A) \[\frac{-1}{2}\]
B) \[\frac{1}{2}\]
C) 1
D) \[-1\]
Correct Answer: D
Solution :
\[\underset{x\to \pi }{\mathop{Lim}}\,\,\,\frac{(1+\cos x)-\sin x}{(1+\cos x)+\sin x}\,=\underset{x\to \pi }{\mathop{Lim}}\,\,\frac{2{{\cos }^{2}}\,\frac{x}{2}-2\sin \,\frac{x}{2}.\cos \frac{x}{2}}{2{{\cos }^{2}}\,\frac{x}{2}\,+2\sin \,\frac{x}{2}\,.\cos \,\frac{x}{2}}\]\[=\underset{x\to \pi }{\mathop{Lim}}\,\tan \left( \frac{\pi }{4}-\frac{\pi }{2} \right)\,\,=-1\] Since, \[f(x)\] is continuous at \[x=\pi ,\] So, \[f(\pi )\,=\underset{x\to \pi }{\mathop{Lim}}\,\,f(x)\,=-1\]You need to login to perform this action.
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