Let $X$ be a random variable with density function $f$ and support $S\subseteq(0,+\infty)$. Assume that $X$ has finite second moment, $\mathbb E X^2<+\infty$. Is it true that $$\lim_{x\to 0^+}xf(x)=0$$
I think that the answer is yes, but I am not sure that I can prove it. I am thinking of the following. Assume that $xf(x)>0$ as $x\to 0^+$. Then, certainly $f(x)\to+\infty$, as $x\to 0^+$. Hence, $$\mathbb EX=\int_{S}uf(u)du= \lim_{x\to0^+}\int_{x}^{+\infty}uf(u)du=+\infty\tag1$$ which contradicts the fact that $\mathbb EX<+\infty$. But, I think that $(1)$ is not justified.
I have also taken several examples and all of them work. For example, if I take $f(x)=1/\sqrt{x}$, then $xf(x)$ goes to $0$. If I take $f(x)=1/x$, then $xf(x)$ goes to $1$, but in this case $\mathbb EX=+\infty$ which is a contradiction to my assumption that $\mathbb EX<+\infty$, and hence $f(x)=1/x$ is not allowed in the first place.
Any help would be appreciated.