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