MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{\tan 2x - x}}{{3x - \sin x}} = $
  • A
    $0$
  • B
    $1$
  • $1/2$
  • D
    $1/3$

Answer

Correct option: C.
$1/2$
c
(c) $\mathop {\lim }\limits_{x \to 0} \,\,\,\frac{{\tan \,\,2x - x}}{{3x - \sin x}} = \mathop {\lim }\limits_{x \to 0} \,\,\,\left\{ {\frac{{\frac{{2\,\tan 2x}}{{2x}} - 1}}{{3 - \frac{{\sin x}}{x}}}} \right\} = \frac{1}{2}.$

Aliter : Apply $L-$ Hospital‘s rule

$\mathop {\lim }\limits_{x \to 0} \,\,\frac{{\tan 2x - x}}{{3x - \sin x}} = \mathop {\lim }\limits_{x \to 0} \,\frac{{2{{\sec }^2}2x - 1}}{{3 - \cos x}} = \frac{{2 - 1}}{{3 - 1}} = \frac{1}{2}.$

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