The value of lim _{x rightarrow 0} frac{int_{ | vedclass

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(A) Let $L = \lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \cos \left(t^{2}\right) d t}{x \sin x}$.Since the limit is in the $\frac{0}{0}$ form, we a

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(A) Let $L = \lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \cos \left(t^{2}\right) d t}{x \sin x}$.Since the limit is in the $\frac{0}{0}$ form, we apply $L$' Hospital's rule.Using the Leibniz rule for differentiation under the integral sign, the derivative of the numerator is $\cos(x^4) \cdot \frac{d}{dx}(x^2) = 2x \cos(x^4)$.The derivative of the denominator $x \sin x$ is $\sin x + x \cos x$.So, $L = \lim _{x \rightarrow 0} \frac{2x \cos(x^4)}{\sin x + x \cos x}$.Divide numerator and denominator by $x$:$L = \lim _{x \rightarrow 0} \frac{2 \cos(x^4)}{\frac{\sin x}{x} + \cos x}$.As $x \rightarrow 0$, $\frac{\sin x}{x} \rightarrow 1$, $\cos(x^4) \rightarrow 1$, and $\cos x \rightarrow 1$.Therefore, $L = \frac{2(1)}{1 + 1} = \frac{2}{2} = 1$.

The value of $\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \cos \left(t^{2}\right) d t}{x \sin x}$ is

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