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

I've seen this problem on a Facebook page yesterday. Can anybody give me some tips to solve it, please? It looks pretty interesting and I have no idea of a proof now.

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HINT: If $\lim_{x\to\infty}f(x)>0$ then there exist $\varepsilon>0$ and $M\in\Bbb{R}$ such that $f(x)>\varepsilon$ for all $x>M$. What does this mean for $F(x)$?

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The statement is not true. There is a counterexample. It is a known fact that $$ \lim_{x\to\infty} \int_0^x \sin(t^2)\ \mathrm dt = \sqrt{\frac\pi 8}=:\frac1{c} $$ So if we define $f(t) = \sin(t^2)-e^{-ct}$, then $$\lim\limits_{x\to\infty}F(x)=\lim\limits_{x\to\infty} \int_0^x \left(\sin(t^2)-e^{-ct}\right) \mathrm dt=\frac1{c}-\frac1{c}=0$$ and $|f(t)|\le 2$ for every $t\ge 0$. But the limit $$ \lim_{t\to\infty} \left(\sin(t^2)-e^{-ct}\right)=\lim_{t\to\infty} \sin(t^2) $$ does not exist.

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I guess your question has a typo. Instead of $f$ being bounded you should have $f'$ being bounded. Otherwise what is need of the hypothesis that $f$ is differentiable.

Assuming that the above typo is fixed, the result is famous and pretty standard.

Lemma: Let $\phi:(a, \infty) \to\mathbb {R} $ be twice differentiable on $(a, \infty) $ and further let second derivative $\phi''$ be bounded on $(a, \infty) $. If $\phi(x) \to L$ as $x\to\infty $ then $\phi' (x) \to 0$ as $x\to\infty$.

Your result follows by taking $\phi=F$ in above lemma. A proof of the above lemma is available here.

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