question_answer

\[\underset{x\to 1}{\mathop{\lim }}\,\frac{1-\sqrt{x}}{{{({{\cos }^{-1}}x)}^{2}}}=\] [AI CBSE 1990]

A)                 1

B)                 \[\frac{1}{2}\]

C)                 \[\frac{1}{4}\]

D)                    Put \[{{\cos }^{-1}}x=y\] and \[x\to 1\,\Rightarrow \,\,y\to 0.\]            \[\underset{x\to 1}{\mathop{\lim }}\,\,\frac{1-\sqrt{x}}{{{({{\cos }^{-1}}x)}^{2}}}=\underset{y\to 0}{\mathop{\lim }}\,\,\frac{1-\sqrt{\cos y}}{{{y}^{2}}}\]            Now rationalizing it, we get \[\underset{y\to 0}{\mathop{\lim }}\,\,\frac{(1-\cos y)}{{{y}^{2}}(1+\sqrt{\cos y})}\]                                 \[=\underset{y\to 0}{\mathop{\lim }}\,\,\frac{1-\cos y}{{{y}^{2}}}\,.\,\underset{y\to 0}{\mathop{\lim }}\,\,\frac{1}{1+\sqrt{\cos y}}=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}.\]