A) 0
B) 1
C) -1
D) ½
Correct Answer: A
Solution :
[a] \[f(x)=\underset{n\to \infty }{\mathop{\lim }}\,{{({{\cos }^{2}}x)}^{n}}\] \[=\left\{ \begin{matrix} 0, & 0\le {{\cos }^{2}}x<1 \\ 1, & {{\cos }^{2}}x=1 \\ \end{matrix}=\left\{ \begin{matrix} 0, & x\ne n\pi ,n\in I \\ 1, & x=n\pi ,n\in I \\ \end{matrix} \right. \right.\] Hence, \[f(x)\] is discontinuous when \[x=n\pi ,n\in I\] For this values of \[\theta ,\sin \theta =0\].
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