Question

The value of $\underset{x\to 0^{+}}{lim}\frac{{\int }_0^{x^2}\sin \sqrt{t}dt}{x^3}$ is ?

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

Answer: (C)

$\frac{2}{3}$

As $x=0$, both the numerator and denominator are 0.
Using L'Hôpital's rule:
$\underset{x\to 0^{+}}{lim}\frac{{\int }_0^{x^2}\sin \sqrt{t}dt}{x^3}=\underset{x\to 0^{+}}{lim}\frac{\frac{d}{dx}{\int }_0^{x^2}\sin \sqrt{t}dt}{\frac{d}{dx}{x}^3}$
$=\underset{x\to 0^{+}}{lim}\frac{\sin \left(\sqrt{x^2}\right)\left(2x\right)}{3x^2}$
$=\underset{x\to 0^{+}}{lim}\frac{2\sin \left(|x|\right)}{3x}$
$=\underset{x\to 0^{+}}{lim}\frac{2\sin \left(x\right)}{3x}$ (As $x>0$)
$=\frac{2}{3}$