How is the sequence $x_{n+1} = \frac{(x_n)^{2} + 5}{ 6}$ going to converge to 1 Having some trouble understanding that if $x_{1} = 4$ and the sequence where n is defined as $x_{n+1} = \frac{(x_n)^{2} + 5}{ 6}$ how is it going to converge to 1. 
I have solved using the L as limit and using the quadratic i get two possibilites that are 5, or 1. 

Since
$$
x_{n+1}=\frac{x_n^2+5}6\tag1
$$
we have
$$
x_{n+2}-x_{n+1}=\frac{x_{n+1}^2-x_n^2}6=(x_{n+1}-x_n)\,\frac{x_{n+1}+x_n}6\tag2
$$
This is a contraction when $\left|\frac{x_{n+1}+x_n}6\right|\lt1$. That is.
$$
-1\lt\frac{(x_n+3)^2-4}{36}\lt1\tag3
$$
Inequality $(3)$ is satisfied when
$$
-3-2\sqrt{10}\lt x_n\lt-3+2\sqrt{10}\tag4
$$
$x_1=4$ is outside the range specified in $(4)$.
