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

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(B) ધારો કે $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

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) ધારો કે $L = \mathop {\lim }\limits_{x 	o \pi /2} \left[ {x	an x - \left( {\frac{\pi }{2}} ight)\sec x} ight]$$L = \mathop {\lim }\limits_{x 	o \pi /2} \left[ {x \frac{\sin x}{\cos x} - \frac{\pi}{2} \frac{1}{\cos x}} ight] = \mathop {\lim }\limits_{x 	o \pi /2} \frac{2x \sin x - \pi}{2 \cos x}$જ્યારે $x 	o \pi/2$ હોય ત્યારે આ $\frac{0}{0}$ સ્વરૂપમાં છે.અંશ અને છેદનું $x$ ની સાપેક્ષમાં વિકલન કરીને $L$-Hospital ના નિયમનો ઉપયોગ કરતા:$L = \mathop {\lim }\limits_{x 	o \pi /2} \frac{\frac{d}{dx}(2x \sin x - \pi)}{\frac{d}{dx}(2 \cos x)} = \mathop {\lim }\limits_{x 	o \pi /2} \frac{2 \sin x + 2x \cos x}{-2 \sin x}$$x = \pi/2$ મુકતા:$L = \frac{2 \sin(\pi/2) + 2(\pi/2) \cos(\pi/2)}{-2 \sin(\pi/2)} = \frac{2(1) + \pi(0)}{-2(1)} = \frac{2}{-2} = -1$

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

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