Let $f : \mathbb R \to \mathbb R$ be absolutely continuous and assume that $f'\in L^2([0,1])$ and that $f(0) = 0.$ Show that the following limit exists and compute its value: $ \lim_{x \to 0} x^{-1/2} f(x).$
My approach since $f$ is absolutely continuous $f'(x)$ exists almost everywhere. And by Mean Value Theorem $f(x) -f(0) =f'(c)(x-0)$ for some $c \in (0, x)$. Which further $f'$ is in $L^2$ of $[0 1]$ which implies the square integral is finite. but from here how can i proceed?
As suggested in a comment by John Dawkins, you need the Cauchy-Schwarz inequality:
$$ f(x)^2 \le \left(\int_0^x |f'(t)|\,dt\right)^2\le \int_0^x 1\,dt \int_0^x |f'(t)|^2\,dt = x \int_0^x |f'(t)|^2\,dt $$ Here the factor $\int_0^x |f'(t)|^2\,dt$ tends to zero as $x\to 0$, and the conclusion $x^{-1}f(x)^2\to0$ follows.