Let [f(x)=_0^x(t+ (1-e^t)) d t, x R].

Question

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

Accepted Answer

Given $$f(x)=\int_\limits0^x\left(t+\sin \left(1-e^t\right)\right) d t$$

Now, $$\lim _\limits{x \rightarrow 0} \frac{f(x)}{x^3}\left(\frac{0}{0} \text { form }\right)$$

$$\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\int_\limits0^x\left(t+\sin \left(1-e^t\right)\right) d t}{x^3} \
& =\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3 x^2}\left(\frac{0}{0}\right) \
& =\lim _{x \rightarrow 0} \frac{1+\cos \left(1-e^x\right)\left(-e^x\right)}{6 x}\left(\frac{0}{0}\right) \
& =\lim _{x \rightarrow 0} \frac{-\sin \left(1-e^x\right)\left(e^x\right)^2+\cos \left(1-e^x\right)\left(-e^x\right)}{6} \
& =-\frac{1}{6}
\end{aligned}$$

Solutions · 1