A simple problem with limit I would like to solve the indeterminacy presents in this limit $\displaystyle \lim_{x \rightarrow -\infty} x F(x)$, in which $F(x)$ is a distribution function of a random variable $X$. The answer is obviously zero, but how to show the steps formally for any continuous distribution? Is it easy? Thank you!

This is actually false. Consider: $$F(x) = \begin{cases} \left|\frac{1}{x+2}\right| & x < -4 \\\ 1 - \frac{1}{32} x^2 & -4 \le x \le 0 \\\ 1 & x > 0 \end{cases}$$

$F(x)$ is a cumulative distribution function (since it is continuously differentiable and nondecreasing, with limit to $-\infty$ at $0$ and limit to $+\infty$ at $1$), but $$\lim_{x \to -\infty} xF(x) = -1 \
e 0$$

You can similarly construct counterexamples that converge to any negative number or that don't converge at all (try $\displaystyle \frac{\arctan x}{\pi} + \frac{1}{2}$ for another counterexample).