A) \[\frac{-1}{2}\]
B) \[0\]
C) \[2\sqrt{2}\]
D) \[2\]
Correct Answer: C
Solution :
\[=\underset{x\to -\pi }{\mathop{\lim }}\,\frac{\left( {{\sin }^{-1}}(\sin x) \right)\cos x}{2(x+\pi )}\left( \frac{0}{0}\text{form} \right)\] \[=\underset{x\to -\pi }{\mathop{\lim }}\,\frac{-(x+\pi )}{2(x+\pi )}\cos x=\frac{1}{2}\]You need to login to perform this action.
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