How to find the limit $\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx$ 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.
 A: 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
$$
\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx=0
$$
A: 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,
$$
\mathop {\lim }\limits_{n \to  + \infty } \int_0^1 {\frac{{x^n e^x }}{{1 + e^x }}dx}  = 0.
$$
A: 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.
