_ x 0 x x - x x^2 x = - Vidyadip

MCQ

$\mathop {\lim }\limits_{x \to 0} \frac{{x\cos x - \sin x}}{{{x^2}\sin x}} = $

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

✓

Answer

Correct option: B.

$ - \frac{1}{3}$

b
(b) $\mathop {\lim }\limits_{x \to 0} \,\,\frac{{x\cos x - \sin x}}{{{x^2}\sin x}}$

$ = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{ - \sin x}}{{2\sin x + x\cos x}}$

(By $L-$ Hospital’s rule)

$ = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{ - \cos x}}{{3\cos x - x\sin x}} = - \frac{1}{3}$,

(Again by $L-$ Hospital’s rule)

$ = - \frac{1}{3}$

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