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

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(B) Let ${\cos ^{ - 1}}x = y$. As $x 	o - 1$,$y 	o \pi$. Then,the limit becomes $\mathop {\lim }\limits_{y 	o \pi } \frac{{\sqrt \pi - \sqrt y }}{{

Vedclass · Accepted Answer

$\frac{1}{{\sqrt {2\pi } }}$

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(B) Let ${\cos ^{ - 1}}x = y$. As $x 	o - 1$,$y 	o \pi$. Then,the limit becomes $\mathop {\lim }\limits_{y 	o \pi } \frac{{\sqrt \pi - \sqrt y }}{{\sqrt {1 + \cos y} }}$. Using the identity $1 + \cos y = 2 \cos^2(y/2)$,we have $\sqrt{1 + \cos y} = \sqrt{2} |\cos(y/2)|$. Since $y 	o \pi$,$\cos(y/2) > 0$,so $\sqrt{1 + \cos y} = \sqrt{2} \cos(y/2)$. Now,$\cos(y/2) = \sin(\pi/2 - y/2)$. So the limit is $\mathop {\lim }\limits_{y 	o \pi } \frac{{\sqrt \pi - \sqrt y }}{{\sqrt{2} \sin(\frac{\pi - y}{2})}}$. Multiply numerator and denominator by $(\sqrt{\pi} + \sqrt{y})$ and use the standard limit $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$: $= \mathop {\lim }\limits_{y 	o \pi } \frac{\pi - y}{\sqrt{2} \sin(\frac{\pi - y}{2}) (\sqrt{\pi} + \sqrt{y})} = \mathop {\lim }\limits_{y 	o \pi } \frac{2 \cdot \frac{\pi - y}{2}}{\sqrt{2} \sin(\frac{\pi - y}{2}) (\sqrt{\pi} + \sqrt{y})} = \frac{2}{\sqrt{2} (2\sqrt{\pi})} = \frac{1}{\sqrt{2\pi}}$.

$\mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt \pi - \sqrt {{{\cos }^{ - 1}}x} }}{{\sqrt {x + 1} }}$ is equal to

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