Limit of a Weird Sequence

Checking the domains, it is clear that $\begin{aligned} \mathrm{Dom}\,\phi & = \; \mathbb{R} \backslash \{2\} \\ \mathrm{Dom}\,\phi^2 & = \; \mathbb{R} \backslash \{2,\phi^{-1}(2)\} \\ \mathrm{Dom}\,\phi^3 & = \; \mathbb{R} \backslash \{2,\phi^{-1}(2),(\phi^{-1})^2(2)\} \\ \cdots & = \; \cdots \\ \mathrm{Dom}\,\phi^n & = \; \mathbb{R} \backslash \{2,\phi^{-1}(2),(\phi^{-1})^2(2),...,(\phi^{-1})^{n-1}(2)\} \\ \end{aligned}$ which means that we must have $x_n \; = \; (\phi^{-1})^{n-1}(2) \hspace{1cm} n \ge 2 \hspace{2cm} x_1 = 2$ In other words. we have the recurrence relation $x_n \; = \;2 + x_{n-1}^{-1} \hspace{1cm} n\ge 2 \hspace{2cm} x_1 = 2$ If we consider $a = 1 + \sqrt{2}$ , then $a = 2 + a^{-1}$ , and hence $x_n - a \; = \; x_{n-1}^{-1} - a^{-1} \; = \; \frac{1}{ax_{n-1}}(a - x_{n-1}) \hspace{2cm} n\ge 2$ It is a simple induction to see that $x_n \ge 2$ for all $n \ge 2$ , and hence $|a_n - a| \; \le \; \tfrac{1}{2a}|x_{n-1}-a| \hspace{2cm} n \ge 2$ and, since $2a > 1$ , we deduce that $\lim_{n \to \infty}x_n = a = \boxed{1 + \sqrt{2}}$ .

It can be proved easily by Mathematical Induction that $\phi^n(x)=\frac{a_{n-1}x+b_{n-1}}{a_{n}x+b_{n}},$ where the sequences $a_n$ and $b_n$ satisfy the recurrence relations $a_{n+2}=-2 a_{n+1}+a_n,$ and $b_{n+2}=-2 b_{n+1}+b_n,$ respectively, and $a_0=0, a_1=1, b_0=1,$ and $b_1=-2.$ Using the theory of Linear Recurrence Relations, it follows that $a_n= -\frac{1}{2\sqrt 2}((-1-\sqrt 2 )^n-(-1+\sqrt 2)^n)$ and $b_n= \frac{1}{4}((2+\sqrt 2)(-1-\sqrt 2 )^n+(2-\sqrt 2)(-1+\sqrt 2)^n)$ Besides that, it can also be proved that $(\mathbb{R}\setminus\text{Dom} \: \phi^n) \setminus (\mathbb{R}\setminus\text{Dom} \: \phi^{n-1})=\{\frac{-b_n}{a_{n}}\}.$ This seems to be obvious, but is not. We would need to verify that the sequences $\{-\frac{b_n}{a_n} \}$ does not have repeated terms, but this can be verified by direct calculation. Then $x_n=-\frac{b_n}{a_n}.$ Using the formulas obtained above for $a_n$ and $b_n$ and making $r=\frac{-1+\sqrt 2}{-1-\sqrt 2}$ , it follows that $x_n=\frac{\sqrt 2}{2}\left(\frac{2+\sqrt 2+(2-\sqrt 2)r^n}{1-r^n}\right)$ Since $|r|<1,$ it is easy to compute that $\lim_{n\rightarrow\infty}x_n= 1+\sqrt{2}\approx \boxed{2.414.}$