1. **State the problem:** Find the value of the limit $$\lim_{x \to \frac{\pi}{2}} \frac{7\sqrt{2}(\sin x + \sin 3x)}{2 \sin(2x) \sin\left(\frac{3x}{2}\right) + \cos\left(\frac{5x}{2}\right) - \left(\sqrt{2} + \sqrt{2} \cos(2x) + \cos\left(\frac{3x}{2}\right)\right)}$$ 2. **Recall important trigonometric values and limits:** - $\sin\left(\frac{\pi}{2}\right) = 1$ - $\cos\left(\frac{\pi}{2}\right) = 0$ - Use substitution $x = \frac{\pi}{2}$ directly if the expression is defined. 3. **Evaluate numerator at $x = \frac{\pi}{2}$:** $$\sin\left(\frac{\pi}{2}\right) + \sin\left(3 \cdot \frac{\pi}{2}\right) = 1 + \sin\left(\frac{3\pi}{2}\right) = 1 + (-1) = 0$$ So numerator is $7\sqrt{2} \times 0 = 0$. 4. **Evaluate denominator at $x = \frac{\pi}{2}$:** Calculate each term: - $2 \sin(2x) \sin\left(\frac{3x}{2}\right) = 2 \sin(\pi) \sin\left(\frac{3\pi}{4}\right) = 2 \times 0 \times \frac{\sqrt{2}}{2} = 0$ - $\cos\left(\frac{5x}{2}\right) = \cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$ - $\sqrt{2} + \sqrt{2} \cos(2x) + \cos\left(\frac{3x}{2}\right) = \sqrt{2} + \sqrt{2} \cos(\pi) + \cos\left(\frac{3\pi}{4}\right) = \sqrt{2} + \sqrt{2} \times (-1) + \left(-\frac{\sqrt{2}}{2}\right) = \sqrt{2} - \sqrt{2} - \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2}$ So denominator is: $$0 + \left(-\frac{\sqrt{2}}{2}\right) - \left(-\frac{\sqrt{2}}{2}\right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0$$ 5. **Since numerator and denominator both approach 0, apply L'Hôpital's Rule:** Differentiate numerator and denominator with respect to $x$. 6. **Differentiate numerator:** $$\frac{d}{dx} \left[7\sqrt{2}(\sin x + \sin 3x)\right] = 7\sqrt{2}(\cos x + 3 \cos 3x)$$ Evaluate at $x = \frac{\pi}{2}$: $$7\sqrt{2} \left(\cos \frac{\pi}{2} + 3 \cos \frac{3\pi}{2}\right) = 7\sqrt{2} (0 + 3 \times 0) = 0$$ 7. **Differentiate denominator:** Denominator is: $$f(x) = 2 \sin(2x) \sin\left(\frac{3x}{2}\right) + \cos\left(\frac{5x}{2}\right) - \left(\sqrt{2} + \sqrt{2} \cos(2x) + \cos\left(\frac{3x}{2}\right)\right)$$ Derivative: $$f'(x) = 2 \left[2 \cos(2x) \sin\left(\frac{3x}{2}\right) + \sin(2x) \cdot \frac{3}{2} \cos\left(\frac{3x}{2}\right)\right] - \frac{5}{2} \sin\left(\frac{5x}{2}\right) - \left(0 - 2\sqrt{2} \sin(2x) - \frac{3}{2} \sin\left(\frac{3x}{2}\right)\right)$$ Simplify: $$f'(x) = 4 \cos(2x) \sin\left(\frac{3x}{2}\right) + 3 \sin(2x) \cos\left(\frac{3x}{2}\right) - \frac{5}{2} \sin\left(\frac{5x}{2}\right) + 2\sqrt{2} \sin(2x) + \frac{3}{2} \sin\left(\frac{3x}{2}\right)$$ Evaluate at $x = \frac{\pi}{2}$: - $\cos(\pi) = -1$ - $\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}$ - $\sin(\pi) = 0$ - $\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$ - $\sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$ Calculate each term: - $4 \times (-1) \times \frac{\sqrt{2}}{2} = -2\sqrt{2}$ - $3 \times 0 \times \left(-\frac{\sqrt{2}}{2}\right) = 0$ - $-\frac{5}{2} \times \left(-\frac{\sqrt{2}}{2}\right) = \frac{5\sqrt{2}}{4}$ - $2\sqrt{2} \times 0 = 0$ - $\frac{3}{2} \times \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{4}$ Sum: $$-2\sqrt{2} + 0 + \frac{5\sqrt{2}}{4} + 0 + \frac{3\sqrt{2}}{4} = -2\sqrt{2} + \frac{8\sqrt{2}}{4} = -2\sqrt{2} + 2\sqrt{2} = 0$$ 8. **Both derivatives at $x=\frac{\pi}{2}$ are zero, apply L'Hôpital's Rule again:** 9. **Second derivative of numerator:** $$\frac{d^2}{dx^2} \left[7\sqrt{2}(\sin x + \sin 3x)\right] = 7\sqrt{2}(-\sin x - 9 \sin 3x)$$ Evaluate at $x=\frac{\pi}{2}$: $$7\sqrt{2}(-1 - 9 \times (-1)) = 7\sqrt{2}(-1 + 9) = 7\sqrt{2} \times 8 = 56\sqrt{2}$$ 10. **Second derivative of denominator:** Differentiate $f'(x)$ again (tedious but necessary): $$f''(x) = \frac{d}{dx} \left[4 \cos(2x) \sin\left(\frac{3x}{2}\right) + 3 \sin(2x) \cos\left(\frac{3x}{2}\right) - \frac{5}{2} \sin\left(\frac{5x}{2}\right) + 2\sqrt{2} \sin(2x) + \frac{3}{2} \sin\left(\frac{3x}{2}\right)\right]$$ Calculate each term: - $\frac{d}{dx} \left[4 \cos(2x) \sin\left(\frac{3x}{2}\right)\right] = 4 \left[-2 \sin(2x) \sin\left(\frac{3x}{2}\right) + \cos(2x) \cdot \frac{3}{2} \cos\left(\frac{3x}{2}\right)\right] = -8 \sin(2x) \sin\left(\frac{3x}{2}\right) + 6 \cos(2x) \cos\left(\frac{3x}{2}\right)$ - $\frac{d}{dx} \left[3 \sin(2x) \cos\left(\frac{3x}{2}\right)\right] = 3 \left[2 \cos(2x) \cos\left(\frac{3x}{2}\right) - \sin(2x) \cdot \frac{3}{2} \sin\left(\frac{3x}{2}\right)\right] = 6 \cos(2x) \cos\left(\frac{3x}{2}\right) - \frac{9}{2} \sin(2x) \sin\left(\frac{3x}{2}\right)$ - $\frac{d}{dx} \left[-\frac{5}{2} \sin\left(\frac{5x}{2}\right)\right] = -\frac{5}{2} \cdot \frac{5}{2} \cos\left(\frac{5x}{2}\right) = -\frac{25}{4} \cos\left(\frac{5x}{2}\right)$ - $\frac{d}{dx} \left[2\sqrt{2} \sin(2x)\right] = 2\sqrt{2} \cdot 2 \cos(2x) = 4\sqrt{2} \cos(2x)$ - $\frac{d}{dx} \left[\frac{3}{2} \sin\left(\frac{3x}{2}\right)\right] = \frac{3}{2} \cdot \frac{3}{2} \cos\left(\frac{3x}{2}\right) = \frac{9}{4} \cos\left(\frac{3x}{2}\right)$ Sum all: $$f''(x) = (-8 - \frac{9}{2}) \sin(2x) \sin\left(\frac{3x}{2}\right) + (6 + 6) \cos(2x) \cos\left(\frac{3x}{2}\right) - \frac{25}{4} \cos\left(\frac{5x}{2}\right) + 4\sqrt{2} \cos(2x) + \frac{9}{4} \cos\left(\frac{3x}{2}\right)$$ Simplify coefficients: $$-8 - \frac{9}{2} = -\frac{16}{2} - \frac{9}{2} = -\frac{25}{2}$$ $$6 + 6 = 12$$ Evaluate at $x=\frac{\pi}{2}$: - $\sin(\pi) = 0$ - $\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}$ - $\cos(\pi) = -1$ - $\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$ - $\cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$ Calculate terms: - $-\frac{25}{2} \times 0 \times \frac{\sqrt{2}}{2} = 0$ - $12 \times (-1) \times \left(-\frac{\sqrt{2}}{2}\right) = 12 \times \frac{\sqrt{2}}{2} = 6\sqrt{2}$ - $-\frac{25}{4} \times \left(-\frac{\sqrt{2}}{2}\right) = \frac{25\sqrt{2}}{8}$ - $4\sqrt{2} \times (-1) = -4\sqrt{2}$ - $\frac{9}{4} \times \left(-\frac{\sqrt{2}}{2}\right) = -\frac{9\sqrt{2}}{8}$ Sum: $$0 + 6\sqrt{2} + \frac{25\sqrt{2}}{8} - 4\sqrt{2} - \frac{9\sqrt{2}}{8} = (6\sqrt{2} - 4\sqrt{2}) + \left(\frac{25\sqrt{2}}{8} - \frac{9\sqrt{2}}{8}\right) = 2\sqrt{2} + \frac{16\sqrt{2}}{8} = 2\sqrt{2} + 2\sqrt{2} = 4\sqrt{2}$$ 11. **Apply L'Hôpital's Rule second time:** $$\lim_{x \to \frac{\pi}{2}} \frac{7\sqrt{2}(-\sin x - 9 \sin 3x)}{f''(x)} = \frac{56\sqrt{2}}{4\sqrt{2}} = 14$$ **Final answer:** 14 **Answer choice:** B) 14