Q:

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!

A:

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 $$\lim_{x \to -\infty} xF(x) = -1 \ne 0$$
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).

