Let f(x) = int_0^x (t + sin(1 - e^t)) dt, x i | vedclass

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(B) Given $f(x) = \int_0^x (t + \sin(1 - e^t)) dt$. We need to find $\lim_{x \rightarrow 0} \frac{f(x)}{x^3}$.Since $f(0) = 0$ and the denominator is

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$-\frac{1}{6}$

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(B) Given $f(x) = \int_0^x (t + \sin(1 - e^t)) dt$. We need to find $\lim_{x ightarrow 0} \frac{f(x)}{x^3}$.Since $f(0) = 0$ and the denominator is $0$ at $x = 0$,we apply $L'H	ext{ôpital's Rule}$:$\lim_{x ightarrow 0} \frac{f'(x)}{3x^2} = \lim_{x ightarrow 0} \frac{x + \sin(1 - e^x)}{3x^2}$.Applying $L'H	ext{ôpital's Rule}$ again:$\lim_{x ightarrow 0} \frac{1 + \cos(1 - e^x) \cdot (-e^x)}{6x}$.Using the Taylor expansion $e^x \approx 1 + x + \frac{x^2}{2}$ and $\cos(	heta) \approx 1 - \frac{	heta^2}{2}$:$1 - e^x \approx -x - \frac{x^2}{2}$.$\cos(1 - e^x) \approx 1 - \frac{(-x - \frac{x^2}{2})^2}{2} \approx 1 - \frac{x^2}{2}$.Substituting back:$\lim_{x ightarrow 0} \frac{1 - (1 - \frac{x^2}{2})(1 + x + \frac{x^2}{2})}{6x} = \lim_{x ightarrow 0} \frac{1 - (1 + x + \frac{x^2}{2} - \frac{x^2}{2})}{6x} = \lim_{x ightarrow 0} \frac{-x}{6x} = -\frac{1}{6}$.

Let $f(x) = \int_0^x (t + \sin(1 - e^t)) dt, x \in R$. Then $\lim_{x \rightarrow 0} \frac{f(x)}{x^3}$ is equal to

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