Evaluate, $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{(\cos x + \cos 3x + \sin 2x)}^2}}}{{{{(\sin x - \sin 3x + 2\cos 2x)}^3}}}$

The given limit,

$ = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{(2\cos 2x\cos x + 2\sin x\cos x)}^2}}}{{{{( - 2\cos 2x\sin x + 2\cos 2x)}^3}}}$

$ = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{\cos }^2}x{{(\cos 2x + \sin x)}^2}}}{{2{{\cos }^3}2x{{( - \sin x + 1)}^3}}}$

$\mathop { = \lim }\limits_{x \to \frac{\pi }{2}} \frac{{(1 + \sin x)(1 - \sin x){{(\cos 2x + \sin x)}^2}}}{{2{{\cos }^3}2x{{( - \sin x + 1)}^3}}}$

$ = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{(1 + \sin x){{(\cos 2x + \sin x)}^2}}}{{2{{\cos }^3}2x{{(1 - \sin x)}^2}}}$

$ = {\left. {\frac{{(1 + \sin x)}}{{2{{\cos }^3}2x}}} \right|_{x = \frac{\pi }{2}}}\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{(\cos 2x + \sin x)}^2}}}{{{{(1 - \sin x)}^2}}}$

$ = ( - 1){\left[ {\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{(\cos 2x + \sin x)}}{{(1 - \sin x)}}} \right]^2}$

$ =  - {\left[ {\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{ - 2\sin 2x + \cos x}}{{ - \cos x}}} \right]^2}$

[Using L.H. Rule as the limit has $\frac {0}{0}$ form]

$ =  - {\left[ {\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{ - 4\sin x\cos x + \cos x}}{{ - \cos x}}} \right]^2}$

$ =  - {\left[ {\mathop {\lim }\limits_{x \to \frac{\pi }{2}} (4\sin x - 1)} \right]^2} =  - 9$