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Find the limit $$ \lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx $$

By intuition, I can guess the answer is $0$, but I have no idea how to start to prove it.

What I have tried is using the mean value theorem. There exists $c \in (0, 1)$: $$ \lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx = \lim_{n \to \infty}\frac{c^ne^c}{1 +e^c} $$ and I don't see how I should continue from here.

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  • $\begingroup$ Try the dominant convergence theorem. $\endgroup$ – CO2 Dec 15 '20 at 14:50
  • $\begingroup$ Welcome to Math.SE! Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 15 '20 at 14:51
  • $\begingroup$ @BiAo the $e^c/(1+e^c)$ is a constant and since $0<c<1$ then $c^n\to 0$ as $n\to\infty$. $\endgroup$ – Pixel Dec 15 '20 at 15:09
  • $\begingroup$ @Pixel The $c$ will tipically depend on $n$. What if $c$ approaches $1$? For example $c \sim 1-\frac{1}{n}$? $\endgroup$ – Gary Dec 16 '20 at 6:49
  • $\begingroup$ @Gary yes that does seem problematic. I tried some substitutions but the carpet still doesn't fit. A $1-\varepsilon$ argument may work as DonAntonio mentioned, but not sure. $\endgroup$ – Pixel Dec 16 '20 at 13:21
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From $$ 0\le \int_0^1{\frac{x^ne^x}{1+e^x}}dx=\int_0^1{\frac{x^n}{1+e^{-x}}}dx\le \int_0^1{\frac{x^n}{1}}dx=\int_0^1{x^n}dx $$ From the squeeze theorem, take limits on both sides and we get $$ 0\le \lim_{n\rightarrow \infty} \int_0^1{\frac{x^ne^x}{1+e^x}}dx=\lim_{n\rightarrow \infty} \int_0^1{\frac{x^n}{1+e^{-x}}}dx\le \lim_{n\rightarrow \infty} \int_0^1{x^n}dx=0 $$ and thus $$ \lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx=0 $$

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We have $$ 0 < \frac{{e^x }}{{1 + e^x }} < 1 $$ for $0<x<1$, thus $$ 0 < \int_0^1 {\frac{{x^n e^x }}{{1 + e^x }}dx} < \int_0^1 {x^n dx} = \frac{1}{{n + 1}}. $$ Consequently, by the squeeze theorem, $$ \mathop {\lim }\limits_{n \to + \infty } \int_0^1 {\frac{{x^n e^x }}{{1 + e^x }}dx} = 0. $$

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Since the integrand is a continuous function, the mean value theorem for integrals tells us that there exists $\;0<c<1\;$ s.t.

$$\int_0^1x^n\frac{e^x}{1+e^x}\,dx=c^n\frac{e^c}{1+e^c}\xrightarrow[n\to\infty]{}0$$

since $\;c^n\to 0\;$ and $\;0\le \cfrac{e^x}{1+e^x}\le 1\;$ is bounded.

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  • $\begingroup$ The $c$ will tipically depend on $n$. What if $c$ approaches $1$? For example $c \sim 1-\frac{1}{n}$? $\endgroup$ – Gary Dec 16 '20 at 6:47
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    $\begingroup$ @Gary Very good point, though taking the upper limit as $\;1-\epsilon\;$ could solve the problem...perhaps. I shall take a second thought on this one. Observe that $\;c_n\to 1\;$ is the only problem here. Any other behaviour, like swinging away from $\;1\;$ or converging to some $\;h\in [0,1)\;$ would yield the same result as above. $\endgroup$ – DonAntonio Dec 16 '20 at 8:49

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