$\lim\limits_{x \rightarrow-\pi / 2} \tan x \log _e \sin x$ is equal to
Answer & explanation
Correct answer: option 1
$\lim\limits_{x \rightarrow \pi / 2} \tan x . \log \sin x$
$=\lim\limits_{x \rightarrow \pi / 2} \frac{\log \sin x}{\cot x}$
Using L'hopital's rule
$=\lim\limits_{x \rightarrow \pi / 2}\frac{\cos x/\sin x}{-cosec^2x}=\lim\limits_{x \rightarrow \pi / 2}-\cos x\sin x$
$=0$