Show that $tE(1/X;X>t)\to0$ when $t\to0+$ Let $X\geqslant 0$ be a random variable，then $$\lim_{t\rightarrow 0+}{ \,\,t\int_{\left [ X> t \right ]} \frac{1}{X} \, {\mathrm{d}\mathbb{P}} }=0$$
I have no idea of how to prove it.
 A: As @Michael suggested, for $t\in(0,1)$ we have $$\{\omega:X(\omega)>t\} = \left\{\omega: t < X(\omega) < t^{\frac12}\right\}\cup\left\{\omega: X(\omega)>t^{\frac12} \right\}. $$
If $t<X(\omega)<t^{\frac12}$ then $t^{-\frac12}<\frac1{X(\omega)}<t^{-1}$, so
\begin{align}
\int_{\left\{\omega\ :\ t < X(\omega) < t^{\frac12}\right\}} t\cdot\frac1{X(\omega)}\,\mathsf d\mathbb P(\omega)&<\int_{\left\{\omega\ :\ t < X(\omega) < t^{\frac12}\right\}} \,\mathsf d\mathbb P(\omega)\\&=\mathbb P\left(t<X<t^{\frac12}\right)\\
&\stackrel{t\to 0}\longrightarrow 0.
\end{align}
If $X(\omega)>t^{\frac12}$, then $\frac1{X(\omega)}<t^{-\frac12}$, so
\begin{align}
\int_{\left\{\omega\ :\ X(\omega)>t^{\frac12} \right\}} t\cdot\frac1{X(\omega)}\,\mathsf d\mathbb P(\omega)&<\int_{\left\{\omega\ :\ X(\omega)>t^{\frac12} \right\}} t^{\frac12}\,\mathsf d\mathbb P(\omega)\\
&=t^{\frac12}\mathbb P\left(X>t^{\frac12}\right)\\
&\stackrel{t\to 0}\longrightarrow 0.
\end{align}
It follows that 
$$\lim_{t\to 0^+}t \int_{\left\{\omega:X(\omega)>t \right\}}\frac1{X(\omega)}\,\mathsf d\mathbb P(\omega)=0. $$
