Prove that $\lim_{n \to \infty} x_n = \sqrt{B},$ understanding the cases. I am trying to prove this question:
Assume $B$ is a positive number. Let $\{x_n\}_{n=1}^{\infty}$ be defined recursively by $x_1 = 1,$ and $$x_{n+1} = \frac{x_n(3B + x_n^2)}{3x_n^2 + B}$$ for all $n \geq 1.$ Prove that $\lim_{n \to \infty} x_n = \sqrt{B}.$
And here is a trial for the solution:





But I am wondering if there is a better and succinct way of solving it? Also, I think I glanced the solution (but with $B = a$ and under a user name Man or Nan, but I am not very sure from the user name ) of this question here on MSE but I lost it quickly and I am unable to get it, could anyone help me in finding the link for it?
Thanks!
EDIT:
1 - I do not see why, in the first case, $x_n = 1,$ could someone show me the detailed calculations for this please?
2- I also, did not understand case 2 and 3, could someone please explain them to me in details?
 A: I intend to give you a glimpse of a possible way to go in the problem.
Let $$g(x)=\frac{x(3b+x^2)}{3x^2+b},\qquad b>0.$$ It follows that $g(x)=x$ just when $x=0$ or $x=\pm\sqrt{b}$.
The given recurrence relation becomes $$x_1=1,\qquad x_{n+1}=g(x_n),\tag{1}$$ and, if $x_n$ converges to some $l$, then $l=g(l)$.
Note that $$g'(x)=3\left(\frac{x^2-b}{b+3x^2}\right)^2> 0,\quad x\neq\pm \sqrt{b},$$ $g'(0)=3$, $g'(\pm \sqrt{b})=0$, and $$\lim_{x\to \pm\infty}g'(x)=\frac{1}{3},$$ and $$g''(x)=48x\frac{(x^2-b)}{(b+3x^2)^3}.$$ As so, you can see that $$0\le g'(x)\leq 3,\qquad 0\leq x\leq\sqrt{b},\tag{2}$$ and $$0\le g'(x)\leq \frac{1}{3},\qquad \sqrt{b}\leq x.\tag{3}$$
It follows from $(1)$ that $x_n>0$ and expressions $(2)$ and $(3)$ means that there exists $0<r<\sqrt{b}$ such that $g'(u)>1$, to all $u\in[0,r)$, $g'(r)=1$, and $g'(u)<1$, to all $u\in(r,+\infty)$.
This enable us to consider two cases:

*

*If some $x_n\leq r$, then $$x_{n+1}=g(x_n)=g(x_n)-g(0)=g'(v)(x_n-0)>x_n,$$ to some $v\in (0,x_n)$. This means that $x_n$ can not converges to $0$.


*If some $x_n>r$, then $$|x_{n+1}-\sqrt{b}|=|g(x_n)-g(\sqrt{b})|=|g'(v)(x_n-\sqrt{b})|<|x_n-\sqrt{b}|,$$ to some $v$ between $x_n$ and $\sqrt{b}$.
This means that $x_n$ converges to $\sqrt{b}$.
A: Definitions. (The Hyper-Trigonometric functions). $\cosh p=(e^p+e^{-p})/2$ and $\sinh p=(e^p-e^{-p})/2.$ If $e^p\ne 1$ then $\coth p=\frac {\cosh p }{ \sinh p}.$ If $p\in \Bbb R$ then $\cosh p \ne 0$ and $\tanh p =\frac {\sinh p}{ \cosh p}.$
Analogous to trigonometric angle-sum formulas, we have
$\cosh (p+q)=\cosh p \cosh q+\sinh p \sinh q.$
$\sinh (p+q)=\sinh p\cosh q+\cosh p\sinh q.$
Applying these with $q=p,$ and again with $q=2p,$ we obtain
$\coth 3p=\dfrac {3\coth p+(\coth p)^3}{3(\coth p)^2+1}\quad$ if $e^p\ne 1.$
$\tanh 3p=\dfrac {3\tanh p+(\tanh p)^3}{3(\tanh p)^2+1}\quad$ if $p\in\Bbb R$.
If $B>0$ and $x_1>\sqrt B,$ let $x_1=\sqrt B\,\coth p$ with  $p>0.$ Then $$x_{n+1}=\sqrt B\, \coth 3^np=\sqrt B \cdot\frac {1+e^{-2\cdot 3^n p}}{1-e^{-2\cdot 3^n p }}.$$
If $B>0$ and $0<x_1<\sqrt B,$ let $x_1=\sqrt B\,\tanh p$ with  $p>0.$ Then $$x_{n+1}=\sqrt B\, \tanh 3^np=\sqrt B \cdot \frac {1-e^{-2\cdot 3^n p}}{1+e^{-2\cdot 3^n p }}.$$
If $B>0$ and $x_1=\sqrt B$ (trivial case) then $x_n=\sqrt B$ for every $n.$
Footnote. In the 1st century CE, Heron (Hero) of Alexandria gave the iteration $y_{n+1}=(y_n+B/y_n)/2,$ converging to $\sqrt B$ when $B$ and $y_1$ are positive. Now if $y_1\ne \sqrt B$ then $y_2>\sqrt B,$ and I once noticed that if $y_2=\sqrt B\coth q$  then $y_{n+2}=\sqrt B\coth 2^n q.$ In particular $y_3=\sqrt B\coth 2q$ and $y_4=\sqrt B\coth 4q,$ and I became curious about $\sqrt B\coth 3q,$ which led me to the iteration in your Q.
