Relationship with uniformly continuous function and its derivative. I'm trying to prove the following.
Let $f:(0,\infty) \to \mathbb{R}$ be a differentiable function. Let also $f'(x)$ be a uniformly continuous function on $(0,\infty)$. 
If $\lim_{x \to \infty}f(x) = L$, then $\lim_{x\to \infty} f'(x) = 0$.
My first instict is to invert the order of limit and derivative because of uniformly continuity, but that seems a bit odd. There is a theorem that uses $[a,\infty)$ to show uniform continuity but here I have an open set.
Any idea of a rigorous proof? 
 A: Proof by contradiction. Assume $\lim_{x\to\infty}f'(x)\neq 0$, then there exists $\epsilon>0$ and an increasing sequence $(x_n)$ with $x_n\to\infty$ such that
$$|f'(x_n)|>2\epsilon,\quad\forall n\in\mathbb{N}.$$
Since $f'$ is uniformly continuous, then there exists $\delta>0$ independent of $n$ such that
$$|f'(x_n)-f'(x)|<\epsilon,\quad\forall |x-x_n|<\delta.$$
This would mean that whenever $|x-x_n|<\delta$ for some $n$, we have
$$|f'(x)|\geq|f'(x_n)|-|f'(x)-f'(x_n)|>\epsilon.$$
As $f'$ is continuous on $(x_n,x_n+\delta)$, this tells that $f'$ does not change sign on $(x_n,x_n+\delta)$ (otherwise by the intermediate value theorem there exists $c\in(x_n,x_n+\delta)$ with $f'(c)=0$, this contradicts $|f'(x)|>\epsilon>0$).
Thus for all $n\in\mathbb{N}$ we have
$$\left|\int_1^{x_n}f'(x)dx-\int_1^{x_n+\delta}f'(x)dx\right|=\int_{x_n}^{x_n+\delta}|f'(x)|dx\geq\int_{x_n}^{x_n+\delta}\epsilon dx=\delta\epsilon.$$
On the other hand by fundamental theorem of calculus
$$\left|\int_1^{x_n}f'(x)dx-\int_1^{x_n+\delta}f'(x)dx\right|=|f(x_n)-f(x_n+\delta)|\to 0$$
as $n\to\infty$ by $\lim_{x\to\infty}f(x)=L.$ This is a contradiction.
