Question

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

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

Answer 1

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

$l$ (say)$=\underset{x\to 1^{-}}{lim}\frac{1-\sqrt{x}}{{\left({\cos }^{-1}x\right)}^2}$ (of the $\left(\frac{0}{0}\right)$)

$l=\underset{x\to 1}{lim}\frac{(-\frac{1}{2}\frac{1}{\sqrt{x}})}{\left(2\left({\cos }^{-1}x\right)\left(\frac{-1}{\sqrt{1-x^2}}\right)\right)}$ (L-Hospital Rule)

$l=\underset{x\to 1}{lim}\frac{1}{4}.\frac{1}{\left({\cos }^{-1}x\right)}.\left(\sqrt{1-x^2}\right).\frac{1}{\sqrt{x}}$

$l=\underset{x\to 1}{lim}\frac{1}{4}.\frac{1.\frac{1}{2}.\frac{1}{\sqrt{1-x^2}}.(-2x)}{\frac{-\sqrt{x}}{\sqrt{1-x^2}}+\left({\cos }^{-1}x\right).\frac{-1}{2}.\frac{1}{\sqrt{x}}}$ (Apply L-Hospital again)

$l=\frac{1}{4}\underset{x\to 1}{lim}\frac{\frac{x}{\sqrt{1-x^2}}.\sqrt{x}}{\frac{2x+{\cos }^{-1}x\sqrt{1-x^2}}{\sqrt{1-x^2}}}=\frac{1}{4}\underset{x\to 1}{lim}\frac{x\sqrt{x}}{2x+{\cos }^{-1}x\sqrt{1-x^2}}$

$l=\frac{1}{4}\left(\frac{1}{2+0}\right)=\frac{1}{4}$