Prove the following Definite Integration

Question:

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

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

  2. (2) 0

  3. (3) $\frac{1}{15}$

  4. (4) $\frac{3}{2}$


Correct Option: 1

Solution:

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

$=\lim _{x \rightarrow 0} \frac{(\sin |x|) 2 x}{3 x^{2}}$

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

$=\frac{2}{3}$

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