# Limits of integral

(CMCS 2010 Longlist) Let $f: [0, + \infty) \rightarrow (0, +\infty)$ be a continuous function such that$\lim\limits_{x \rightarrow \infty} \int_{0}^{x}f(t)\,dt=l\in \mathbb{R}$. Prove that $$\lim\limits_{x \rightarrow \infty}\frac{1}{\sqrt {x}}\int_{0}^{x}\sqrt {f (t)}\,dt = 0.$$

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By-parts integration ($u=t,~dv=fdt$) and then L'Hôpital's, in that order, gives
$$\triangle=\frac{1}{x}\int_0^x tf(t)dt=\int_0^xf(t)dt-\frac{1}{x}\int_0^x \left[\int_0^tf(\tau)d\tau\right]dt \xrightarrow{x\to\infty} l-l=0.$$
Equating the above with the result of direct application of L'Hôpital's to $\triangle$ gives
$$\lim_{t\to\infty} tf(t)=0.$$
Now apply L'Hôpital's to the original problem, noting $\sqrt{\cdot}$ is continuous on $[0,\infty)$.