Mathematics - Limits Question with Solution | TestHub

$\underset{x\to \frac{\pi }{2}}{\lim }\frac{8\sqrt{2 }\sin 2x·\cos x}{\cos \frac{x}{2}-\cos \frac{7x}{2}+\cos \frac{5x}{2}-\left(\sqrt{2}·2{\cos }^{2}x+\cos \frac{3x}{2}\right)}$

$=\underset{x\to \frac{\pi }{2}}{\lim }\frac{8\sqrt{2 }\sin 2x·\cos x}{\left(\cos \frac{x}{2}-\cos \frac{3x}{2}\right)+\left(\cos \frac{5x}{2}-\cos \frac{7x}{2}\right)-\left(2\sqrt{2}{\cos }^{2}x\right)}$

$=\underset{x\to \frac{\pi }{2}}{\lim }\frac{16\sqrt{2}\sin x\cos x·\cos x}{2\sin x\sin \frac{x}{2}+2\sin 3x\sin \frac{x}{2}-2\sqrt{2}{\cos }^{2}x}$

$=\underset{x\to \frac{\pi }{2}}{\lim }\frac{16\sqrt{2}\sin x\cos x·\cos x}{2\sin \frac{x}{2}\left(\sin x+\sin 3x\right)-2\sqrt{2}{\cos }^{2}x}$

$=\underset{x\to \frac{\pi }{2}}{\lim }\frac{16\sqrt{2}\sin x {\cos }^{2}x}{2\sin \frac{x}{2}\left(2\sin 2x·\cos x\right)-2\sqrt{2}{\cos }^{2}x}$

$=\underset{x\to \frac{\pi }{2}}{\lim }\frac{16\sqrt{2}\sin x}{2\sin {\displaystyle \frac{x}{2}}.4\sin x-2\sqrt{2}}$

$=\frac{16\sqrt{2}}{4\sqrt{2}-2\sqrt{2}}=\frac{16\sqrt{2}}{2\sqrt{2}}=8$.