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

The Correct Option is B

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,

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