This is an old qualifying exam problem, and I'm wondering if my proof is correct. (I'm still not too comfortable with absolute continuity.)
We know that $$f'(x) = \frac{\sin(\sqrt{x})}{2\sqrt{x}} + \frac{\cos(\sqrt{x})}{2}$$ and this is defined for all $x \in (0,1].$
We want to check that $f'(x)$ is Lebesgue integrable on $[0,1].$ First, note that for any $\epsilon > 0,$ we have that $f(x)$ is Lipschitz continuous on $[\epsilon,1].$ Indeed, for any $x \in [\epsilon,1],$ we know that $$ |f'(x)| = \frac{1}{2}|x^{-\frac{1}{2}}\sin(\sqrt{x}) + \cos(\sqrt{x})| \leq \frac{1}{2}(|x|^{-\frac{1}{2}} + 1) \leq \frac{1}{2}(\epsilon^{-\frac{1}{2}}+1).$$ In particular, $f(x)$ is absolutely continuous on $[\epsilon,1],$ so $f'(x)$ is Lebesgue integrable on $[\epsilon,1].$
Next, note that \begin{align*} \int_0^1 f'(x)\,dx &= \lim_{\epsilon \to 0+} \int_\epsilon^1 f'(x) \, dx \\ &= \lim_{\epsilon \to 0+} (f(1) - f(\epsilon)) \\ &= \sin{1} - \lim_{\epsilon \to 0+} \sqrt{\epsilon}\sin(\sqrt\epsilon). \end{align*} But by L'Hospital, \begin{align*} \lim_{\epsilon \to 0+} \sqrt{\epsilon}\sin(\sqrt\epsilon) &= \lim_{\epsilon \to 0+} \frac{\frac{\cos(\sqrt{\epsilon})}{2\sqrt{\epsilon}}}{\frac{-1}{2\epsilon^{3/2}}} \\ &= \lim_{\epsilon \to 0+} -\epsilon\cos(\sqrt{\epsilon})\\ &= 0. \end{align*} Hence, $$\int_0^1 f'(x) \, dx = \sin{1} < \infty,$$ and so $f'(x)$ is integrable on $[0,1].$
By the same procedure as above, we know that $$\int_0^x f'(t)\,dt = f(x) = f(x) - f(0),$$ and so this implies that $f$ is absolutely continuous.
I feel like I'm only using the defintion of absolute continuity, and basic facts about integration, which is why I was wondering if I didn't make any errors.