Limits, Continuity & Differentiability
L'Hôpital Rule — Limit Involving Integral of Differentiable Function
nta_pyq_2024_jan
Grade 12
Question:
Let $f:\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\to\mathbb{R}$ be a differentiable function such that $f(0)=\dfrac{1}{2}$. If $\displaystyle\lim_{x\to0}\dfrac{x\int_0^x f(t)\,dt}{e^{x^2}-1}=\alpha$, then $8\alpha^2$ is equal to:
Step-by-Step Solution
Key Concept: Rewrite the limit as $\lim_{x\to0}\frac{\int_0^x f(t)dt}{x}\cdot\frac{x^2}{e^{x^2}-1}$. The second factor $\to1$. The first factor $\to f(0)$ by L'Hôpital (or fundamental theorem of calculus).
$\alpha=f(0)=1/2$. $8\alpha^2=8\times1/4=2$.
Correct Answer: 2