MCQ
$\mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{\sin x - x + \frac{{{x^3}}}{6}}}{{{x^5}}}} \right\} = $
- ✓$1/120$
- B$-1/120$
- C$1/20$
- DNone of these
Aliter : Apply $L-$ Hospital’s rule
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x + \frac{{{x^3}}}{6}}}{{{x^5}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x - 1 + \frac{{3{x^2}}}{6}}}{{5{x^4}}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{ - \sin x + \frac{{6x}}{6}}}{{20{x^3}}} = \mathop {\lim }\limits_{x \to 0} \,\frac{{ - \cos x + 1}}{{60{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{120\,x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{120}} = \frac{1}{{120}}.$
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