The value of mathop {lim }limits_{x to 0} lef | vedclass

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(C) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{\int_0^{x^2} \sec^2 t \, dt}{x \sin x}$.Since this is a $\frac{0}{0}$ form,we apply $L'	ext{H\^op

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$1$

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(C) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{\int_0^{x^2} \sec^2 t \, dt}{x \sin x}$.Since this is a $\frac{0}{0}$ form,we apply $L'	ext{H\^opital's rule}$:$L = \mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}(\int_0^{x^2} \sec^2 t \, dt)}{\frac{d}{dx}(x \sin x)}$.Using the $Leibniz$ rule for differentiation under the integral sign:$L = \mathop {\lim }\limits_{x 	o 0} \frac{\sec^2(x^2) \cdot 2x}{\sin x + x \cos x}$.Divide numerator and denominator by $x$:$L = \mathop {\lim }\limits_{x 	o 0} \frac{2 \sec^2(x^2)}{\frac{\sin x}{x} + \cos x}$.Substituting $x = 0$:$L = \frac{2 \cdot \sec^2(0)}{1 + \cos(0)} = \frac{2 \cdot 1}{1 + 1} = \frac{2}{2} = 1$.

The value of $\mathop {\lim }\limits_{x \to 0} \left( \frac{\int_0^{x^2} \sec^2 t \, dt}{x \sin x} \right)$ is

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