mathop {lim }limits_{x to 1} (1 - x)tan left( | vedclass

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(C) Let $L = \mathop {\lim }\limits_{x 	o 1} (1 - x)	an \left( {\frac{{\pi x}}{2}} ight)$.Substitute $y = 1 - x$,so as $x 	o 1$,$y 	o 0$ and $x

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$\frac{2}{\pi }$

Vedclass · Answer

(C) Let $L = \mathop {\lim }\limits_{x 	o 1} (1 - x)	an \left( {\frac{{\pi x}}{2}} ight)$.Substitute $y = 1 - x$,so as $x 	o 1$,$y 	o 0$ and $x = 1 - y$.Then $L = \mathop {\lim }\limits_{y 	o 0} y 	an \left( {\frac{{\pi (1 - y)}}{2}} ight) = \mathop {\lim }\limits_{y 	o 0} y 	an \left( {\frac{\pi }{2} - \frac{{\pi y}}{2}} ight)$.Using the identity $	an \left( {\frac{\pi }{2} - 	heta } ight) = \cot 	heta$,we get:$L = \mathop {\lim }\limits_{y 	o 0} y \cot \left( {\frac{{\pi y}}{2}} ight) = \mathop {\lim }\limits_{y 	o 0} \frac{y}{	an \left( {\frac{{\pi y}}{2}} ight)}$.Multiply and divide by $\frac{\pi }{2}$:$L = \mathop {\lim }\limits_{y 	o 0} \frac{1}{\frac{\pi }{2}} \cdot \frac{{\frac{{\pi y}}{2}}}{{	an \left( {\frac{{\pi y}}{2}} ight)}} = \frac{2}{\pi } \cdot 1 = \frac{2}{\pi }$.

$\mathop {\lim }\limits_{x \to 1} (1 - x)\tan \left( {\frac{{\pi x}}{2}} \right) = $

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