Question

The value of $\underset{x\to 1^{-}}{\lim }\frac{1-\sqrt{x}}{{\left({\cos }^{-1}x\right)}^2}$ is

• A. 4
• B. 1 / 2
• C. 2
• D. 1 / 4

Answer 1

Answer: (D)

We have
$\underset{x\to 1}{\lim }\frac{1-\sqrt{x}}{{\left({\cos }^{-1}x\right)}^2}$
$=\underset{x\to 1}{\lim }\frac{(1-\sqrt{x})(1+\sqrt{x})}{{\left({\cos }^{-1}x\right)}^2(1+\sqrt{x})}$
$=\underset{x\to 1}{\lim }\frac{1-x}{{\left({\cos }^{-1}x\right)}^2(1+\sqrt{x})}$
$=\underset{\theta \to 0}{\lim }\frac{1-\cos \theta }{{\theta }^2(1+\sqrt{\cos \theta })}1$,
where $x=\cos \theta$
$[\because x\to 1$ or $\cos \theta \to 1$ or $\theta \to 0]$
$=\underset{\theta \to 0}{\lim }\frac{1-\cos \theta }{{\theta }^2}\frac{1}{(1+\sqrt{\cos \theta })}$
$=\underset{\theta \to 0}{\lim }\frac{2{\sin }^2\frac{\theta }{2}}{{\theta }^2}\left(\frac{1}{1+\sqrt{\cos \theta }}\right)$
$=\frac{1}{2}\underset{\theta \to 0}{\lim }{\left(\frac{\sin \frac{\theta }{2}}{\frac{\theta }{2}}\right)}^2\frac{1}{(1+\sqrt{\cos \theta })}$
$=\frac{1}{2}(1{)}^2\frac{1}{(1+1)}=\frac{1}{4}$