lim _{x rightarrow 0} frac{int_{0}^{x^{2}}(si | vedclass

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(A) To evaluate the limit $L = \lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \sin \sqrt{t} dt}{x^{3}}$,we observe that it is of the form $\frac{0}{0}

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$2/3$

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(A) To evaluate the limit $L = \lim _{x ightarrow 0} \frac{\int_{0}^{x^{2}} \sin \sqrt{t} dt}{x^{3}}$,we observe that it is of the form $\frac{0}{0}$.Applying $L$'$H$ôpital's rule and the Leibniz integral rule,we differentiate the numerator and the denominator with respect to $x$:Numerator derivative: $\frac{d}{dx} \int_{0}^{x^{2}} \sin \sqrt{t} dt = (\sin \sqrt{x^{2}}) \cdot \frac{d}{dx}(x^{2}) = (\sin x) \cdot (2x)$.Denominator derivative: $\frac{d}{dx}(x^{3}) = 3x^{2}$.Now,the limit becomes:$L = \lim _{x ightarrow 0} \frac{2x \sin x}{3x^{2}} = \lim _{x ightarrow 0} \frac{2 \sin x}{3x}$.Using the standard limit $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$,we get:$L = \frac{2}{3} 	imes 1 = \frac{2}{3}$.

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

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