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Question: \[\lim_{x \rightarrow 0}\frac{\tan x - \sin x}{x^{3}}\]...

lim⁡x→0tan⁡x−sin⁡xx3\lim_{x \rightarrow 0}\frac{\tan x - \sin x}{x^{3}}

A

12\frac{1}{2}

B

−12- \frac{1}{2}

C

23\frac{2}{3}

D

None of these

Answer

12\frac{1}{2}

Explanation

Solution

lim⁡x→0tan⁡x−sin⁡xx3=lim⁡x→0sin⁡x−sin⁡xcos⁡xx3cos⁡x\lim_{x \rightarrow 0}\frac{\tan x - \sin x}{x^{3}} = \lim_{x \rightarrow 0}\frac{\sin x - \sin x\cos x}{x^{3}\cos x}

=lim⁡x→0sin⁡x(2sin⁡2x2)x3cos⁡x= \lim_{x \rightarrow 0}\frac{\sin x\left( 2\sin^{2}\frac{x}{2} \right)}{x^{3}\cos x} =lim⁡x→0[sin⁡xx.2cos⁡x.sin⁡2x2(x2)2.14]=12= \lim_{x \rightarrow 0}\left\lbrack \frac{\sin x}{x}.\frac{2}{\cos x}.\frac{\sin^{2}\frac{x}{2}}{\left( \frac{x}{2} \right)^{2}}.\frac{1}{4} \right\rbrack = \frac{1}{2}