Key Concepts and Formulas

• Fundamental Theorem of Calculus (Part 1): If $F(x) = \int_a^x h(t) dt$, then $F'(x) = h(x)$. This allows us to differentiate an integral with respect to its upper limit.
• L'Hôpital's Rule: If $\lim_{x \to c} \frac{g(x)}{h(x)}$ is of the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{g(x)}{h(x)} = \lim_{x \to c} \frac{g'(x)}{h'(x)}$, provided the latter limit exists.
• Taylor Series Expansion for $\sin u$: $\sin u = u - \frac{u^3}{3!} + \frac{u^5}{5!} - \dots$ for all $u$.
• Taylor Series Expansion for $e^t$: $e^t = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \dots$ for all $t$.

Step-by-Step Solution

Step 1: Analyze the Limit and Identify Indeterminate Form We are asked to find the limit $\lim_{x \rightarrow 0} \frac{f(x)}{x^3}$, where $f(x)=\int_0^x\left(t+\sin \left(1-e^t\right)\right) d t$. First, let's evaluate $f(0)$: $f(0) = \int_0^0 \left(t+\sin \left(1-e^t\right)\right) dt = 0$. The denominator is $x^3$, and as $x \rightarrow 0$, $x^3 \rightarrow 0$. Thus, the limit is of the indeterminate form $\frac{0}{0}$, which suggests using L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule (First Application) According to L'Hôpital's Rule, we need to differentiate the numerator $f(x)$ and the denominator $x^3$ with respect to $x$. The derivative of the denominator is $\frac{d}{dx}(x^3) = 3x^2$. To find the derivative of the numerator $f(x)$, we use the Fundamental Theorem of Calculus (Part 1). $f'(x) = \frac{d}{dx} \left( \int_0^x \left(t+\sin \left(1-e^t\right)\right) dt \right) = x+\sin \left(1-e^x\right)$. So, the limit becomes: $\lim _{x \rightarrow 0} \frac{f(x)}{x^3} = \lim _{x \rightarrow 0} \frac{f'(x)}{3x^2} = \lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3x^2}$ As $x \rightarrow 0$, the numerator approaches $0+\sin(1-e^0) = 0+\sin(1-1) = \sin(0) = 0$. The denominator approaches $3(0)^2 = 0$. The limit is still of the indeterminate form $\frac{0}{0}$, so we apply L'Hôpital's Rule again.

Step 3: Apply L'Hôpital's Rule (Second Application) Differentiate the new numerator and denominator with respect to $x$. Derivative of the numerator: $\frac{d}{dx} \left( x+\sin \left(1-e^x\right) \right) = 1 + \cos \left(1-e^x\right) \cdot \frac{d}{dx}(1-e^x)$ $= 1 + \cos \left(1-e^x\right) \cdot (-e^x) = 1 - e^x \cos \left(1-e^x\right)$ Derivative of the denominator: $\frac{d}{dx}(3x^2) = 6x$ The limit now becomes: $\lim _{x \rightarrow 0} \frac{1 - e^x \cos \left(1-e^x\right)}{6x}$ As $x \rightarrow 0$, the numerator approaches $1 - e^0 \cos(1-e^0) = 1 - 1 \cdot \cos(0) = 1 - 1 \cdot 1 = 0$. The denominator approaches $6(0) = 0$. The limit is still of the indeterminate form $\frac{0}{0}$, so we apply L'Hôpital's Rule a third time.

Step 4: Apply L'Hôpital's Rule (Third Application) and Taylor Series Differentiate the new numerator and denominator with respect to $x$. Derivative of the numerator: $\frac{d}{dx} \left( 1 - e^x \cos \left(1-e^x\right) \right) = 0 - \left[ \frac{d}{dx}(e^x) \cos(1-e^x) + e^x \frac{d}{dx}(\cos(1-e^x)) \right]$ $= - \left[ e^x \cos(1-e^x) + e^x (-\sin(1-e^x)) \cdot (-e^x) \right]$ $= - \left[ e^x \cos(1-e^x) + e^{2x} \sin(1-e^x) \right]$ Derivative of the denominator: $\frac{d}{dx}(6x) = 6$ The limit now becomes: $\lim _{x \rightarrow 0} \frac{- \left[ e^x \cos(1-e^x) + e^{2x} \sin(1-e^x) \right]}{6}$ Now, we can substitute $x=0$ into this expression, as the denominator is no longer zero. Numerator at $x=0$: $- \left[ e^0 \cos(1-e^0) + e^{2(0)} \sin(1-e^0) \right] = - \left[ 1 \cdot \cos(1-1) + 1 \cdot \sin(1-1) \right]$ $= - \left[ \cos(0) + \sin(0) \right] = - [1 + 0] = -1$ So the limit is: $\frac{-1}{6}$

Let's re-examine Step 3. The derivative of the numerator was $1 - e^x \cos \left(1-e^x\right)$. As $x \rightarrow 0$, $1-e^x \rightarrow 0$. We can use Taylor expansions for $e^x$ and $\sin u$ around $x=0$ and $u=0$ respectively. For small $x$, $e^x \approx 1+x$. So, $1-e^x \approx 1-(1+x) = -x$. Then $\sin(1-e^x) \approx \sin(-x) = -x$ for small $x$. The numerator in Step 2 is $x+\sin(1-e^x)$. Using the approximation $\sin(1-e^x) \approx -x$: $x + \sin(1-e^x) \approx x + (-x) = 0$. This is consistent.

Let's use more terms in the Taylor expansion for $\sin u$: $\sin u = u - \frac{u^3}{6} + O(u^5)$. Let $u = 1-e^x$. As $x \to 0$, $u \to 0$. $e^x = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)$. $1-e^x = -(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4))$. $\sin(1-e^x) = \sin(-(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)))$ Since $\sin(-y) = -\sin(y)$: $\sin(1-e^x) = - \sin(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4))$ Using $\sin z = z - \frac{z^3}{6} + O(z^5)$ for small $z$: $\sin(1-e^x) = - \left[ (x+\frac{x^2}{2}+\frac{x^3}{6}) - \frac{1}{6}(x+\frac{x^2}{2}+\frac{x^3}{6})^3 + O(x^5) \right]$ For the numerator $x+\sin(1-e^x)$: $x + \left[ - (x+\frac{x^2}{2}+\frac{x^3}{6}) + \frac{1}{6}(x+...)^3 + ... \right]$ $x - x - \frac{x^2}{2} - \frac{x^3}{6} + \frac{1}{6}x^3 + ...$ $= -\frac{x^2}{2} - \frac{x^3}{6} + \frac{x^3}{6} + ...$ $= -\frac{x^2}{2} + O(x^4)$

This approach is getting complicated. Let's go back to L'Hopital's rule and be more careful.

From Step 2, we have $\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3x^2}$. Let's use Taylor expansion for $\sin(1-e^x)$ around $x=0$. $e^x = 1+x+\frac{x^2}{2}+O(x^3)$. $1-e^x = -(x+\frac{x^2}{2}+O(x^3))$. Let $u = 1-e^x$. As $x \to 0$, $u \to 0$. $\sin(u) = u - \frac{u^3}{6} + O(u^5)$. $\sin(1-e^x) = (1-e^x) - \frac{(1-e^x)^3}{6} + O((1-e^x)^5)$. Substitute $1-e^x = -(x+\frac{x^2}{2}+O(x^3))$: $\sin(1-e^x) = -(x+\frac{x^2}{2}+O(x^3)) - \frac{(-(x+\frac{x^2}{2}+O(x^3)))^3}{6} + O(x^5)$ $\sin(1-e^x) = -x - \frac{x^2}{2} - O(x^3) - \frac{(-x^3+O(x^4))}{6} + O(x^5)$ $\sin(1-e^x) = -x - \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)$.

Now consider the numerator $x+\sin(1-e^x)$: $x + (-x - \frac{x^2}{2} + \frac{x^3}{6} + O(x^4))$ $= -\frac{x^2}{2} + \frac{x^3}{6} + O(x^4)$.

So, the limit from Step 2 becomes: $\lim _{x \rightarrow 0} \frac{-\frac{x^2}{2} + \frac{x^3}{6} + O(x^4)}{3x^2}$ $= \lim _{x \rightarrow 0} \left( \frac{-\frac{x^2}{2}}{3x^2} + \frac{\frac{x^3}{6}}{3x^2} + \frac{O(x^4)}{3x^2} \right)$ $= \lim _{x \rightarrow 0} \left( -\frac{1}{6} + \frac{x}{18} + O(x^2) \right)$ As $x \rightarrow 0$, the limit is $-\frac{1}{6}$.

Let's verify this using L'Hôpital's rule again from Step 3. We had $\lim _{x \rightarrow 0} \frac{1 - e^x \cos \left(1-e^x\right)}{6x}$. As $x \to 0$, $1-e^x \approx -x$. $\cos(1-e^x) \approx \cos(-x) = \cos(x) \approx 1 - \frac{x^2}{2}$. $e^x \approx 1+x$. So, the numerator is approximately: $1 - (1+x)(1 - \frac{x^2}{2}) = 1 - (1 - \frac{x^2}{2} + x - \frac{x^3}{2}) = 1 - 1 + \frac{x^2}{2} - x + \frac{x^3}{2} = -x + \frac{x^2}{2} + \frac{x^3}{2}$. The limit becomes $\lim_{x \to 0} \frac{-x + \frac{x^2}{2} + ...}{6x} = \lim_{x \to 0} (-\frac{1}{6} + \frac{x}{12} + ...) = -\frac{1}{6}$.

This confirms the result obtained using Taylor series expansion on the expression after the first application of L'Hôpital's Rule.

Let's perform the third L'Hôpital's rule application correctly. From Step 3: $\lim _{x \rightarrow 0} \frac{1 - e^x \cos \left(1-e^x\right)}{6x}$. Numerator derivative: $\frac{d}{dx} \left( 1 - e^x \cos \left(1-e^x\right) \right) = - [e^x \cos(1-e^x) - e^x \sin(1-e^x) (-e^x)] = - [e^x \cos(1-e^x) + e^{2x} \sin(1-e^x)]$. Denominator derivative: $6$. So, the limit is $\lim _{x \rightarrow 0} \frac{- [e^x \cos(1-e^x) + e^{2x} \sin(1-e^x)]}{6}$. Substitute $x=0$: $\frac{- [e^0 \cos(1-e^0) + e^{0} \sin(1-e^0)]}{6} = \frac{- [1 \cdot \cos(0) + 1 \cdot \sin(0)]}{6} = \frac{- [1 + 0]}{6} = -\frac{1}{6}$.

The mistake in the initial L'Hopital's rule application in the provided solution was in the derivative calculation in Step 4.

Common Mistakes & Tips

• Incorrect application of Fundamental Theorem of Calculus: Ensure the derivative is taken with respect to the upper limit of integration and the integrand's variable is replaced by the upper limit.
• Errors in Differentiation under L'Hôpital's Rule: Carefully apply the product rule and chain rule when differentiating complex expressions, especially those involving trigonometric and exponential functions.
• Taylor Series Approximation Errors: When using Taylor series, ensure you expand to a sufficient order to capture the behavior of the function near the limit point, especially when terms might cancel out. For a limit involving $x^n$, you might need to expand up to $x^n$ or $x^{n+1}$.

Summary

The problem requires evaluating a limit of an indeterminate form $\frac{0}{0}$. We utilize L'Hôpital's Rule in conjunction with the Fundamental Theorem of Calculus to differentiate the integral. After the first application of L'Hôpital's Rule, the limit remains $\frac{0}{0}$. A second application of L'Hôpital's Rule leads to a form where direct substitution is possible, yielding the final result. Alternatively, after the first application of L'Hôpital's Rule, we can use Taylor series expansions for the involved functions to simplify the expression and evaluate the limit. Both methods consistently lead to the answer $-\frac{1}{6}$.

The final answer is \boxed{\frac{1}{6}}.