Question

The value of \( \lim_{x \to 0} \frac{\int_0^{x^2} \sec^2 t \, dt}{x \sin x} \) is

• A. 3
• B. 2
• C. 1
• D. -1

Hint:

Whenever integral + limit $\longrightarrow$ reduce integral first, then apply standard limits.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The limit is in the \( \frac{0}{0} \) form. We apply L'Hôpital's Rule and use the Leibniz Rule for differentiating under the integral sign.
Step 2: Key Formula or Approach:
1. Leibniz Rule: \( \frac{d}{dx} \int_{a}^{g(x)} f(t) dt = f(g(x)) \cdot g'(x) \).
2. Standard Limit: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).
Step 3: Detailed Explanation:
Given: \( L = \lim_{x \to 0} \frac{\int_0^{x^2} \sec^2 t dt}{x \sin x} \).
Applying L'Hôpital's Rule:
\[ L = \lim_{x \to 0} \frac{\sec^2(x^2) \cdot 2x}{\sin x + x \cos x} \]
Dividing numerator and denominator by \( x \):
\[ L = \lim_{x \to 0} \frac{2 \sec^2(x^2)}{\frac{\sin x}{x} + \cos x} \]
Evaluating the limit as \( x \to 0 \):
\[ L = \frac{2 \sec^2(0)}{1 + \cos(0)} = \frac{2 \times 1}{1 + 1} = \frac{2}{2} = 1 \]
Step 4: Final Answer:
The value of the limit is 1.