Question

The value of the limit
$\underset{x\to \frac{\pi }{2}}{\lim }\frac{4\sqrt{2}(\sin 3x+\sin x)}{\left(2\sin 2x\sin \frac{3x}{2}+\cos \frac{5x}{2}\right)-\left(\sqrt{2}+\sqrt{2}\cos 2x+\cos \frac{3x}{2}\right)}$ is _______.

Answer 1

Answer: 8

$\underset{x\to \frac{\pi }{2}}{\lim }\left(\frac{4\sqrt{2}\cdot 2\sin 2x\cos x}{2\sin 2x\sin \frac{3x}{2}+\left(\cos \frac{5x}{2}-\cos \frac{3x}{2}\right)-\sqrt{2}(1+\cos 2x)}\right)$
$\Rightarrow \underset{x\to \frac{\pi }{2}}{\lim }\left(\frac{16\sqrt{2}\sin x{\cos }^{2}x}{8\sin x\sin \frac{x}{2}\cdot {\cos }^{2}x-2\sqrt{2}{\cos }^{2}x}\right)$
$\Rightarrow \underset{x\to \frac{\pi }{2}}{\lim }\frac{16\sqrt{2}\sin x}{8\sin x\sin \frac{x}{2}-2\sqrt{2}}=8$