Question

Let $f : \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \rightarrow \mathbb{R}$ be a differentiable function such that $f(0) = \frac{1}{2}$. If the $\lim_{x \to 0} \frac{\int_{0}^{x} f(t) \, dt}{e^{x^2} - 1} = \alpha$, then $8\alpha^2$ is equal to:

• A. 16
• B. 2
• C. 1
• D. 4

Answer 1

Answer: (B)

2

Answer 2

Solution: Rewrite the limit as follows:

$\lim_{x \to 0} \frac{\int_0^x f(t) \, dt}{e^{x^2} - 1} = \lim_{x \to 0} \left( \frac{\int_0^x f(t) \, dt}{x} \times \frac{x}{e^{x^2} - 1} \right)$

Evaluate each part separately:

For the first part, use L'Hôpital's Rule:

$\lim_{x \to 0} \frac{\int_0^x f(t) \, dt}{x} = \lim_{x \to 0} f(x) = f(0) = \frac{1}{2}$

For the second part, apply the Taylor series expansion $e^{x^2} \approx 1 + x^2$ near $x = 0$:

$\lim_{x \to 0} \frac{x}{e^{x^2} - 1} = \lim_{x \to 0} \frac{x}{x^2} = \lim_{x \to 0} \frac{1}{x} = 1$

So, $\alpha = \frac{1}{2}$. Then,

$8\alpha^2 = 8 \times \left( \frac{1}{2} \right)^2 = 2$