<s>[INST] 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. 
 [/INST] By induction, $1< x_n\le 4$ for all $n$. Indeed, this is true for $n=1$, and if $1< x_n\le 4$, then $1< x_n^2\le 16$ and so $x_{n+1}=\frac{x_n^2+5}{6}$ is $> \frac{1+5}6=1$ and $\le \frac{16+5}6<4$.
With these bounds in mind, we have
$$x_{n+1}-x_n=\frac{x_n^2-6x_n+5}6=\frac{(x_n-1)(x_n-5)}6<0$$
because the first factor in the numerator is positive and the second is negative.
So $\{x_n\}_n$ is bounded and strictly decreasing, hence convergent. You already found out that the only possible limits are $1$ and $5$. However, $5$ can be excluded according to the above observations.
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