_ x 0 x - x x^3 = - Vidyadip

MCQ

$\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{x^3}}} = $

✓
$\frac{1}{2}$
B
$ - \frac{1}{2}$
C
$\frac{2}{3}$
D
None of these

✓

Answer

Correct option: A.

$\frac{1}{2}$

a
(a) $\mathop {\lim }\limits_{x \to 0} \,\frac{{\tan x - \sin x}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \,\frac{{\sin x - \sin x\,\cos x}}{{{x^3}\cos x}}$

$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{\sin x\,\left( {2\,\,{{\sin }^2}\frac{x}{2}} \right)}}{{{x^3}\,\cos x}} $

$= \mathop {\lim }\limits_{x \to 0} \,\left[ {\frac{{\sin x}}{x}.\frac{2}{{\cos x}}.\frac{{{{\sin }^2}\frac{x}{2}}}{{{{\left( {\frac{x}{2}} \right)}^2}}}.\frac{1}{4}} \right] = \frac{1}{2}$.

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