mathop {lim }limits_{x to pi /2} tan x log si | vedclass

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Question

(A) Let $L = \mathop {\lim }\limits_{x 	o \frac{\pi }{2}} 	an x \log \sin x$.This is an indeterminate form of type $\infty 	imes 0$.We can rewrite

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(A) Let $L = \mathop {\lim }\limits_{x 	o \frac{\pi }{2}} 	an x \log \sin x$.This is an indeterminate form of type $\infty 	imes 0$.We can rewrite the expression as $L = \mathop {\lim }\limits_{x 	o \frac{\pi }{2}} \frac{\log \sin x}{\cot x}$.Applying $L'	ext{Hospital's rule}$,we differentiate the numerator and denominator with respect to $x$:$L = \mathop {\lim }\limits_{x 	o \frac{\pi }{2}} \frac{\frac{d}{dx}(\log \sin x)}{\frac{d}{dx}(\cot x)} = \mathop {\lim }\limits_{x 	o \frac{\pi }{2}} \frac{\frac{1}{\sin x} \cos x}{-\csc^2 x}$.Simplifying the expression:$L = \mathop {\lim }\limits_{x 	o \frac{\pi }{2}} \frac{\cot x}{-\csc^2 x} = \mathop {\lim }\limits_{x 	o \frac{\pi }{2}} (-\cos x \sin x)$.As $x 	o \frac{\pi }{2}$,$\cos x 	o 0$ and $\sin x 	o 1$.Therefore,$L = -(0 	imes 1) = 0$.

$\mathop {\lim }\limits_{x \to \pi /2} \tan x \log \sin x = $

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