Question:
$\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}}(\sin \sqrt{t}) d t}{x^{3}}$ is equal to :
Correct Option: 1
Solution:
$\lim _{x \rightarrow 0^{+}} \frac{\int_{0}^{x^{2}} \sin \sqrt{t} d t}{x^{3}}=\lim _{x \rightarrow 0^{+}} \frac{(\sin x) 2 x}{3 x^{2}}$
$=\lim _{x \rightarrow 0^{+}}\left(\frac{\sin x}{x}\right) \times \frac{2}{3}=\frac{2}{3}$