Let H(x) = int_{x^2}^{x^3} (x + 1) sin(t^3) d | vedclass

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(C) Given $H(x) = (x + 1) \int_{x^2}^{x^3} \sin(t^3) dt$.Since $H(1) = (1 + 1) \int_{1}^{1} \sin(t^3) dt = 0$,the limit $\lim_{x 	o 1} \frac{H(x)}{x

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$2\sin(1)$

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(C) Given $H(x) = (x + 1) \int_{x^2}^{x^3} \sin(t^3) dt$.Since $H(1) = (1 + 1) \int_{1}^{1} \sin(t^3) dt = 0$,the limit $\lim_{x 	o 1} \frac{H(x)}{x - 1}$ is of the form $\frac{0}{0}$.Applying $L'	ext{Hôpital's rule}$,we have $\lim_{x 	o 1} \frac{H'(x)}{1} = H'(1)$.Using the product rule and Leibniz integral rule:$H'(x) = \frac{d}{dx}(x + 1) \cdot \int_{x^2}^{x^3} \sin(t^3) dt + (x + 1) \cdot \frac{d}{dx} \left( \int_{x^2}^{x^3} \sin(t^3) dt ight)$.$H'(x) = 1 \cdot \int_{x^2}^{x^3} \sin(t^3) dt + (x + 1) \left[ \sin((x^3)^3) \cdot 3x^2 - \sin((x^2)^3) \cdot 2x ight]$.Evaluating at $x = 1$:$H'(1) = \int_{1}^{1} \sin(t^3) dt + (1 + 1) \left[ 3(1)^2 \sin(1^9) - 2(1) \sin(1^6) ight]$.$H'(1) = 0 + 2 [3 \sin(1) - 2 \sin(1)] = 2 \sin(1)$.

Let $H(x) = \int_{x^2}^{x^3} (x + 1) \sin(t^3) dt$. Then $\lim_{x \to 1} \frac{H(x)}{x - 1}$ is equal to:

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