how I found the problem stated, "$f$ bounded" and I also believe that it is not correct. – razvanelda Mar 12 '19 at 13:45
1

There was likely a typo. See This – Mark Viola Mar 12 '19 at 13:52

If $f$ is bounded and $\lim\limits_{x\to+\infty} F(x)=0$ then $\lim\limits_{x\to+\infty} f(x)=0$

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