Solve this following


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


  1. $\frac{2}{3}$

  2. $\frac{3}{2}$

  3. 0

  4. $\frac{1}{15}$

Correct Option: 1


$\lim _{x \rightarrow 0^{+}} \frac{\int_{0}^{x^{2}} \sin \sqrt{t} d t}{x^{3}}=\lim _{x \rightarrow 0^{+}} \frac{(\sin x) 2 x}{3 x^{2}}$

$=\lim _{x \rightarrow 0^{+}}\left(\frac{\sin x}{x}\right) \times \frac{2}{3}=\frac{2}{3}$


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