Let f(x) = \(\int_0^x\) \(\left(t+\sin \left(1-e^t\right)\right)\) dt, x ∈ ℝ. Then \(\lim _{x \rightarrow 0} \frac{f(x)}{x^3}\) is equal to

1
\(\frac{1}{6}\)
2
-\(\frac{1}{6}\)
3
-\(\frac{2}{3}\)
4
\(\frac{2}{3}\)

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