limx→π2 1−tan⁡x2[1−sin⁡x]1+tan⁡x2[π−2x]3 is

=Limx→π21−tan⁡x2(1−sin⁡x)1+tan⁡x2(π−2x)3=Limx→π2tan⁡π4−x2(1−sin⁡x)(π−2x)3=Limx→π2(sin⁡x−1)tan⁡π4−x2(2x−π)3=Limh→0−(cosh−1)tan⁡h2(2h)3 =Limh→0(1−cosh)tan⁡h28h3 =18Limh→01−coshh2tan⁡h2h =18Limh→02sin2h24h22tan⁡h2h2×2 =132

limx→π2 1−tan⁡x2[1−sin⁡x]1+tan⁡x2[π−2x]3 is

=Limx→π21−tan⁡x2(1−sin⁡x)1+tan⁡x2(π−2x)3=Limx→π2tan⁡π4−x2(1−sin⁡x)(π−2x)3=Limx→π2(sin⁡x−1)tan⁡π4−x2(2x−π)3=Limh→0−(cosh−1)tan⁡h2(2h)3 =Limh→0(1−cosh)tan⁡h28h3 =18Limh→01−coshh2tan⁡h2h =18Limh→02sin2h24h22tan⁡h2h2×2 =132

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$lim_{x \to \frac{π}{2}} \frac{[1 - \tan (\frac{x}{2})] [1 - \sin x]}{1 + \tan (\frac{x}{2})] [π - 2 x]^{3}} is$

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$\infty$

$\frac{1}{8}$

$\frac{1}{32}$

answer is D.

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Detailed Solution

$\begin{aligned} = \underset{x \to \frac{π}{2}}{L i m} \frac{(1 - \tan \frac{x}{2}) (1 - \sin x)}{(1 + \tan \frac{x}{2}) (π - 2 x)^{3}} \\ = \underset{x \to \frac{π}{2}}{L i m} \frac{\tan (\frac{π}{4} - \frac{x}{2}) (1 - \sin x)}{(π - 2 x)^{3}} \end{aligned}$

$= \underset{x \to \frac{π}{2}}{L i m} \frac{(\sin x - 1) \tan (\frac{π}{4} - \frac{x}{2})}{(2 x - π)^{3}}$

$= \underset{h \to 0}{L i m} \frac{- (\cosh - 1) \tan (\frac{h}{2})}{(2 h)^{3}} = \underset{h \to 0}{L i m} \frac{(1 - \cosh) \tan (\frac{h}{2})}{8 h^{3}} = \frac{1}{8} \underset{h \to 0}{L i m} (\frac{1 - \cosh}{h^{2}}) \frac{\tan (\frac{h}{2})}{h} = \frac{1}{8} \underset{h \to 0}{L i m} (\frac{2 \sin^{2} \frac{h}{2}}{4 {(\frac{h}{2})}^{2}}) \frac{\tan (\frac{h}{2})}{\frac{h}{2} \times 2} = \frac{1}{32}$