mathop {Limit}limits_{x to frac{pi }{2}} ,fra | vedclass

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Question

(A) We know that $\sin(3x) = 3\sin x - 4\sin^3 x$. Therefore,$\frac{1}{4}(3\sin x - \sin 3x) = \sin^3 x$. The expression becomes $\mathop {Limit}\limi

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$\frac{2}{\pi }$

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(A) We know that $\sin(3x) = 3\sin x - 4\sin^3 x$. Therefore,$\frac{1}{4}(3\sin x - \sin 3x) = \sin^3 x$. The expression becomes $\mathop {Limit}\limits_{x 	o \frac{\pi }{2}} \frac{\sin x}{\cos^{-1}[\sin^3 x]}$. As $x 	o \frac{\pi}{2}$,$\sin x 	o 1$,so $\sin^3 x 	o 1$. Since $\sin^3 x Thus,the limit is $\mathop {Limit}\limits_{x 	o \frac{\pi}{2}^-} \frac{\sin x}{\cos^{-1}(0)} = \frac{1}{\frac{\pi}{2}} = \frac{2}{\pi}$. For $x > \frac{\pi}{2}$,$\sin x The limit is $\frac{2}{\pi}$ from both sides.

$\mathop {Limit}\limits_{x \to \frac{\pi }{2}} \,\frac{{\sin x}}{{{{\cos }^{ - 1}}\left[ {\frac{1}{4}\,(3\sin x\, - \,\sin 3x)} \right]}}\,$,where $[ \cdot ]$ denotes the greatest integer function,is

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