Question

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

• A. $\frac{-1}{4}$
• B. $1/2$
• C. $2$
• D. None of these

Answer 1

Answer: (A)

$\frac{-1}{4}$

${lim}_{x\to 1}\left(\frac{1-\sqrt{x}}{(co{s}^{-1}x{)}^2}\right){\Rightarrow lim}_{x\to 1}1-\sqrt{x}=0,{\Rightarrow lim}_{x\to 1}(co{s}^{-1}x{)}^2=0{lim}_{x\to 1}\left(\frac{1-\sqrt{x}}{(co{s}^{-1}x{)}^2}\right)=\frac{0}{0}Nowusing{l}^{\prime }hospita{l}^{\prime }srule{lim}_{x\to 1}\left(\frac{1-\sqrt{x}}{(co{s}^{-1}x{)}^2}\right)={lim}_{x\to 1}\left(\frac{\frac{d(1-\sqrt{x})}{dx}}{\frac{d(co{s}^{-1}x{)}^2}{dx}}\right)={lim}_{x\to 1}\frac{1-\frac{1}{2\sqrt{x}}}{2co{s}^{-1}x\ast \frac{-1}{\sqrt{1-x^2}}}Now{lim}_{x\to 1}1-\frac{1}{2\sqrt{x}}=1-1/2=1/2And{lim}_{x\to 1}\frac{-2co{s}^{-1}x}{\sqrt{1-x^2}}=\frac{0}{0}Againusing{l}^{\prime }hospita{l}^{\prime }srule{lim}_{x\to 1}\frac{-2co{s}^{-1}x}{\sqrt{1-x^2}}={lim}_{x\to 1}\frac{-2d\left(co{s}^{-1}x\right)/dx}{d\left(\sqrt{1-x^2}\right)/dx}={lim}_{x\to 1}\frac{-2\left(\frac{-1}{\sqrt{1-x^2}}\right)}{\frac{-2x}{2\sqrt{1-x^2}}}={lim}_{x\to 1}\frac{-2}{x}=-2Therefore,{lim}_{x\to 1}\frac{1-\frac{1}{2\sqrt{x}}}{2co{s}^{-1}x\ast \frac{-1}{\sqrt{1-x^2}}}=\frac{1/2}{-2}=-1/4$