Suppose that $\eta_n\xrightarrow{P}a$ and $0<\xi_n\xrightarrow{P}b>0$ with assumption that $\vert\frac{\eta_n}{\xi_n}\vert>0$. I want to show that $E(\frac{\eta_n}{\xi_n})\to\frac{a}{b}$.
How should I deal with expected value in this problem? At first I thought that it may be helpful to treat it as $E(\frac{\eta_n}{\xi_n}) = \int_\mathbb{R} x\frac{\eta_n}{\xi_n}dx$, but have no idea how to proceed further.
This is wrong even in the case where $\xi_n=1$ almost surely. Take for example $\eta_n$ such that $\mathbb P\left(\eta_n=n^2\right)=1/n$, $\mathbb P\left(\eta_n=1\right)=1-1/n$.