Question

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

• A. $4$
• B. $\frac{1}{2}$
• C. $2$
• D. $\frac{1}{4}$

Answer 1

The correct option is D $\frac{1}{4}$

$\underset{x\to 1}{lim}\frac{1-\sqrt{x}}{{\left(co{s}^{-1}x\right)}^2}$
$=\underset{x\to 1}{lim}\frac{(1-\sqrt{x})(1+\sqrt{x})}{{\left({\cos }^{-1}x\right)}^2(1+\sqrt{x})}$
$=\underset{x\to 1}{lim}\frac{(1-x)}{{\left({\cos }^{-1}x\right)}^2(1+\sqrt{x})}$
Let ${\cos }^{-1}x=\theta$ Then $x=\cos \theta$
As $x\to 1\Rightarrow \theta \to 0$
$\underset{\theta \to 0}{lim}\frac{1-\cos \theta }{{\theta }^2(1+\sqrt{\cos \theta })}$
$=\underset{\theta \to 0}{lim}\frac{1-\cos \theta }{{\theta }^2}\frac{1}{(1+\sqrt{\cos \theta })}$
$=\underset{\theta \to 0}{lim}\frac{2{\sin }^2\frac{\theta }{2}}{4\frac{{\theta }^2}{4}\left(\frac{1}{1+\sqrt{\cos \theta }}\right)}$
$=\frac{1}{2}\underset{\theta \to 0}{lim}{\left(\frac{\sin \frac{\theta }{2}}{\frac{\theta }{2}}\right)}^2\frac{1}{1+\sqrt{\cos \theta }}$
$=\frac{1}{4}$