Absolutely Continuous but not Holder continuous

$$f(x)=\begin{cases} 1/\log x \quad &\text{if } x\in (0,1/2] \\ 0 &\text{if }x=0\end{cases}$$ Consider the behavior of $f(x)/x^\alpha$ at zero.

Also notice that $f'$ is bounded on $[1/n,1/2]$ for all $n=3,4,\dots$.

I think the above is a good example, but if you want to find some function $f$ such that $f$ is absolutely continuous, but not $\alpha\ -H\ddot{o}ler$ continuous, where $0<\alpha<1$, then just take$$f(x):=x^\beta\quad where \ \ 0<\beta<\alpha<1$$ Since $f(x)=\int_0^x \beta\cdot x^{\beta-1}\mathrm{d}x$, $f$ is absolutely sontinuous on $[\ 0,1\ ]$, but $\lim_{x\to 0}\frac{x^\beta}{x^\alpha}$ is infinity.