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 $\lim\sup_{x\rightarrow \infty} xf'(x) = 0$.