Does $x_n$ converge where $x_{n+1}=-16+ 6x_n+\frac{12}{x_n}$ 
Define $x_{n+1}=-16+ 6x_n+\frac{12}{x_n} \hspace{0.2cm} \forall n\in \mathbb{N}$ where $x_1\neq 2$. I need to find whether it diverges or converges?

I tried proving $\{x_n\}$ is  increasing/ decreasing and bounded, but nothing really worked. I guess that it should diverge!
 A: We shall first show that for $x_1 >2$ the series diverges.
We first solve for fixed $\epsilon > 0$ the equation $x_{n+1} = -16 + 6x_n + \frac{12}{x_n} > (1+\epsilon)x_n$.
\begin{eqnarray}
0 &<& -16 + 6 x_n +\frac{12}{x_n} - (1+ \epsilon) x_n \\
&=&(5- \epsilon)x_n -16 + \frac{12}{x_n}
\end{eqnarray}
Multiplying both sides by $x_n$ yields
$$ 0< (5- \epsilon) x_n^2 -16x_n + 12.$$
This gives as positive root $x_n > \frac{8 + \sqrt{4+ \epsilon}}{5-\epsilon}$.
So given $x_1 > 2$, we can choose an $\epsilon > 0$ such that $x_2 > (1+ \epsilon) x_1$, and we can therefore show that $x_n > ( 1+ \epsilon)^n x_1$. And as $\epsilon > 0$, we have that $x_n$ diverges.
Now, let us see for which $x_1 < 2$ we have $x_2 > 2$.
This gives us $$2 < -16 + 6x_1 + \frac{12}{x_1} \implies 0 < 6x_1^2 -18x_1 + 12 = 6(x_1 - 1)(x_1 - 2)$$
So for $x_1 \in [1,2]$ we have $x_2 <2$ and for $x <1$ we have $x_2 > 2$. So for $x_1 < 1$, we have $x_2 > 2$, so we can find a $\epsilon > 0$ such that $x_3 > (1+\epsilon)x_2$, so we can, as before, conclude that $x_n$ diverges.
Now, we are left with $x_1 \in [1,2]$. However, I cannot find an rigorous proof for which values it converges and for which values it diverges. This since we can plot the function $f(x)=-16+6x+\frac{12}{x}$ and $y=x$ and see how the series behaves.
To do so, pick an $x_1$ on the $x$-axis, go up till you hit $f(x_1)$, then horizontal till you hit $y=x$, which gives $x_2$, go vertical till you hit the $f(x_2)$, and horizontal till you hit $y=x$, which gives $x_3$ and repeat.
However, when we zoom in around $x_1 \in [1.4,1.5)$, we see that $f(x_1) <1$, so then we diverge.
It might be possible to give a rigorous proof for the points in $[1,2]$ for which the function converges, and for which the function diverges, but at the moment, I do not see such proof.
A: Outline :
Let us assume that it converges, then we can set $x_{n+1} = x_n$ and solve for x. 
We get the quadratic $(x-2)(5x - 6) = 0$ so we get $x = 2$ and $x = \frac{6}{5}$.
So definitely for $x_1 = \frac{6}{5}$ it is a constant sequence and it converges. We cannot let $x_1 = 2$ because it is specifically excluded from the problem. 
But we could have for some value of $n$, $n \gt 1$, $x_n = 2$ or $x_n = \frac{6}{5}$. 
So we now take $\frac{6}{5} = -16 + 6x_n + \frac{12}{x_n}$. Solving this we get $x_n = \frac{5}{3}$. So if we let $x_1 = \frac{5}{3}$, we have yet another solution. We can continue this chain.
Similarly, take the other branch, $x_n = 2$ for $n > 1$. 
Solve $2 = -16 + 6x_n + \frac{12}{x_n}$ and you can $x = 1$. Continue the chain and you will get $\frac{4}{3}$, $\frac{3}{2}$. Continuing this, you get some irrational answers like $\frac{35 \pm \sqrt(73)}{24}$
At this point, we can show that there are many values of $x_1$ for which it converges,
for eg. $1, 1.2, \frac{35 - \sqrt(73)}{24} = 1.102333, \frac{35 + \sqrt(73)}{24} =1.814333, \frac{4}{3}, \frac{5}{3}$ and so on.
Disclaimer :
I'm not very happy with the missing rigour in this answer, and it is also not a complete answer, but at the same time it hopefully provides more insight than putting it in short in a comment. Hope that someone would post a more rigorous and a complete answer.
