$$\lim_{x \to \frac{\pi}{2}} \left(\frac{\int_{x^3}^{(\pi/2)^3} \left(\sin(2t^{1/3}) + \cos(t^{1/3})\right) dt}{\left(x - \frac{\pi}{2}\right)^2}\right)$$ is equal to

Evaluate $$\lim_{x \to \pi/2} \frac{\int_{x^3}^{(\pi/2)^3} (\sin(2t^{1/3}) + \cos(t^{1/3})) \, dt}{(x - \pi/2)^2}$$.

As $$x \to \pi/2$$, the upper limit equals the lower limit so the numerator approaches 0 while the denominator also approaches 0, resulting in a $$\frac{0}{0}$$ form and prompting the use of L'Hopital's Rule.

Next, let $$N(x) = \int_{x^3}^{(\pi/2)^3} (\sin(2t^{1/3}) + \cos(t^{1/3})) \, dt = -\int_{(\pi/2)^3}^{x^3} (\sin(2t^{1/3}) + \cos(t^{1/3})) \, dt$$, and by the Leibniz rule (chain rule for integrals) we find $$N'(x) = -(\sin(2(x^3)^{1/3}) + \cos((x^3)^{1/3})) \cdot 3x^2 = -(\sin(2x) + \cos x) \cdot 3x^2$$.

At $$x = \pi/2$$ this gives $$N'(\pi/2) = -(\sin \pi + \cos(\pi/2)) \cdot 3(\pi/2)^2 = 0$$. Since the derivative of the denominator is $$2(x - \pi/2)$$ which also vanishes at $$x = \pi/2$$, the expression remains of the form $$\frac{0}{0}$$, so L'Hopital's Rule is applied a second time.

Applying the rule again yields $$N''(x) = -\frac{d}{dx}[(\sin 2x + \cos x) \cdot 3x^2] = -[3x^2(2\cos 2x - \sin x) + 6x(\sin 2x + \cos x)]$$. Evaluating at $$x = \pi/2$$ gives $$N''(\pi/2) = -\left[3 \cdot \frac{\pi^2}{4}(2\cos \pi - \sin(\pi/2)) + 6 \cdot \frac{\pi}{2}(\sin \pi + \cos(\pi/2))\right] = \frac{9\pi^2}{4}$$, while the second derivative of the denominator is $$2$$.

Since the limit equals the ratio of these second derivatives, it becomes $$\frac{N''(\pi/2)}{2} = \frac{9\pi^2/4}{2} = \frac{9\pi^2}{8}$$.

The correct answer is Option (2): $$\frac{9\pi^2}{8}$$.