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

Explanation -

\(\lim _{x \rightarrow 0} \frac{f(x)}{x^3} \) = 0/0

Using L Hopital Rule.

\(\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{3 x^2}=\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3 x^2} \) (Again L Hopital)

Using L.H. Rule

= \(\lim _{x \rightarrow 0} \frac{-\left[\sin \left(1-e^x\right)\left(-e^x\right) \cdot e^x+\cos \left(1-e^x\right) \cdot e^x\right]}{6} \)

= \(-\frac{1}{6} \)