# Finding the limit of $f$, knowing the limit of its primitive $F$.

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.

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 $$F(x)$$?
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.
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.
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.