I have been studying for my real analysis qualifying exam, and I have noticed a trend of questions similar to the following:
Suppose that $f$ is absolutely continuous, $f'\in L^3$, and $f(0)=0$. Show that the following limit exists and compute its value: $$\lim_{x\to 0^+} x^{-2/3}f(x).$$
My best guess is that absolute continuity is just there to guarantee a derivative a.e., and that we are supposed to proceed by L'Hôpital's rule, but I have no idea what the point of the $L^3$ restriction on $f'$ is supposed to be about. I suspect that it the $3$ is a red herring and you only need maybe $L^{5/3}$ but this is wild speculation on my part.
I think I am in part hampered by my inability to find an exact theorem statement of L'Hôpital for a.e.-defined functions. It is not in baby or big Rudin, or in Stein & Shakarchi IV, and it did not come up for me in some simple google searches; these are my only references on hand
What is a statement of L'Hôpital that works in this context?
Is there a simpler way to go about things that makes the $L^3$ restriction more obvious?
I asked a similar question earlier
Since $f$ is absolutely continuous it's absolutely continuous on any closed interval, so for a given $x>0$ we can write: $f(x)-0=f(x)-f(0)=\int_0^x f'$ by the fundamental theorem of calculus. Then we bound from above and show that the bound goes to zero: $$\left|\frac{f(x)}{x^{2/3}}\right|\leq\frac{\int_0^x |f'\cdot 1|}{x^{2/3}}=\frac{||f'\cdot 1||_{L^1([0,x])}}{x^{2/3}}\overset{\text{Holders}}{\leq}\frac{||f'\cdot 1||_{L^3([0,x])}||1||_{L^{3/2}([0,x])}}{x^{2/3}}=\frac{||f'\cdot 1||_{L^3([0,x])} (\int_0^x 1^{3/2} dt)^{2/3}}{x^{2/3}}=||f'\cdot 1||_{L^3([0,x])}\to 0\text{, as } x\to 0^+$$ The limit is zero by the Dominated Convergence Theorem because $\chi_{(0,x)}\cdot |f'|^3\leq |f'|^3\in L^1$ and as $x\to 0$ we get $\chi_{(0,x)}|f'|^3\to 0$ a.e.