Question

$\displaystyle\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \sin \sqrt{t} d t}{x^{3}}$ is equal to

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $0$
• D. $\infty$

Answer 1

Answer: (A)

$\displaystyle\lim _{x \rightarrow 0} \frac{\int\limits_{0}^{x^{2}} \sin \sqrt{t} d t}{x^{3}}$
$\left(\text { form } \frac{0}{0}\right)$
$=\displaystyle\lim _{x \rightarrow 0} \frac{\sin \sqrt{x^{2}} \times 2 x}{3 x^{2}}$
(using L' Hospital's rule)
$=\displaystyle\lim _{x \rightarrow 0} \frac{2}{3} \frac{\sin x}{x}=\frac{2}{3} \times 1=\frac{2}{3}$