Does $\limsup\limits_{x\to\infty}f'(x)=\infty$ imply the nonuniform continuity of $f$ on $(0,\infty)$?

Question

Suppose $f$ is differentiable on $(0,\infty)$ and $\limsup\limits_{x\to\infty}f'(x)=\infty$. Does this imply the function $f$ is nonumiform continuous on $(0,\infty)$?

I know and can prove the following conclusions:

(1) If $|f'(x)|\leq M$, then $f(x)$ is umiform continuous on $(0,\infty)$, and

(2) If $\lim\limits_{x\to\infty}f'(x)=\infty$, then $f$ is nonumiform continuous on $(0,\infty)$.

For the proof of (1) and (2), just use the Lagrange MVT.

What I want to know is that: Does $\limsup\limits_{x\to\infty}f'(x)=\infty$ imply the nonuniform continuity of $f$ on $(0,\infty)$?

Any help or hints will welcome! Thanks a lot!

Answer 1

No. See for instance $\frac1x\sin(x^3)$.

Answer 2

The answer is no because you can easily construct a differentiable function $f$ on $\Bbb R$ such that $f'(n)=n$ and $\lim_\infty f=0.$