Prove that the recursively defined sequence is Cauchy. Define a sequence recursively by $x_1=1$, $x_{n+1}=\frac{5+5x_n}{5+x_n}$. Prove that this sequence is Cauchy. Then find the limit.
We call a sequence $\{a_n\}$ Cauchy, if for any $\epsilon>0$, there exists $N\in\mathbb{N}$ such that for all $n,m\geq M$, $|a_n-a_m|<\epsilon$.

$$
x_{n+1} - \sqrt 5 = \frac{5+5x_n}{5+x_n} - \sqrt{5} = \frac{(5-\sqrt{5})(x_n-\sqrt 5)}{5+x_n}
$$
$$
x_{n+1} + \sqrt 5 = \frac{5+5x_n}{5+x_n} + \sqrt{5} = \frac{(5+\sqrt{5})(x_n+\sqrt 5)}{5+x_n}
$$
Therefore
$$
\frac{x_{n+1}-\sqrt 5}{x_{n+1}+\sqrt 5}=\frac{5-\sqrt 5}{5+\sqrt 5} \cdot \frac{x_n-\sqrt 5}{x_n+\sqrt 5}\\
\Rightarrow \frac{x_n-\sqrt 5}{x_n+\sqrt 5}=\left(\frac{5-\sqrt 5}{5+\sqrt 5}\right)^{n-1} \cdot \frac{1-\sqrt{5}}{1+\sqrt{5}}
$$
then you can get a closed form solution for $x_n$ and the rest should be straightforward.
Please check this post: Recursive sequence depending on the parameter
I just learned about Möbius transformation the other day.
