The value of mathop {lim }limits_{x to frac{p | vedclass

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(C) We know that $\frac{1 - 	an(x/2)}{1 + 	an(x/2)} = 	an(\frac{\pi}{4} - \frac{x}{2})$.So,the limit becomes $\mathop {\lim }\limits_{x 	o \frac{\

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$\frac{1}{32}$

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(C) We know that $\frac{1 - 	an(x/2)}{1 + 	an(x/2)} = 	an(\frac{\pi}{4} - \frac{x}{2})$.So,the limit becomes $\mathop {\lim }\limits_{x 	o \frac{\pi }{2}} \frac{{	an \left( {\frac{\pi }{4} - \frac{x}{2}} ight)\,(1 - \sin x)}}{{{{(\pi - 2x)}^3}}}$.Let $x = \frac{\pi }{2} + y$,where $y 	o 0$. Then $\pi - 2x = \pi - 2(\frac{\pi}{2} + y) = -2y$.Also,$1 - \sin x = 1 - \sin(\frac{\pi}{2} + y) = 1 - \cos y = 2\sin^2(\frac{y}{2})$.Substituting these,we get $\mathop {\lim }\limits_{y 	o 0} \frac{{	an \left( {\frac{\pi}{4} - (\frac{\pi}{4} + \frac{y}{2})} ight)\,(2\sin^2 \frac{y}{2})}}{{{{( - 2y)}^3}}}$.$= \mathop {\lim }\limits_{y 	o 0} \frac{{-	an(\frac{y}{2}) \cdot 2\sin^2(\frac{y}{2})}}{-8y^3} = \mathop {\lim }\limits_{y 	o 0} \frac{1}{4} \cdot \frac{	an(y/2)}{y} \cdot \left(\frac{\sin(y/2)}{y}ight)^2$.$= \frac{1}{4} \cdot \frac{1}{2} \cdot (\frac{1}{2})^2 = \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{32}$.

The value of $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\left[ {1 - \tan \left( {\frac{x}{2}} \right)} \right]\,[1 - \sin x]}}{{\left[ {1 + \tan \left( {\frac{x}{2}} \right)} \right]\,{{[\pi - 2x]}^3}}}$ is

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