QuestionReportQuestion: \(f(\pi) = 1f\left( \frac{\pi}{2} \right) = - 1\)=...f(π)=1f(π2)=−1f(\pi) = 1f\left( \frac{\pi}{2} \right) = - 1f(π)=1f(2π)=−1=A1B2C0DDoes not existAnswer0ExplanationSolution = sin∞∞\frac { \sin \infty } { \infty }∞sin∞ = = 0
= ∞sin∞ = 
= 0
= ∞sin∞ = = 0