Question

Evaluate: \[ \lim_{x \to 0} \frac{\int_0^{x^2} \sin\sqrt{t}\,dt}{x^3} \]

• A. \(\frac{2}{3}\)
• B. \(\frac{1}{3}\)
• C. 0
• D. \(\infty\)

Hint:

For limits with integrals, differentiate using Leibniz rule or use small-angle approximations.

Answer 1

The Correct Option is A

Concept: Use the approximation \(\sin\sqrt{t} \approx \sqrt{t}\) for small \(t\), or apply L'Hôpital's rule.

Step 1: Apply L'Hôpital's rule. The limit is of the form \(\frac{0}{0}\). Differentiate numerator and denominator with respect to \(x\): \[ \frac{d}{dx} \int_0^{x^2} \sin\sqrt{t}\,dt = \sin\sqrt{x^2} \cdot 2x = \sin|x| \cdot 2x \] For \(x>0\) near 0, \(|x| = x\), so: \[ \frac{d}{dx} \left( \int_0^{x^2} \sin\sqrt{t}\,dt \right) = 2x \sin x \] Denominator: \(\frac{d}{dx}(x^3) = 3x^2\).

Step 2: Apply L'Hôpital again. \[ \lim_{x \to 0} \frac{2x \sin x}{3x^2} = \lim_{x \to 0} \frac{2 \sin x}{3x} = \frac{2}{3} \cdot 1 = \frac{2}{3} \] Alternative method using approximation: \[ \int_0^{x^2} \sqrt{t}\,dt = \frac{2}{3} (x^2)^{3/2} = \frac{2}{3}x^3 \] \[ \lim_{x \to 0} \frac{\frac{2}{3}x^3}{x^3} = \frac{2}{3} \]