lim _{x rightarrow 1}left((1-x) tan left(frac | vedclass

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(D) Let $L = \lim _{x ightarrow 1}(1-x) 	an \left(\frac{\pi x}{2}ight)$.This is an indeterminate form of type $0 	imes \infty$.We can rewrite th

Vedclass · Accepted Answer

$\frac{2}{\pi}$

Vedclass · Answer

(D) Let $L = \lim _{x ightarrow 1}(1-x) 	an \left(\frac{\pi x}{2}ight)$.This is an indeterminate form of type $0 	imes \infty$.We can rewrite the limit as:$L = \lim _{x ightarrow 1} \frac{1-x}{\cot \left(\frac{\pi x}{2}ight)}$.Applying $L$'Hospital's rule by differentiating the numerator and denominator with respect to $x$:$L = \lim _{x ightarrow 1} \frac{\frac{d}{dx}(1-x)}{\frac{d}{dx}(\cot \left(\frac{\pi x}{2}ight))}$$L = \lim _{x ightarrow 1} \frac{-1}{-\csc^2 \left(\frac{\pi x}{2}ight) \cdot \frac{\pi}{2}}$$L = \frac{1}{\frac{\pi}{2} \csc^2 \left(\frac{\pi}{2}ight)}$Since $\csc \left(\frac{\pi}{2}ight) = 1$,we get:$L = \frac{1}{\frac{\pi}{2} \cdot 1^2} = \frac{2}{\pi}$.Thus,the correct option is $D$.

$\lim _{x \rightarrow 1}\left((1-x) \tan \left(\frac{\pi x}{2}\right)\right)=$

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