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

L'Hôpital's Rule ($$\frac{0}{0}$$ form):

$$L = \lim_{x \to 0} \frac{f'(x)}{3x^2}$$

Using Leibnitz Rule: $$f'(x) = x + \sin(1 - e^x)$$.

Apply L'Hôpital again:

$$L = \lim_{x \to 0} \frac{1 + \cos(1 - e^x) \cdot (-e^x)}{6x}$$

Apply L'Hôpital one last time:

$$L = \lim_{x \to 0} \frac{-\sin(1-e^x)(-e^x)(-e^x) + \cos(1-e^x)(-e^x)}{6}$$

At $$x=0$$: $$L = \frac{-\sin(0)(1)(1) + \cos(0)(-1)}{6} = \frac{0 - 1}{6} = -\frac{1}{6}$$.

Correct Option: A