Question

$\underset{x\to 0}{lim}\frac{1}{x^3}{\int }_0^{x^2}\sin \sqrt{t}dt$ is equal to

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

Answer 1

The correct option is B $\frac{2}{3}$

$\underset{x\to 0}{lim}\frac{1}{x^3}{\int }_0^{x^2}\sin \sqrt{t}dt=\underset{x\to 0}{lim}\frac{{\int }_0^{x^2}\sin \sqrt{t}dt}{x^3}$

$\left(\frac{0}{0}form\right)(\because {\int }_0^{x^2}\sin \sqrt{t}dt=0atx=0)$

$=\underset{x\to 0}{lim}\frac{{\left[\sin \sqrt{t}\right]}_{t=x^2}\frac{d}{dx}\left(x^2\right)}{3x^2}$ (Using L'Hospital rule)

$=\underset{x\to 0}{lim}\frac{\left(\sin x\right)\left(2x\right)}{3x^2}=\frac{2}{3}$