67th Romanian Mathematical Olympiad

Existence of $c_n$ follows from the (first) mean-value theorem (each integrand is continuous), and uniqueness follows from injectivity of each integrand.

To evaluate the required limit, let $I_n={\int }_0^1(1+x^n{)}^{-1}dx$, and notice that $|I_n-1|=\left|{\int }_0^1\frac{-x^n}{1+x^n}dx\right|\le {\int }_0^1{x}^{n}dx=\frac{1}{n+1}$, so ${\lim }_{n\to \infty }{I}_n=1$. Since $n(c_n{)}^n=n(1/I_n-1)$, it is sufficient to evaluate ${\lim }_{n\to \infty }n(1-I_n)$. To this end, write $n(1-I_n)=n{\int }_0^1\frac{x^n}{1+x^n}dx={\int }_0^{1}x\cdot (\log (1+x^n){)}^{\prime }dx=\log 2-{\int }_0^1\log (1+x^n)dx$. Since $0\le {\int }_0^1\log (1+x^n)dx\le {\int }_0^1{x}^{n}dx=\frac{1}{n+1}$, it follows that ${\lim }_{n\to \infty }{\int }_0^1\log (1+x^n)dx=0$, so ${\lim }_{n\to \infty }n(c_n{)}^n=\log 2$.