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

To solve the problem, we need to evaluate the following limit:

\[\lim_{x \to 0} \frac{\int_{0}^{x} f(t) \, dt}{e^{x^2} - 1} = \alpha\]

Given that \(f(0) = \frac{1}{2}\) and \(f(x)\) is differentiable, we will use L'Hôpital's Rule. L'Hôpital's Rule is applicable since both the numerator and the denominator approach 0 as \(x \to 0\).

First, we differentiate the numerator and the denominator with respect to \(x\):

• The derivative of the numerator \(\int_{0}^{x} f(t) \, dt\) with respect to \(x\) is \(f(x)\), according to the Fundamental Theorem of Calculus.
• The derivative of the denominator \(\left(e^{x^2} - 1\right)\) is \(2xe^{x^2}\).

Using L'Hôpital's Rule, we have:

\[\lim_{x \to 0} \frac{f(x)}{2x e^{x^2}} = \alpha\]

Substituting \(f(0) = \frac{1}{2}\), we focus on the expression at \(x = 0\):

\[\lim_{x \to 0} \frac{f(x)}{2x e^{x^2}} = \lim_{x \to 0} \frac{\frac{1}{2} + \frac{d}{dx}(f(x) - \frac{1}{2})}{2x e^{x^2}}\]

Notice that the higher order terms vanish faster than \(x\), hence:

\[\lim_{x \to 0} \frac{\frac{1}{2}}{2x} = \frac{1}{4}\]

Thus, \(\alpha = \frac{1}{4}\).

Next, we find \(8\alpha^{2}\):

\[8 \alpha^2 = 8 \left(\frac{1}{4}\right)^2 = 8 \times \frac{1}{16} = \frac{1}{2} \times 2 = 2\]

Hence, the correct answer is 2.

Answer 2

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,

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