A recursive problem

Let $\lim_{n \to \infty} x_n =a$ . From $x_{n+1}=\frac{5}{6-x_n}$ , we have $a=\frac{5}{6-a}$ , which means that $a^2-6a+5=0$ . Solving the equation we obtain $a=1$ or $a=5$ .

As $x_1=2<3$ , we have $x_2=\frac{5}{6-x_1}<\frac{5}{3}<3$ . By similar argument, $x_n<3$ for all $n\ge 1$ . Hence $a \neq 5$ and therefore $a=1$ .