Question:
$\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\frac{1}{\left(x-\frac{\pi}{2}\right)^2}\int_{x^3}^{\left(\frac{\pi}{2}\right)^3}\cos\left(t^{1/3}\right)dt\right)$ is equal to
$\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\frac{1}{\left(x-\frac{\pi}{2}\right)^2}\int_{x^3}^{\left(\frac{\pi}{2}\right)^3}\cos\left(t^{1/3}\right)dt\right)$ is equal to
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The limit is of the form $0/0$: apply L'Hospital's rule (with the Leibniz rule for the integral) twice.
Updated On: Oct 26, 2026
- $\dfrac{3\pi^2}{4}$
- $\dfrac{3\pi^2}{8}$
- $\dfrac{3\pi}{4}$
- $\dfrac{3\pi}{8}$
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The Correct Option is B
Solution and Explanation
At $x=\frac{\pi}{2}$ the integral limits coincide, so the expression is of the form $\frac{0}{0}$. By L'Hospital's rule and the Leibniz rule,
$\displaystyle L=\lim_{x\to\pi/2}\frac{-3x^2\cos x}{2\left(x-\frac{\pi}{2}\right)}$ (using $\frac{d}{dx}\int_{x^3}^{c}\cos t^{1/3}dt=-3x^2\cos x$).
This is again $\frac{0}{0}$; differentiating once more, $L=\dfrac{-3\left(2x\cos x-x^2\sin x\right)}{2}\Big|_{x=\pi/2}=\dfrac{-3\left(0-\frac{\pi^2}{4}\right)}{2}=\dfrac{3\pi^2}{8}$.
Option (2).
$\displaystyle L=\lim_{x\to\pi/2}\frac{-3x^2\cos x}{2\left(x-\frac{\pi}{2}\right)}$ (using $\frac{d}{dx}\int_{x^3}^{c}\cos t^{1/3}dt=-3x^2\cos x$).
This is again $\frac{0}{0}$; differentiating once more, $L=\dfrac{-3\left(2x\cos x-x^2\sin x\right)}{2}\Big|_{x=\pi/2}=\dfrac{-3\left(0-\frac{\pi^2}{4}\right)}{2}=\dfrac{3\pi^2}{8}$.
Option (2).
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