Question

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

• A. \( \frac{9\pi^2}{8} \)
• B. \( \frac{11\pi^2}{10} \)
• C. \( \frac{3\pi^2}{2} \)
• D. \( \frac{5\pi^2}{9} \)

Answer 1

The Correct Option is A

To evaluate the limit:

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

We apply L'Hôpital's rule due to the indeterminate form \(\frac{0}{0}\) as \(x \to \frac{\pi}{2}\).

Differentiate the numerator and denominator with respect to \(x\):

Numerator Derivative:

Using the Fundamental Theorem of Calculus and Leibniz rule:

\[ \frac{d}{dx} \left[ \int_{x^3}^{\left(\frac{\pi}{2}\right)^3} \left( \sin\left(2t^{1/3}\right) + \cos\left(t^{1/3}\right) \right) \, dt \right] = -\left( \sin\left(2(x^3)^{1/3}\right) + \cos\left((x^3)^{1/3}\right) \right) \cdot 3x^2 = -\left( \sin(2x) + \cos(x) \right) \cdot 3x^2 \]

Denominator Derivative:

\[ \frac{d}{dx} ((x - \frac{\pi}{2})^2) = 2(x - \frac{\pi}{2}) \]

Apply L'Hôpital's Rule:

\[ \lim_{x \to \frac{\pi}{2}} \frac{-\left( \sin(2x) + \cos(x) \right) \cdot 3x^2}{2(x - \frac{\pi}{2})} \]

Evaluate the limit by substituting \(x = \frac{\pi}{2}\):

\[ \frac{-\left( \sin(\pi) + \cos(\frac{\pi}{2}) \right) \cdot 3(\frac{\pi}{2})^2}{2(\frac{\pi}{2} - \frac{\pi}{2})} = \frac{-(0+0) \cdot 3\frac{\pi^2}{4}}{0} \]

This expression indicates that further application of L'Hôpital's rule is needed, or direct substitution into the result of differentiation was prematurely simplified. Let's re-evaluate the limit expression after differentiation:

\[ \lim_{x \to \frac{\pi}{2}} \frac{-3x^2(\sin(2x) + \cos(x))}{2(x - \frac{\pi}{2})} \]

This limit is still of the form \(\frac{0}{0}\). Applying L'Hôpital's rule again:

Derivative of numerator: \(-3 [2x(\sin(2x) + \cos(x)) + x^2(2\cos(2x) - \sin(x))] \)\

Derivative of denominator: \(2\)

\[ \lim_{x \to \frac{\pi}{2}} \frac{-3 [2x(\sin(2x) + \cos(x)) + x^2(2\cos(2x) - \sin(x))]}{2} \]

Substitute \(x = \frac{\pi}{2}\):

\[ \frac{-3 [2(\frac{\pi}{2})(\sin(\pi) + \cos(\frac{\pi}{2})) + (\frac{\pi}{2})^2(2\cos(\pi) - \sin(\frac{\pi}{2}))]}{2} \]

\[ \frac{-3 [\pi(0+0) + \frac{\pi^2}{4}(2(-1) - 1)]}{2} \]

\[ \frac{-3 [\frac{\pi^2}{4}(-3)]}{2} \]

\[ \frac{-3 (-\frac{3\pi^2}{4})}{2} \]

\[ \frac{\frac{9\pi^2}{4}}{2} = \frac{9\pi^2}{8} \]

The limit evaluates to:

\[ \boxed{\frac{9\pi^2}{8}} \]