# proving existence of limit

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be absolutely continuous and assume that $f^{'} \in L^3(\mathbb{R})$ and that $f(0) = 0$. Show that the following limit exists and compute its value $\displaystyle lim_{x\rightarrow 0+} \ x^{−2/3}f(x)$.

Absolute continuity allows me to use the $2^{nd}$ fundamental theorem of calculus i.e., $f(b) - f(a) = \int_{a}^{b} f^{'}(x)dx$. I am thinking along the lines of using L'Hospital's rule on $\frac{f(x)}{x^{2/3}}$ as both the numerator and denominator approach $0$ when $x \rightarrow 0^+$. How do I proceed from here on? Alternate solutions are also welcome.

Use the fundamental theorem and apply Holder's inequality: for $x > 0$ you have $$|f(x)| = |f(x) - f(0)| \le \int_0^x |f'(t)| \, dt \le \left(\int_0^x \, dt\right)^{2/3} \left( \int_0^x |f'(t)|^3 \, dt\right)^{1/3}$$ so that $$|x^{-2/3} f(x)| \le \left( \int_0^x |f'(t)|^3 \, dt \right)^{1/3},\quad x > 0.$$ But $f' \in L^3$ implies $$\lim_{x \to 0^+} \int_0^x |f'(t)|^3 \, dt = 0.$$