MCQ
$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} |(1 - \sin x)\tan x$ is
  • A
    $\frac{\pi }{2}$
  • B
    $1$
  • $0$
  • D
    $\infty $

Answer

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

Apply $L-$ Hospital’s rule, we get

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

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