Question

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

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. 0
• D. $\frac{1}{15}$

Answer 1

Answer: (A)

$\displaystyle\lim _{x \rightarrow 0^{+}} \frac{\int\limits_{0}^{x^{2}} \sin \sqrt{t} d t}{x^{3}}=\displaystyle\lim _{x \rightarrow 0^{+}} \frac{(\sin x) 2 x}{3 x^{2}}$
$=\displaystyle\lim _{x \rightarrow 0^{+}}\left(\frac{\sin x}{x}\right) \times \frac{2}{3}=\frac{2}{3}$