sequences and series - Evaluating $sqrt{6+sqrt{6+cdots}}$

Tough as introduction to analysis for beginners (Dutch handbook - I'm Belgian). Again
($n$) means index $n$, $x_1 = \sqrt6$, $x_{n+1} = \sqrt{6+x_n}$

1. Question:

$$|x_{n+1} - 3| \le 1/5 \cdot |x_n - 3|$$

For me this means that $3$ as a 'limit', we need to find that the distance between $x_{n+1}$ and the 'limit' is $1/5$ the distance between the $x_n$ and the limit.
Where does the $1/5$ come from?

1. Prove that $|x_n - 3|\le (1/5)^{n-1}$




2. prove that the sequence converges to $3$.

ps:
When I studied maths in 1980. we went quickly towards metric spaces, so these calculus minded times are nothing compared to those times. But still, as I didn't pass then, I'd like to restart on a new basis.
Thanks for all the help.
If you know where maths can be studied in community on the net, always welcome.