Limit Evaluation Involving Trigonometric Functions

To evaluate the limit $\lim_{x \to \frac{\pi}{2}} \frac{4\sqrt{2}(\sin 3x + \sin x)}{\left(2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2}\right) - \left(\sqrt{2} + \sqrt{2} \cos 2x + \cos \frac{3x}{2}\right)}$

we simplified the trigonometric terms in the denominator using product-to-sum identities to obtain: $2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2} - \cos \frac{3x}{2} = 2 \sin \frac{x}{2} (\sin 3x + \sin x)$

Substituting this back into the denominator yields: $D = 2 \sin \frac{x}{2} (\sin 3x + \sin x) - \sqrt{2}(1 + \cos 2x)$

By setting $x = \frac{\pi}{2} + t$ as $t \to 0$, we expand the trigonometric terms up to second order in $t$: $\sin 3x + \sin x = 2 \sin 2t \sin t \approx 4t^2$ $\sin \frac{x}{2} = \sin\left(\frac{\pi}{4} + \frac{t}{2}\right) \approx \frac{1}{\sqrt{2}}$ $1 + \cos 2x = 1 - \cos 2t = 2 \sin^2 t \approx 2t^2$

Substituting these approximations into the numerator $N$ and denominator $D$: $N \approx 4\sqrt{2} (4t^2) = 16\sqrt{2} t^2$ $D \approx 2 \cdot \frac{1}{\sqrt{2}} (4t^2) - \sqrt{2}(2t^2) = 4\sqrt{2} t^2 - 2\sqrt{2} t^2 = 2\sqrt{2} t^2$

Taking the limit as $t \to 0$: $\lim_{t \to 0} \frac{16\sqrt{2} t^2}{2\sqrt{2} t^2} = 8$

The value of the limit is $8$.