Existence of limit for sequence $x_n=frac12left(x_{n-1}+frac8{x_{n-2}}right)$ with initial values $x_0=5,x_1=10$

As in my comments, let $y_n=x_n/sqrt8=1+z_n$. Then $$z_{n+1}=frac12(z_n-z_{n-1}+frac{z_{n-1}^2} {1+z_{n-1}})\ =frac14left(-z_{n-1}-z_{n-2}+2frac{z_{n-1}^2}{1+z_{n-1}}+frac{z_{n-2}^2}{1+z_{n-2}}right)$$ So if $|z_{n-1}|$ and $|z_{n-2}|$ are both at most $c$ which is less than $1/4$ then $|z_{n+1}| le frac c4+frac c4 +frac{3c^2}{4(1-c)}lt frac34c$

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: $$frac{1}{2}(-A-epsilon+frac{8}{A(1+frac{epsilon}{A})})<x_{n}-x_{n-1}<frac{1}{2}(-A+epsilon+frac{8}{A(1-frac{epsilon}{A})})$$

$$frac{1}{2}(-A-epsilon+frac{8}{A}(1-frac{epsilon}{A}))<x_{n}-x_{n-1}<frac{1}{2}(-A+epsilon+frac{8}{A}(1+frac{epsilon}{A}))$$

$$frac{1}{2}(-A+frac{8}{A}-epsilon-8frac{epsilon}{A^{2}})<x_{n}-x_{n-1}<frac{1}{2}(-A+frac{8}{A}+epsilon+8frac{epsilon}{A^{2}})$$

Also, we know that $|a+b|<|a|+|b|$ and $|a-b|<|a|+|b|$, such that

$$|x_{n}-x_{n-1}|<frac{1}{2}(|-A+frac{8}{A}|+|epsilon+8frac{epsilon}{A^{2}}|)$$

if we set $A=sqrt{8}$

we get: $$|x_{n}-x_{n-1}|<frac{1}{2}(|-sqrt{8}+frac{8}{sqrt{8}}|+|epsilon+8frac{epsilon}{8}|)$$

$$|x_{n}-x_{n-1}|<|epsilon|$$

which means that $x_{n}$ converge to 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 $|d_{n-1}|<sqrt{a}/3$, $$|d_{n+1}|le begin{cases}tfrac{1}{2}|d_n|,&d_{n-1}d_n>0\ frac{1}{2}|d_n|+frac{3}{4}|d_{n-1}|,&d_{n-1}d_n<0end{cases}$$ Since the worst case cannot happen twice in succession, we must have $$|d_{n+2}|letfrac{1}{4}|d_n|+tfrac{3}{8}|d_{n-1}|$$

This recurrence inequality can be solved, $|d_n|le A|r_1|^n+B|r_2|^n+C|r_3|^nto0$ since $r_1approx0.84$, $|r_2|=|r_3|approx0.67$.

Hence, as long as some $d_k$ comes close enough to $sqrt{a}$, $x_ntosqrt{a}$. (In fact, the sequence may converge to $-sqrt{a}$, e.g. $x_0=x_1=-1$ for $a=8$. )