# Limit property of derivative of bounded monotone function

I am trying to prove the following or find a counterexample. Suppose a function $$f: \mathbb R \rightarrow \mathbb R$$ is bounded, increasing, and continuously differentiable. Then $$\lim_{x\rightarrow \infty}xf'(x)=0$$.

So far, I have shown that $$\lim\inf_{x\rightarrow \infty} xf'(x)= 0$$: Suppose $$\lim \inf_{x\rightarrow \infty }xf'(x)>0$$. Then exists a $$\delta>0$$ and $$\epsilon >0$$ such that when $$x>\delta$$, $$x f'(x)>\epsilon$$. Then for $$x>\delta$$, $$f'(x)>\epsilon /x$$. The antiderivative of $$\epsilon /x$$ is $$\epsilon \ln x$$ which converges to $$\infty$$ as $$x\rightarrow \infty$$. This contradicts the boundedness of $$f$$. Therefore $$\lim\inf_{x\rightarrow \infty} xf'(x)\leq 0$$. Since $$f$$ is increasing, it must be that $$\lim\inf_{x\rightarrow \infty} xf'(x) \geq 0$$. Combining these conditions yields $$\lim\inf_{x\rightarrow \infty} xf'(x)= 0$$.

Also from $$f$$ increasing, we know $$\lim\sup_{x\rightarrow \infty} xf'(x) \geq 0$$. So it remains to show that $$\lim\sup_{x\rightarrow \infty} xf'(x) = 0$$.

Counter-example: Let $$g(x)=n$$ for $$n \leq x \leq n+\frac 1 {n2^{n}},g(x)=0$$ for $$x >n+\frac 2 {n2^{n}}$$ and $$0$$ for $$x with a straight line graph between $$n$$ and $$n+\frac 2 {n2^{n}}$$ as well as between $$n-\frac 1 {n2^{n}}$$ and $$n$$ (for each $$n$$). Let $$f(x)=\int_0^{x}g(t)dt$$. You can easily check that this is a counter-example. (In fact $$f'(n)\to \infty$$).