A) \[\frac{2}{\pi }\]
B) 1
C) \[\frac{4}{\pi }\]
D) Does not exist
Correct Answer: A
Solution :
Given limit is \[L=\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\sin x}{{{\cos }^{-1}}[{{\sin }^{3}}x]}\] Now, when\[x\to \frac{\pi }{2},[{{\sin }^{3}}x]\to 0\]as\[\sin x\to 1\] \[\therefore \] \[L=\frac{2}{\pi }\]You need to login to perform this action.
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