Let f(x)= _x^0(t+ (1-e^t))dt,x R.

Question

Let $f(x)={\int }_{x}^{0}(t+\sin (1-{e}^{t}))dt,x\in R$. Then, $\underset{x	o 0}{lim}\frac{f(x)}{{x}^{3}}$ is equal to

Accepted Answer

Given $f(x)=\int _{0}^{x}(t+\sin (1-{e}^{t}))dt$

Now, $\underset{x	o 0}{lim}\frac{f(x)}{{x}^{3}}(\frac{0}{0})$

$\begin{gathered} =\underset{x	o 0}{lim}\frac{\int _{0}^{x}(t+\sin (1-{e}^{t}))dt}{{x}^{3}} \ =\underset{x	o 0}{lim}\frac{x+\sin (1-{e}^{x})}{3{x}^{2}}(\frac{0}{0}) \ =\underset{x	o 0}{lim}\frac{1+\cos (1-{e}^{x})(-{e}^{x})}{6x}(\frac{0}{0}) \ =\underset{x	o 0}{lim}\frac{-\sin (1-{e}^{x}){({e}^{x})}^{2}+\cos (1-{e}^{x})(-{e}^{x})}{6} \ =-\frac{1}{6} \end{gathered}$

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