A) \[\frac{\pi }{2}\]
B) 1
C) 0
D) \[\infty \]
Correct Answer: C
Solution :
\[\underset{x\to \pi /2}{\mathop{\lim }}\,\,\left\{ (1-\sin x)\tan x \right\}=\underset{x\to \pi /2}{\mathop{\lim }}\,\,\frac{\sin x-{{\sin }^{2}}x}{\cos x}\] Apply L-Hospital?s rule, we get \[\underset{x\to \pi /2}{\mathop{\lim }}\,\,\frac{\cos x-\sin 2x}{-\sin x}=0\].
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