Existence of limit for sequence $x_n=\frac12\left(x_{n-1}+\frac8{x_{n-2}}\right)$ with initial values $x_0=5,x_1=10$ Let $x_0=5,x_1=10,$ and for all integers $n\ge2$ let $x_n=\frac12\left(x_{n-1}+\frac8{x_{n-2}}\right).$ By induction, we have $\forall m\in\mathbb Z_{\ge0}\enspace x_m>0,$ so we can avoid division by $0$ and the sequence is well-defined.
According to a Math GRE practice problem, the limit exists. How can we prove that? Note that, if we assume the limit exists, then we can show it equals $\sqrt8,$ but finding the value of the limit is not my goal here.

My work: We can compute $x_2=5.8,x_3=3.3,$ which are strictly between $4/3$ and $6,$ and then, assuming an inductive hypothesis, for all integers $n\ge4$ we have $4/3<x_{n-1}<6$ and $4/3<8/x_{n-2}<6,$ so that $4/3<x_n<6.$ We can probably compute more values of $x_n$ to get tighter bounds, but I don't see how to actually show convergence.
 A: Let $$x_{n+1}=\tfrac{1}{2}(x_n+\frac{a}{x_{n-1}})$$
Then for $d_n=x_n-\sqrt{a}$, \begin{align}
x_{n+1}-\sqrt{a}&=\tfrac{1}{2}(x_n-\sqrt{a})+\frac{a}{2}\left(\frac{1}{x_{n-1}}-\frac{1}{\sqrt{a}}\right)\\
d_{n+1}&=\tfrac{1}{2}d_n-\frac{\sqrt{a}}{2}\frac{d_{n-1}}{d_{n-1}+\sqrt{a}}=\tfrac{1}{2}d_n-\frac{1}{2}\frac{d_{n-1}}{\frac{d_{n-1}}{\sqrt{a}}+1}\\
\end{align}
So if $|d_{n-1}|<\sqrt{a}/3$, $$|d_{n+1}|\le \begin{cases}\tfrac{1}{2}|d_n|,&d_{n-1}d_n>0\\
\frac{1}{2}|d_n|+\frac{3}{4}|d_{n-1}|,&d_{n-1}d_n<0\end{cases}$$
Since the worst case cannot happen twice in succession, we must have $$|d_{n+2}|\le\tfrac{1}{4}|d_n|+\tfrac{3}{8}|d_{n-1}|$$
This recurrence inequality can be solved, $|d_n|\le A|r_1|^n+B|r_2|^n+C|r_3|^n\to0$ since $r_1\approx0.84$, $|r_2|=|r_3|\approx0.67$.
Hence, as long as some $d_k$ comes close enough to $\sqrt{a}$, $x_n\to\sqrt{a}$. (In fact, the sequence may converge to $-\sqrt{a}$, e.g. $x_0=x_1=-1$ for $a=8$. )
A: As in my comments, let $y_n=x_n/\sqrt8=1+z_n$.  Then
$$z_{n+1}=\frac12(z_n-z_{n-1}+\frac{z_{n-1}^2} {1+z_{n-1}})\\
=\frac14\left(-z_{n-1}-z_{n-2}+2\frac{z_{n-1}^2}{1+z_{n-1}}+\frac{z_{n-2}^2}{1+z_{n-2}}\right)$$
So if $|z_{n-1}|$ and $|z_{n-2}|$ are both at most $c$ which is less than $1/4$ then $|z_{n+1}| \le \frac c4+\frac c4 +\frac{3c^2}{4(1-c)}\lt \frac34c$
A: it is easy to show that $|x_{n}-A|<\epsilon$ for all the $x_{n}$ for a given $A$.
such that
$$A-\epsilon<x_{n}<A+\epsilon$$
$$-A-\epsilon<-x_{n}<-A+\epsilon$$
$$\frac{8}{A+\epsilon}<\frac{8}{x_{n}}<\frac{8}{A-\epsilon}$$
use equation $$x_{n}=\frac{1}{2}(x_{n-1}+\frac{8}{x_{n-2}})$$
we have $$x_{n}-x_{n-1}=\frac{1}{2}(-x_{n-1}+\frac{8}{x_{n-2}})$$
using inequatity (2)(3) we have
$$\frac{1}{2}(-A-\epsilon+\frac{8}{A+\epsilon})<x_{n}-x_{n-1}<\frac{1}{2}(-A+\epsilon+\frac{8}{A-\epsilon})$$
since $\epsilon<<A$ we use the expansion and keep the first term we get:
$$\frac{1}{1+\epsilon}=1-\epsilon+O(\epsilon^{2})$$
we get:
$$\frac{1}{2}(-A-\epsilon+\frac{8}{A(1+\frac{\epsilon}{A})})<x_{n}-x_{n-1}<\frac{1}{2}(-A+\epsilon+\frac{8}{A(1-\frac{\epsilon}{A})})$$
$$\frac{1}{2}(-A-\epsilon+\frac{8}{A}(1-\frac{\epsilon}{A}))<x_{n}-x_{n-1}<\frac{1}{2}(-A+\epsilon+\frac{8}{A}(1+\frac{\epsilon}{A}))$$
$$\frac{1}{2}(-A+\frac{8}{A}-\epsilon-8\frac{\epsilon}{A^{2}})<x_{n}-x_{n-1}<\frac{1}{2}(-A+\frac{8}{A}+\epsilon+8\frac{\epsilon}{A^{2}})$$
Also, we know that $|a+b|<|a|+|b|$ and $|a-b|<|a|+|b|$, such that
$$|x_{n}-x_{n-1}|<\frac{1}{2}(|-A+\frac{8}{A}|+|\epsilon+8\frac{\epsilon}{A^{2}}|)$$
if we set $A=\sqrt{8}$
we get:
$$|x_{n}-x_{n-1}|<\frac{1}{2}(|-\sqrt{8}+\frac{8}{\sqrt{8}}|+|\epsilon+8\frac{\epsilon}{8}|)$$
$$|x_{n}-x_{n-1}|<|\epsilon|$$
which means that $x_{n}$ converge to A
