Limits

Relevant wiki: Half Angle Tangent Substitution

$\begin{aligned} L & = \lim_{x \to \pi / 2} \frac {\left(1-\tan \frac x2\right)(1-\sin x)}{\left(1+\tan \frac x2\right)(\pi - 2x)^3} \\ & = \lim_{x \to \pi / 2} \frac {\left(1-\tan \frac x2\right)(1-\sin x)}{8\left(1+\tan \frac x2\right){\color{#3D99F6} \left(\frac \pi 2 - x\right)}^3} & \small \color{#3D99F6} \text{Let }u = \frac \pi 2 - x \implies x = \frac \pi 2 - u \\ & = \lim_{\color{#3D99F6} u \to 0} \frac {{\color{#D61F06}\left(1-\tan \left(\frac \pi 4 - \frac u2 \right) \right)}\left(1-\sin \left(\frac \pi 2 - u \right)\right)}{8{\color{#D61F06}\left(1+\tan \left(\frac \pi 4 - \frac u2 \right) \right)}{\color{#3D99F6} u}^3} \\ & = \lim_{u \to 0} \frac {{\color{#D61F06}\tan \frac u2} \left(1-{\color{#3D99F6}\cos u} \right)}{8u^3} & \small \color{#3D99F6} \text{By half-angle tangent substitution} \\ & = \lim_{u \to 0} \frac {\tan \frac u2 \left(1-{\color{#3D99F6}\frac {1-\tan^2 \frac u2}{1+\tan^2 \frac u2}} \right)}{8u^3} \\ & = \lim_{u \to 0} \frac {\tan^2 \frac u2 \sin u}{8u^3} \\ & = \lim_{u \to 0} \frac 1{32} \left(\frac {\tan \frac u2}{\frac u2}\right)^2 \frac {\sin u}u \\ & = \frac 1{32} = \boxed{0.03125} \end{aligned}$