A) \[\frac{1}{2}\]
B) \[-\frac{1}{2}\]
C) \[\frac{2}{3}\]
D) None of these
Correct Answer: A
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\tan x-\sin x}{{{x}^{3}}}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sin x-\sin x\,\cos x}{{{x}^{3}}\cos x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sin x\,\left( 2\,\,{{\sin }^{2}}\frac{x}{2} \right)}{{{x}^{3}}\,\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\,\left[ \frac{\sin x}{x}.\frac{2}{\cos x}.\frac{{{\sin }^{2}}\frac{x}{2}}{{{\left( \frac{x}{2} \right)}^{2}}}.\frac{1}{4} \right]=\frac{1}{2}\].
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