<s>[INST] 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.
 [/INST] 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 $a=8$. )
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