A) \[-1\]
B) \[1/2\]
C) 2
D) \[-2\]
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x-\sin x}{{{x}^{3}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x-\sin x\cos x}{\cos x.{{x}^{3}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x(1-\cos x)}{{{x}^{3}}.\cos x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x\left( 2{{\sin }^{2}}\frac{x}{2} \right)}{{{x}^{3}}.\cos x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\sin x}{x}.\frac{2}{\cos x}.\frac{{{\sin }^{2}}\frac{x}{2}}{{{\left( \frac{x}{2} \right)}^{2}}}.\frac{1}{4} \right]\] \[=1.2.1.\frac{1}{4}.\frac{1}{2}\] \[\left[ \because \underset{x\to \infty }{\mathop{\lim }}\,\frac{\sin x}{x}=1 \right]\]
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