MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x}}{{{x^3}}} = $
- A$\frac{1}{3}$
- B$ - \frac{1}{3}$
- C$\frac{1}{6}$
- ✓$ - \frac{1}{6}$
Expand $sin\ x$, then
$ = \mathop {\lim }\limits_{x \to 0} \frac{{ - \frac{{{x^3}}}{{3\,!}} + \frac{{{x^5}}}{{5\,!}} - ...}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \left[ { - \frac{1}{{3\,!}} + \frac{{{x^2}}}{{5\,!}} - ...} \right] $
$= \frac{{ - 1}}{{3\,!}} = \frac{{ - 1}}{6}$.
Aliter : Apply $L-$ Hospital’s rule.
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