Let $X$ be a non-negative random variable. Show that $$ \lim_{x\to\infty} xE\left[X_{\{ X > x \}}^{-1}\right] = \lim_{x\downarrow 0} xE\left[X^{-1}1_{\{ X > x \}}\right] =0. $$
I was given the hint to not assume that $E[X^{-1}]<\infty$.
We replace $x$ by $n$ to make a sequence of random variables, $\lim_{n\to\infty} nE[X_{\left\{ X > n \right\}}^{-1}]$, where our sequence is defined by $X_n=\frac{1}{X}1_{\left\{ X > n \right\}}$. Then, $X_n=\frac{1}{X}1_{\left\{ X > n \right\}} \to \emptyset$ as $n\to\infty$ almost surely.
But, I have that extra copy of $n$ in my problem. So, I should be defining my $X_n = \frac{n}{X}1_{\left\{ X > n \right\}}$, and I do not how to proceed.
Note that $$\left|x \cdot X^{-1} 1_{\{X>x\}}\right| = \left| \frac{x}{X} 1_{\{x/X<1\}} \right| \leq 1,$$
i.e. the constant function $1$ is an (integrable) dominating function for the expression on the left-hand side, uniformly for all $x$. On the other hand, it is not difficult to see that
$$x \cdot X^{-1} 1_{\{X>x\}} \xrightarrow[]{x \to 0} 0 \cdot X^{-1} 1_{\{X>0\}} = 0$$
and
$$ \left| x \cdot X^{-1} 1_{\{X>x\}} \right| \leq 1_{\{X>x\}} \xrightarrow[]{x \to \infty} 1_{\{X=\infty\}}.$$
Because of the dominating integrable function which we found at the beginning, this means that we can apply the dominated convergence theorem to conclude that
$$ \mathbb{E}(x X^{-1} 1_{\{X>x\}}) \xrightarrow[]{x \to 0} 0$$
and
$$\mathbb{E}(x X^{-1} 1_{\{X>x\}}) \xrightarrow[]{x \to \infty} \mathbb{P}(X=\infty)=0.$$