Limit of the product of X and the Survival Function

Question

Given a continuous random variable $X$ whose support is defined on $[0, \infty)$, prove that $$\lim_{x\rightarrow\infty}x^\alpha\overline{F}(x)=0$$ where $\alpha > 0$ and $\overline{F}(x)=1-F(x)$ where $F(x)$ is the CDF of $X$.

I stumbled across this the other day but have been unable to find a proof. I'm not entirely sure that it's even always true for that matter. Can you prove it?

Answer 1

I don't think the statement is true as stated. Consider a R.V. $X$ such that, $\bar{F}(x) = \dfrac{1}{(x+1)^\alpha}$.

For the result to hold, a sufficient condition would be $\mathbb{E}[X^\alpha] < \infty$.