Romanian Mathematical Olympiad - District Round

The limit is equal to $0$.
Let $x_n = (2a_n - a_1 - a_2)(2a_n - a_2 - a_3) \cdots (2a_n - a_{n-2} - a_{n-1})(2a_n - a_{n-1} - a_1)$. The sequence $(a_n)_{n \ge 1}$ is convergent; let $L = \lim_{n \to \infty} a_n$. Since $a_n \le L$ for all $n \ge 1$ we have $2a_n - a_k - a_{k+1} \le 2L - a_k - a_{k+1} \le 2(L - a_k)$, $k = 1, 2, \dots, n-2$, so $0 \le x_n \le 2^{n-1}(L - a_1)(L - a_2) \cdots (L - a_{n-2})(L - a_1) = y_n$.
Suppose $y_n > 0$, otherwise $y_n = 0$ and then $x_n = 0$. Since $\frac{y_{n+1}}{y_n} = 2(L - a_n) \to 0$, we infer that $\lim_{n \to \infty} y_n = 0$ and finally $\lim_{n \to \infty} x_n = 0$.