With scant information, how to prove this probability limit tends to zero?

Given $X$ a nonnegative r.v. with $E(|X|^\alpha)<\infty$, then how to prove that
$$\lim_{x\to\infty}x^\alpha P(|X|>x)=0.$$
If $X$ has a PDF, then it seems this can be proved using some integration by parts. But now there is nothing mentioned about the regularity of its CDF. So how shall I proceed?

Best regards.

Answer

You have as $x\rightarrow \infty$:
$$x^\alpha P(|X|>x) \leq E(|X|^\alpha 1_{|X|>x}) \rightarrow 0 $$
by dominated convergence.