A) 0
B) \[\infty \]
C) 1
D) None
Correct Answer: C
Solution :
Period of \[{{\sin }^{2m}}=\frac{\pi }{\sqrt{k}}\] According to question,\[=\frac{\pi }{\sqrt{k}}=\pi \Rightarrow k=1\] \[\therefore \]\[\underset{n\to \infty }{\mathop{Lt}}\,{{k}^{n}}=\underset{n\to \infty }{\mathop{Lt}}\,{{1}^{n}}=1\]You need to login to perform this action.
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