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$\lim _{x \rightarrow \pi / 2} \tan x \log \sin x=$
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$\begin{aligned} & \lim _{x \rightarrow \frac{\pi}{2}} \tan x \log \sin x=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\log \sin x}{\cot x} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\frac{1}{\sin x} \cos x}{-\operatorname{cosec}^2 x}=0 \quad \text {(Applying L-Hospital's)}\end{aligned}$
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