Suppose $x$ is a random variable with support on $(0,\infty)$ and finite expectation $E(x)$ and PDF $f(x)$. I believe this means that it must be true that $\lim_{x \to \infty} x * f(x) = 0$; otherwise $\int_0^\infty x*f(x) dx$ would not give a finite value.
I'm hoping to prove something further, though, specifically given $y > 0$ and $z >0$, I want to be able to show that:
$$\lim_{x \to 1^-} f\left(\frac{y}{1-x}\right)* \frac{z}{(1-x)^2} =0$$
Based on simulating every distribution I can think of, this appears to be true, but I'm not really sure where to start.
Consider the function $f(x)$ that equals $\frac{C}{n}$ over the intervals $\left(n-\frac{1}{n^2},n+\frac{1}{n^2}\right)$, with $n\in\mathbb{N}^*$.
For a suitable choice of the $C$ constant ($\frac{1}{2\zeta(3)}$) it is the the probability density function of a random variable with finite expectation ($\frac{\pi^2}{6\zeta(3)}$), but the given limit does not exist: $$\limsup_{x\to +\infty}x\cdot f(x) = C>0,\qquad \liminf_{x\to +\infty}x\cdot f(x) = 0.$$