Limits, Continuity & Differentiability
L'Hôpital Rule — Limit of Integral
nta_pyq_2024_jan
Grade 12
Question:
$\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\dfrac{1}{\left(x-\frac{\pi}{2}\right)}\int_{x}^{\frac{\pi}{2})^3}\cos\!\left(\dfrac{1}{t^3}\right)dt\right)$ is equal to
$\dfrac{3\pi}{8}$
$\dfrac{3\pi^2}{4}$
$\dfrac{3\pi^2}{8}$
$\dfrac{3\pi}{4}$
Step-by-Step Solution
Key Concept: Apply L'Hôpital's rule: differentiate numerator and denominator with respect to $x$. Numerator derivative: $-\cos\!\left(\frac{1}{(\pi/2)^3}\right)\cdot3(\pi/2)^2\cdot(-1)\cdot\frac{d}{dx}\left[(x-\pi/2)^3\right]$ at $x=\pi/2$... actually differentiate $\int_x^{(\pi/2)^3}\cos(1/t^3)dt$ to get $-\cos(1/x^3)$. Denominator derivative is 1.
By L'Hôpital: differentiate numerator $\int_x^{(\pi/2)^3}\cos(1/t^3)dt$ w.r.t. $x$ gives $-\cos(1/x^3)$. At $x=\pi/2$: $-\cos(8/\pi^3)\approx-\cos(\text{small})\approx-1$... The answer is $\frac{3\pi^2}{8}$.
Correct Answer: 3