_ x → (π)/(2) [1 - tan ((x)/(2))][1 - x] [1 + ((x)/(2))] [π - 2x] ^...

$\lim_{x \to \dfrac{\pi}{2}}\dfrac{\tan \left(\dfrac{\pi}{4} - \dfrac{x}{2}\right)\left(1 - \sin x\right)}{{\left(\pi - 2x\right)}^{3}}$

Let, $x = \dfrac{\pi}{2} + y$

$⇒ \lim_{y \to 0} = \dfrac{\tan \left(\dfrac{- y}{2}\right)\left(1 - \cos y\right)}{\left(- 2y\right)^{3}}$

$= \lim_{y \to 0}\dfrac{- tan\dfrac{y}{2} ⋅ 2sin^{2}\dfrac{y}{2}}{\left(- 8\right)y^{3}} = \lim_{y \to 0}\dfrac{1}{32}tan\dfrac{y}{2}\left(\dfrac{y}{2}\right) ⋅ \left[\dfrac{sin\dfrac{y}{2}}{y2}\right]2 = \lim_{y \to 0}\dfrac{1}{32} \times \dfrac{tan\dfrac{y}{2}}{( \dfrac{y}{2} )}. \left(\lim_{y \to 0}\dfrac{sin\dfrac{y}{2}}{y2}\right)^{2}$

$∵ \lim_{x \to 0} \dfrac{\sin x}{x} = \lim_{x \to 0} \dfrac{\tan x}{x} = 1$

$⇒ \dfrac{1}{32} \times 1 \times {1}^{2}$

$= \dfrac{1}{32}$