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.

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
