mathop {lim }limits_{x to pi /2} left[ {xtan | vedclass

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(B) Let $L = \mathop {\lim }\limits_{x 	o \pi /2} \left[ {x	an x - \left( {\frac{\pi }{2}} ight)\sec x} ight]$$L = \mathop {\lim }\limits_{x 	o

Vedclass · Accepted Answer

$-1$

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(B) Let $L = \mathop {\lim }\limits_{x 	o \pi /2} \left[ {x	an x - \left( {\frac{\pi }{2}} ight)\sec x} ight]$$L = \mathop {\lim }\limits_{x 	o \pi /2} \left[ {x \frac{\sin x}{\cos x} - \frac{\pi}{2} \frac{1}{\cos x}} ight] = \mathop {\lim }\limits_{x 	o \pi /2} \frac{2x \sin x - \pi}{2 \cos x}$This is in the indeterminate form $\frac{0}{0}$ as $x 	o \pi/2$.Applying $L$-Hospital's rule by differentiating the numerator and denominator with respect to $x$:$L = \mathop {\lim }\limits_{x 	o \pi /2} \frac{\frac{d}{dx}(2x \sin x - \pi)}{\frac{d}{dx}(2 \cos x)} = \mathop {\lim }\limits_{x 	o \pi /2} \frac{2 \sin x + 2x \cos x}{-2 \sin x}$Substituting $x = \pi/2$:$L = \frac{2 \sin(\pi/2) + 2(\pi/2) \cos(\pi/2)}{-2 \sin(\pi/2)} = \frac{2(1) + \pi(0)}{-2(1)} = \frac{2}{-2} = -1$

$\mathop {\lim }\limits_{x \to \pi /2} \left[ {x\tan x - \left( {\frac{\pi }{2}} \right)\sec x} \right] = $

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