The value of the limit limx → π242(sin3x+sinx)(2sin2xsin3x2+cos5x2)−(2+2cos2x+cos3x2) is

Ltx → π242(sin3x+sinx)(2sin2x . sin3x2+cos5x2)−(2+2cos2x+cos3x2)=Ltx → π282(sin3x+sinx)(cosx2−cos7x2+cos5x2)−(2 . 2cos2x+cos3x2)=Ltx → π2162sinx.cos2x2sinx.sinx2+2sin3x.sinx2−22cos2x=Ltx → π2162sinx.cos2x8sinx2sinx−22=1628(12)(1)−22=8

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The value of the limit $\lim_{x \to \frac{π}{2}} \frac{4 \sqrt{2} (\sin 3 x + \sin x)}{(2 \sin 2 x \sin \frac{3 x}{2} + \cos \frac{5 x}{2}) - (\sqrt{2} + \sqrt{2} \cos 2 x + \cos \frac{3 x}{2})}$ is

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answer is 8.

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Detailed Solution

$\underset{x \to \frac{π}{2}}{L t} \frac{4 \sqrt{2} (\sin 3 x + \sin x)}{(2 \sin 2 x . \sin \frac{3 x}{2} + \cos \frac{5 x}{2}) - (\sqrt{2} + \sqrt{2} \cos 2 x + \cos \frac{3 x}{2})}$

$= \underset{x \to \frac{π}{2}}{L t} \frac{8 \sqrt{2} (\sin 3 x + \sin x)}{(\cos \frac{x}{2} - \cos \frac{7 x}{2} + \cos \frac{5 x}{2}) - (\sqrt{2} . 2 \cos^{2} x + \cos \frac{3 x}{2})}$

$= \underset{x \to \frac{π}{2}}{L t} \frac{16 \sqrt{2} \sin x . \cos^{2} x}{2 \sin x . \sin \frac{x}{2} + 2 \sin 3 x . \sin \frac{x}{2} - 2 \sqrt{2} \cos^{2} x} = \underset{x \to \frac{π}{2}}{L t} \frac{16 \sqrt{2} \sin x . \cos^{2} x}{8 \sin \frac{x}{2} \sin x - 2 \sqrt{2}} = \frac{16 \sqrt{2}}{8 (\frac{1}{\sqrt{2}}) (1) - 2 \sqrt{2}} = 8$