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