What is the limit of $U_{n+1} = \dfrac{2U_n + 3}{U_n + 2}$ and $U_0 = 1$?
I need the detail, and another way than using the solution of $f(x)=x$, as $f(x) = \frac{2x+3}{x+2}$ because I can't show that $f(I) \subseteq I$ as $I = ]-\infty; -2[~\cup~ ]-2; +\infty[$.