mathop {lim }limits_{x to 1} {left( {frac{4}{ | vedclass

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(D) Let $L = \mathop {\lim }\limits_{x 	o 1} {\left( {\frac{4}{\pi }{{	an }^{ - 1}}x} ight)^{\frac{1}{{({x^2} - 1)}}}}$Since the form is $1^{\inft

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$e^{\frac{1}{\pi}}$

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(D) Let $L = \mathop {\lim }\limits_{x 	o 1} {\left( {\frac{4}{\pi }{{	an }^{ - 1}}x} ight)^{\frac{1}{{({x^2} - 1)}}}}$Since the form is $1^{\infty}$,we use the formula $\mathop {\lim }\limits_{x 	o a} f(x)^{g(x)} = e^{\mathop {\lim }\limits_{x 	o a} g(x)(f(x) - 1)}$.$L = e^{\mathop {\lim }\limits_{x 	o 1} \frac{1}{x^2 - 1} \left( \frac{4}{\pi} 	an^{-1} x - 1 ight)}$$L = e^{\mathop {\lim }\limits_{x 	o 1} \frac{4}{\pi} \frac{	an^{-1} x - 	an^{-1} 1}{x^2 - 1}}$Using $L$'Hopital's rule or the derivative of $	an^{-1} x$ at $x=1$:$L = e^{\frac{4}{\pi} \mathop {\lim }\limits_{x 	o 1} \frac{\frac{1}{1+x^2}}{2x}} = e^{\frac{4}{\pi} \cdot \frac{1/2}{2}} = e^{\frac{4}{\pi} \cdot \frac{1}{4}} = e^{\frac{1}{\pi}}$

$\mathop {\lim }\limits_{x \to 1} {\left( {\frac{4}{\pi }{{\tan }^{ - 1}}x} \right)^{\frac{1}{{({x^2} - 1)}}}}$ is equal to -

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