The value of \( \lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \sin \sqrt{t}}{x^{3}} d t \) is

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The value of \( \lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \sin \sqrt{t}}{x^{3}} d t \) is

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Options:

A \( 0 \)
B \( 2 / 9 \)
C \( 1 / 3 \)
D \( 2 / 3 \)

Solution:

2046 Upvotes Verified Answer

The correct answer is: \( 2 / 3 \)

Let $L = \lim_{x \to 0} \frac{\int_{0}^{x^{2}} \sin \sqrt{t}}{x^{3}} dt$
$= lim_{x \to 0} \frac{(sin x) \cdot 2 x}{3 x^{2}}$ (by Leibnitz rule)
$= \frac{2}{3} lim_{x \to 0} \frac{sin x}{x}$

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

$= \frac{2}{3}$

Hence, $L = \frac{2}{3}$ .

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