A) \[\frac{1}{\sqrt{\pi }}\]
B) \[\frac{1}{\sqrt{2\pi }}\]
C) 1
D) 0
Correct Answer: B
Solution :
Put \[{{\cos }^{-1}}x=y.\] So if \[x\to -1,\,\,y\to \pi \] \[\therefore \,\,\,\underset{x\to -1}{\mathop{\lim }}\,\,\frac{\sqrt{\pi }-\sqrt{{{\cos }^{-1}}x}}{\sqrt{x+1}}=\underset{y\to \pi }{\mathop{\lim }}\,\,\frac{\sqrt{\pi }-\sqrt{y}}{\sqrt{1+\cos y}}\] \[=\underset{y\to \pi }{\mathop{\lim }}\,\,\frac{\sqrt{\pi }-\sqrt{y}}{\sqrt{2}\,\cos \,(y/2)}\,=\underset{y\to \pi }{\mathop{\lim }}\,\,\,\frac{\sqrt{\pi }-\sqrt{y}}{\sqrt{2}\,\sin \,\left( \frac{\pi }{2}-\frac{y}{2} \right)}\frac{\left( \frac{\pi }{2}-\frac{y}{2} \right)}{\left( \frac{\pi }{2}-\frac{y}{2} \right)}\] \[=\underset{y\to \pi }{\mathop{\lim }}\,\,\frac{1}{\frac{\sqrt{2}}{2}(\sqrt{\pi }+\sqrt{y})}.\frac{1}{\frac{\sin \,\left( \frac{\pi }{2}-\frac{y}{2} \right)}{\left( \frac{\pi }{2}-\frac{y}{2} \right)}}=\frac{1}{\sqrt{2\pi }}.\]
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