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

The value of the limit lim_(x→π/2) (4√2(sin 3x+sin x))/((2sin 2xsin3x/2+cos5x/2)-(√2+√2cos 2x+cos3x/2)) is

$\lim_{x\to\frac{\pi}{2}} \dfrac{4\sqrt{2}\,(\sin 3x+\sin x)}{\left(2\sin 2x\sin\frac{3x}{2}+\cos\frac{5x}{2}\right)-\left(\sqrt{2}+\sqrt{2}\cos 2x+\cos\frac{3x}{2}\right)}$

$\text{Find the value of this limit } \left(\tfrac{0}{0}\ \text{form}\right)$

$\text{1. Put } \theta = x - \dfrac{\pi}{2},\ \text{so } \theta \to 0$

$\text{2. Rewrite } N \text{ and } D \text{ in } \theta$

$\text{3. Expand for small } \theta, \text{ then take the ratio}$

$x = \dfrac{\pi}{2} + \theta,\qquad \theta \to 0$

$\lim_{x\to\pi/2} = \lim_{\theta\to 0}$

$\sin 3x + \sin x = 2\sin 2x\cos x$

$N = 4\sqrt{2}\cdot 2\sin 2x\cos x = 8\sqrt{2}\,\sin 2x\cos x$

$\sin 2x = -\sin 2\theta,\qquad \cos x = -\sin\theta$

$N = 8\sqrt{2}\,\sin 2\theta\,\sin\theta \approx 16\sqrt{2}\,\theta^{2}$

$\sin 2x = -\sin 2\theta$

$\sin\frac{3x}{2} = \frac{\sqrt{2}}{2}\left(\cos\frac{3\theta}{2}-\sin\frac{3\theta}{2}\right)$

$\cos\frac{5x}{2} = \frac{\sqrt{2}}{2}\left(-\cos\frac{5\theta}{2}+\sin\frac{5\theta}{2}\right)$

$\cos 2x = -\cos 2\theta$

$\cos\frac{3x}{2} = -\frac{\sqrt{2}}{2}\left(\cos\frac{3\theta}{2}+\sin\frac{3\theta}{2}\right)$

$D = -\sqrt{2}\sin 2\theta\left(\cos\frac{3\theta}{2}-\sin\frac{3\theta}{2}\right) - \sqrt{2}(1-\cos 2\theta)$

$+ \frac{\sqrt{2}}{2}\left(-\cos\frac{5\theta}{2}+\sin\frac{5\theta}{2}\right) + \frac{\sqrt{2}}{2}\left(\cos\frac{3\theta}{2}+\sin\frac{3\theta}{2}\right)$

$D \approx 2\sqrt{2}\,\theta^{2}$

$\lim_{\theta\to 0} \dfrac{N}{D} = \dfrac{16\sqrt{2}\,\theta^{2}}{2\sqrt{2}\,\theta^{2}} = 8$

$\boxed{\text{Answer: } 8}$