$\lim\limits_{x \rightarrow-\pi / 2} \tan x \log _e \sin x$ is equal to

0

$\lim\limits_{x \rightarrow \pi / 2} \tan x . \log  \sin x$

$=\lim\limits_{x \rightarrow \pi / 2} \frac{\log  \sin x}{\cot x}$

Using L'hopital's rule

$=\lim\limits_{x \rightarrow \pi / 2}\frac{\cos x/\sin x}{-cosec^2x}=\lim\limits_{x \rightarrow \pi / 2}-\cos x\sin x$

$=0$