$\lim_{x \to \frac{π}{2}} \frac{(1 - \sin x) (8 x^{3} - π^{3}) \cos x}{(π - 2 x)^{4}}$

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$- \frac{π^{2}}{16}$

$\frac{3 π^{2}}{16}$

$\frac{π^{2}}{16}$

$- \frac{3 π^{2}}{16}$

answer is D.

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Detailed Solution

Let $f (x) = \frac{(1 - \sin x) (8 x^{3} - π^{3}) \cos x}{(π - 2 x)^{4}} = \frac{(1 - \sin x) \cos x (2 x - π) (4 x^{2} + 2 πx + π^{2})}{(2 x - π)^{4}}$

$= \frac{(1 - \sin x) \cos x (4 x^{2} + 2 πx + π^{2})}{(2 x - π)^{3}} Therefore \lim_{x \to \frac{π}{2}} f (x) = \lim_{x \to \frac{π}{2}} \frac{(1 - \sin x) \cos x}{(2 x - π)^{3}} \cdot (3 π^{2})$

$\lim_{x \to \frac{π}{2}} f (x) = \lim_{x \to \frac{π}{2}} \frac{(1 - \sin x) \cos x}{(2 x - π)^{3}} \cdot (3 π^{2})$

Put $2 x - π = y$ so that $y \to 0$ as $x \to π / 2$ .

. Therefore now $\begin{array}{r} \frac{(1 - \sin x) \cos x}{(2 x - π)^{3}} = \frac{[1 - \sin (\frac{π + y}{2})] \cos (\frac{π + y}{2})}{y^{3}} = \frac{(1 - \cos \frac{y}{2}) (- \sin \frac{y}{2})}{y^{3}} = - (\frac{2 \sin^{2} \frac{y}{4}}{y^{2}}) (\frac{\sin \frac{y}{2}}{y}) \\ = - 2 {(\frac{\sin \frac{y}{4}}{y / 4})}^{2} \cdot \frac{1}{16} \cdot (\frac{\sin \frac{y}{2}}{y / 2}) \cdot \frac{1}{2} = \frac{- 1}{16} {(\frac{\sin \frac{y}{4}}{y / 4})}^{2} (\frac{\sin \frac{y}{2}}{y / 2}) \end{array}$