We have $f:\mathbb R\to \mathbb R$ a function that has a primitive $F:\mathbb R\to (0,\infty)$. Show that for every $\epsilon>0$, there exists $x_{\epsilon}$ such that $|f(x_{\epsilon})|<\epsilon$.
I supposed that there exists an $\epsilon>0$ such that for every $x$ real number we have $|f(x)|>\epsilon$. So $f(x)\in (-\infty,-\epsilon)\cup(\epsilon,\infty)$. But f has Darboux property because it has $F$ as a primitive. So $f(x)<-\epsilon$ or $f(x)>\epsilon$ for every $x$ real number. In the case when $f(x)<-\epsilon$ for every $x$, we have $f(x)<0$ for every $x$ so $F$ is strictly decreasing and it has limits at $\pm\infty$ (I don't know if this helps). I tried using Lagrange theorem. If we take $a<b$ then $F(a)>F(b)>0$ and by Lagrange theorem we get that $\frac{F(b)-F(a)}{b-a}=f(c)<-\epsilon$. I dont know how to finish the problem and how to get to a contradiction. I feel like I need to use some limit or derivative close to $\epsilon$ to get a contradiction.
Consider the case where $f>\epsilon$ everywhere. Then $F(x)-\epsilon x$ is an increasing function, since it has a positive derivative. Hence, we find $$F(-F(0)/\epsilon)-\epsilon(-F(0)/\epsilon)<F(0),$$ or equivalently, $F(-F(0)/\epsilon)<0$. This yields the desired contradiction.