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I would like to find the following limit without using any Taylor expansion.

$$\lim _{n\to \infty }n \int_{-1}^0(x + e^x)^{n}dx = \: ?$$

I know the answer is $\frac{1}{2}$. I tried to write $-1 < x < 0$ and $0 < e^x < 1$, sum up the relations and then integrate, but I found nothing conclusive.

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We are looking for $\lim_{n\to +\infty} n\int_{0}^{1}(e^{-x}-x)^n\,dx $. With simple estimations we notice that the contribution given by the integral over $(1/2,1)$ is irrelevant, then notice that over the interval $(0,1/2)$ the integrand function is bounded between $(1-2x)^n$ and $e^{-2nx}$. Since an explicit computation gives

$$ \lim_{n\to +\infty}n\int_{0}^{1/2}(1-2x)^n\,dx = \color{red}{\frac{1}{2}} = \lim_{n\to +\infty}n\int_{0}^{1/2}e^{-2nx}\,dx$$ the claim follows by squeezing.

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  • $\begingroup$ Can you explain, please, why "With simple estimations we notice that the contribution given by the integral over (1/2,1)(1/2,1) is irrelevant" $\endgroup$ – Liviu Apr 13 '17 at 11:47
  • $\begingroup$ @Liviu: over the interval $(1/2,1)$ the integrand function is bounded in absolute value by $\left(\frac{2}{3}\right)^n$, hence such sub-interval of the integration range does not contribute to the limit. $\endgroup$ – Jack D'Aurizio Apr 13 '17 at 11:50

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