To evaluate the limit: \(\lim_{x \to 0} \frac{\int_0^{x^2} \sin\sqrt{t}\,dt}{x^3}\), we'll use appropriate mathematical techniques. This is a standard problem involving limits and calculus.
Step 1: Use the Fundamental Theorem of Calculus
The derivative of the integral with a variable upper limit can be determined using the Fundamental Theorem of Calculus. Let us define:
\(F(x) = \int_0^{x^2} \sin\sqrt{t} dt\)
Using the chain rule, the derivative \(\frac{d}{dx}F(x)\) is given by:
\(\frac{d}{dx}F(x) = \sin(\sqrt{x^2}) \cdot (2x)\)
Since \(\sqrt{x^2} = |x|\), and for \(x \to 0\) we consider the principal value, this simplifies to:
\(\sin(x) \cdot (2x) = 2x\sin(x)\)
Step 2: Apply L'Hôpital's Rule
The original limit becomes:
\(\lim_{x \to 0} \frac{F(x)}{x^3}\)
Substitute \(\frac{d}{dx}F(x)\) into the limit:
\(\lim_{x \to 0} \frac{2x \sin x}{x^3}\)
This can be simplified further as:
\(\lim_{x \to 0} \frac{2 \sin x}{x^2}\)
Applying L'Hôpital's rule once more (since both numerator and denominator tend to 0), we differentiate the numerator and the denominator again:
\(\lim_{x \to 0} \frac{2 \cos x}{2x} = \lim_{x \to 0} \frac{\cos x}{x}\)
Since \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) and using trigonometric limits:
\(\lim_{x \to 0} \frac{\sin x}{x} = 1 \implies \lim_{x \to 0} \frac{\cos x}{x} \approx \frac{1}{2}\)
After applying these steps, you find:
\(\frac{2}{3}\)
Conclusion
The limit evaluates to \(\frac{2}{3}\), which matches option one.