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

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

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

We will use L'Hôpital's rule, which applies to limits of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). First, let's verify that the expressions in the numerator and denominator both approach zero as \(x \to 0\):

• The numerator: since \(\int_{0}^{x^2}\sin(\sqrt{t}) \, dt\) represents the integral from 0 to \(x^2\), when \(x \to 0\), this integral approaches 0.
• The denominator: clearly \(x^3 \to 0\) as \(x \to 0\).

Both the numerator and the denominator approach 0, so we can apply L'Hôpital's Rule:

First, we find the derivatives of the numerator and the denominator:

• The derivative of the denominator \(x^3\) with respect to \(x\) is \(3x^2\).
• For the numerator \(\int_{0}^{x^2}\sin(\sqrt{t}) \, dt\), we differentiate using the Fundamental Theorem of Calculus. By substitution, set \(u = x^2\), hence \(\frac{du}{dx} = 2x\). The derivative is then \(\sin(\sqrt{x^2}) \times 2x = \sin(x) \times 2x\).

Using L'Hôpital's rule, the limit becomes:

\(\lim_{x \to 0} \frac{2x \sin(x)}{3x^2}\)

Simplify the expression:

\(\lim_{x \to 0} \frac{2 \sin(x)}{3x}\)

Recognizing the standard limit \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\), we apply it here:

\(\lim_{x \to 0} \frac{2 \sin(x)}{3x} = \frac{2}{3} \times 1 = \frac{2}{3}\)

Therefore, the given limit is \(\frac{2}{3}\).

The correct answer is 2/3.