\[\underset{x\to -1}{\mathop{\lim }}\,\frac{\sqrt{\pi }-\sqrt{{{\cos }^{-1}}x}}{\sqrt{x+1}}\]is given by

\[\underset{x\to -1}{\mathop{\lim }}\,\frac{\sqrt{\pi }-\sqrt{{{\cos }

1.	\[\underset{x\to -1}{\mathop{\lim }}\,\frac{\sqrt{\pi }-\sqrt{{{\cos }^{-1}}x}}{\sqrt{x+1}}\]is given by
A.	\[\frac{1}{\sqrt{\pi }}\]
B.	\[\frac{1}{\sqrt{2\pi }}\]
C.	1
D.	0
Answer» C. 1

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\[\underset{x\to -1}{\mathop{\lim }}\,\frac{\sqrt{\pi }-\sqrt{{{\cos }^{-1}}x}}{\sqrt{x+1}}\]is given by

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