A) \[1/2\]
B) \[1/3\]
C) \[1/6\]
D) \[1/12\]
Correct Answer: C
Solution :
\[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{-1}}x-x}{{{x}^{3}}\cos x}\] \[\left( \frac{0}{0}\,form \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{1}{\sqrt{1-{{x}^{2}}}}-1}{{{x}^{3}}(-\sin x)+3{{x}^{2}}\cos x}\] (using L? Hospital?s rule) \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\sqrt{1-{{x}^{2}}}}{\sqrt{1-{{x}^{2}}}.{{x}^{2}}(-x\,\sin \,x+3\,\cos \,x)}\] \[\times \frac{1+\sqrt{1-{{x}^{2}}}}{1+\sqrt{1-{{x}^{2}}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-1+{{x}^{2}}}{\left[ \begin{align} & {{x}^{2}}\sqrt{1-{{x}^{2}}}(1+\sqrt{1-{{x}^{2}}}) \\ & (-x\,\sin \,x+3\,\cos \,x) \\ \end{align} \right]}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\left[ \begin{align} & \sqrt{1-{{x}^{2}}}(1+\sqrt{1-{{x}^{2}}}) \\ & (-x\,\sin \,x+3\,\cos \,x) \\ \end{align} \right]}\] \[=\frac{1}{1(1+1)\,(3)}\] \[=\frac{1}{6}\]
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