If $\int\limits_0^1\frac{x^n}{1+x}\,\mathrm{d}x=\xi_n^n\,\ln 2,$ prove that $\xi_n^n \to 0$ 
By mean value theorem for integrals, for all $n$ there exists $\xi_n\in [0,1]$ such that:
  $$\int\limits_0^1\frac{x^n}{1+x}\,\mathrm{d}x=\xi_n^n\,\ln 2.$$
  Prove that $\xi_n^n \to 0$ as $n\to +\infty.$

Attempt. It is not hard to see that $\xi_n\in [0,1)$. But how can we be sure that $\xi_n$ is not, for example, $1-\frac{1}{n}$ and therefore $\xi_n^n\to \frac{1}{e}$?
Thanks in advance for the help.
 A: This is equivalent to the problem of showing $$\lim_{n\rightarrow\infty}\int_0^1\frac {x^n}{1+x}\;dx = 0$$ Use the compound inequality $$0\leq\int_0^1\frac {x^n}{1+x}\;dx\leq\int_0^1 x^n\;dx =\frac{1}{n+1}$$ and we must have $$0\leq\lim_{n\rightarrow\infty}\int_0^1\frac {x^n}{1+x}\;dx\leq\lim_{n\rightarrow\infty}\frac{1}{n+1} = 0 $$
$$\therefore\;\lim_{n\rightarrow\infty}\int_0^1\frac {x^n}{1+x}\;dx=0=\lim_{n\rightarrow\infty}\xi_n^n\cdot ln(2) $$ $$\therefore \lim_{n\rightarrow\infty}\xi_n^n = 0$$
A: Hint:(to calculate the integral explicitly)
Let $$f(n)=\displaystyle \int_0^1 \frac{x^n}{1+x}dx$$
So $$f(n)=\displaystyle \int_0^1 \frac{x^n+x^{n-1}-x^{n-1}}{1+x}dx$$
$$ =\int_0^1 x^{n-1}-\int_0^1\frac{x^{n-1}}{1+x}dx=\frac{1}{n}-f(n-1)$$
$$\implies f(n)=\frac{1}{n}-f(n-1)$$
A: \begin{align*}
\int_{0}^{1}\dfrac{x^{n}}{1+x}dx\leq\int_{0}^{1}\dfrac{x^{n}}{2\sqrt{x}}dx=\dfrac{1}{2}\int_{0}^{1}x^{n-1/2}dx=\dfrac{1}{2}\dfrac{1}{n+1/2}\rightarrow 0.
\end{align*}
A: For $0<x <1$ and $n>0$ we have
$$\frac{x^n}{2}<\frac{x^n}{1+x} < x^n \implies \int_{0}^{1} \frac{x^n}{2} dx < I_n= \int_{0}^{1} \frac{x^n}{1+x} dx < \int_{0}^{1} x^n ~ dx$$ $$\implies \frac{1}{2(n+1)}< I_n< \frac{1}{1+n}$$
Hence by squeez law $$ \lim_{n \rightarrow \infty} I_n= 0$$
