Functions Question 172
Question: $ \underset{x\to -1}{\mathop{\lim }},\frac{\sqrt{\pi }-\sqrt{{{\cos }^{-1}}x}}{\sqrt{x+1}} $ is given by
Options:
A) $ \frac{1}{\sqrt{\pi }} $
B) $ \frac{1}{\sqrt{2\pi }} $
C) 1
D) 0
Show Answer
Answer:
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 ,( \frac{\pi }{2}-\frac{y}{2} )}\frac{( \frac{\pi }{2}-\frac{y}{2} )}{( \frac{\pi }{2}-\frac{y}{2} )} $ $ =\underset{y\to \pi }{\mathop{\lim }}\frac{1}{\frac{\sqrt{2}}{2}(\sqrt{\pi }+\sqrt{y})}.\frac{1}{\frac{\sin ,( \frac{\pi }{2}-\frac{y}{2} )}{( \frac{\pi }{2}-\frac{y}{2} )}}=\frac{1}{\sqrt{2\pi }}. $
BETA
