1. We need

$$\lim_{x\to 0}\frac{f(x)}{x^3},\qquad f(x)=\int_0^x\left(t+\sin(1-e^t)\right)dt.$$

So first simplify the integrand near $t=0$.

2. Expand $e^t$ about $t=0$:

$$e^t=1+t+\frac{t^2}{2}+\frac{t^3}{6}+\cdots$$

Hence,

$$1-e^t= -t-\frac{t^2}{2}-\frac{t^3}{6}+\cdots$$

3. Now use

$$\sin u = u - \frac{u^3}{6}+\cdots$$

with

$$u=1-e^t = -t-\frac{t^2}{2}-\frac{t^3}{6}+\cdots$$

Then up to terms needed,

$$\sin(1-e^t)=\sin u = u+O(u^3).$$

Since $u=O(t)$, we get

$$\sin(1-e^t)= -t-\frac{t^2}{2}+O(t^3).$$

Therefore,

$$t+\sin(1-e^t)= t + \left(-t-\frac{t^2}{2}+O(t^3)\right) = -\frac{t^2}{2}+O(t^3).$$

4. Integrate from $0$ to $x$:

$$f(x)=\int_0^x\left(-\frac{t^2}{2}+O(t^3)\right)dt = -\frac{1}{2}\cdot \frac{x^3}{3}+O(x^4) = -\frac{x^3}{6}+O(x^4).$$

5. Divide by $x^3$ and take the limit:

$$\frac{f(x)}{x^3}= -\frac{1}{6}+O(x).$$

So,

$$\lim_{x\to 0}\frac{f(x)}{x^3}=-\frac{1}{6}.$$

6. Checking options:

• A: $\frac16$ ✗
• B: $-\frac16$ ✓
• C: $\frac23$ ✗
• D: $-\frac23$ ✗

Hence the correct option is B.