Let $f \in L^2[0,\infty)$ be a continuous, positive and decreasing function, then $\lim_{x\to \infty}\sqrt{x}f(x)=0$

Question

Let $f \in L^2[0,\infty)$ be a continuous, positive and decreasing function, then $\lim_{x\to \infty}\sqrt{x}f(x)=0$

This feels intuitively true but I am not sure how to prove it. I know that if $f$ is uniformly continuous then $\lim_{x \to \infty}f(x)=0$ but I am not sure how to handle this. I am not sure how the hint is helpful.

Answer 1

$\int_a^{2a}f(x)^2dx<\epsilon$ for $a$ large enough and $af(2a)^2 \le \int_a^{2a}f(x)^2dx$ because $f$ is decreasing.