mathop {lim }limits_{x to 1} frac{{1 - sqrt x | vedclass

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(C) Let ${\cos ^{ - 1}}x = y$. As $x 	o 1$,$y 	o 0$. Also,$x = \cos y$.$\mathop {\lim }\limits_{x 	o 1} \frac{{1 - \sqrt x }}{{{{({{\cos }^{ - 1}}x

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(C) Let ${\cos ^{ - 1}}x = y$. As $x 	o 1$,$y 	o 0$. Also,$x = \cos y$.$\mathop {\lim }\limits_{x 	o 1} \frac{{1 - \sqrt x }}{{{{({{\cos }^{ - 1}}x)}^2}}} = \mathop {\lim }\limits_{y 	o 0} \frac{{1 - \sqrt {\cos y} }}{{{y^2}}}$Rationalizing the numerator:$= \mathop {\lim }\limits_{y 	o 0} \frac{{(1 - \sqrt {\cos y} )(1 + \sqrt {\cos y} )}}{{{y^2}(1 + \sqrt {\cos y} )}} = \mathop {\lim }\limits_{y 	o 0} \frac{{1 - \cos y}}{{{y^2}(1 + \sqrt {\cos y} )}}$Using the standard limit $\mathop {\lim }\limits_{y 	o 0} \frac{{1 - \cos y}}{{{y^2}}} = \frac{1}{2}$:$= \left( \mathop {\lim }\limits_{y 	o 0} \frac{{1 - \cos y}}{{{y^2}}} ight) 	imes \left( \mathop {\lim }\limits_{y 	o 0} \frac{1}{{1 + \sqrt {\cos y} }} ight)$$= \frac{1}{2} 	imes \frac{1}{{1 + \sqrt 1 }} = \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4}$.

$\mathop {\lim }\limits_{x \to 1} \frac{{1 - \sqrt x }}{{{{({{\cos }^{ - 1}}x)}^2}}} = $

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