Let \(f(x)={\int }_{0}^{x}\left(t+\sin \left(1-{e}^{t}\right)\right)dt,x\in R\). Then, \(\lim _{x\to 0}\frac{f(x)}{{x}^{…

Let \(L=\lim_{x\to 0}\frac{f(x)}{x^3}\).

Since \(f(x)=\int_0^x\left(t+\sin(1-e^t)\right)dt\),

we have \(f(0)=0\).

Using L'Hospital's rule,

\(L=\lim_{x\to 0}\frac{x+\sin(1-e^x)}{3x^2}\).

Again it is of the form \( \frac{0}{0} \),

so applying L'Hospital's rule again,

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

Again it is of the form \( \frac{0}{0} \),

so applying L'Hospital's rule once more,

\(L=\lim_{x\to 0}\frac{-e^x\cos(1-e^x)-e^{2x}\sin(1-e^x)}{6}\).

Putting \(x=0\), we get

\(L=\frac{-1\cdot \cos 0-1\cdot \sin 0}{6}=-\frac{1}{6}\).