Question

\( \displaystyle \lim_{x \to 0} \frac{\int_{0}^{x^2} \sin(\sqrt{t}) \, dt}{x^3} \) is equal to :

• A. 2/3
• B. 3/2
• C. 1/15
• D. 0

Hint:

Leibniz Rule: $\frac{d}{dx} \int_{g(x)}^{h(x)} f(t) dt = f(h(x))h'(x) - f(g(x))g'(x)$. It is essential for limits involving integrals.

Answer 1

The Correct Option is A

Step 1: Apply L'Hospital's Rule: $\lim_{x \to 0} \frac{\frac{d}{dx} \int_0^{x^2} \sin \sqrt{t} dt}{3x^2}$.
Step 2: Use Leibniz Rule: $\frac{d}{dx} \int_0^{x^2} \sin \sqrt{t} dt = \sin(\sqrt{x^2}) \cdot 2x = 2x \sin x$.
Step 3: $\lim_{x \to 0} \frac{2x \sin x}{3x^2} = \frac{2}{3} \lim_{x \to 0} \frac{\sin x}{x} = \frac{2}{3}$.