<s>[INST] prove conjecture; the limit of iterating is $\sqrt{z^2 - 2}$ $$\lim_{n \to \infty} f_n(x)=x-\frac{1}{nx}\;\;\; g(x)=f_n^{on}(x)$$
The conjecture is for values of $|x|>\sqrt{2}$: $g(x) = \sqrt{z^2 - 2}$
This question comes from another matstack question/answer.
If $n=2^m$, then convergence is much quicker by starting with $f_n(x)=x-\frac{1}{nx}$ and then iterating the Taylor/Laurent series for $f(x) \mapsto f(f(x))$ m times.  I start by generating the formal Taylor series after moving the fixed point from infinity to zero, $1/f(\frac{1}{x})\;$.  Then we start iterating with $f_n$ 
$$f_n(x) = \frac{x}{1 - x^2/n} =  x + \frac{x^3}{n} + \frac{x^5}{n^2} + \frac{x^7}{n^3} + \frac{x^9}{n^4} + \frac{x^{11}}{n^5} ...$$
Using this speedup with $m=\log_2(n)$, one can iterate the Taylor series for $f(x) \mapsto f(f(x))$ m times, rather then iterating $f^{on}$, but the two are of course identical.  Its just that otherwise convergence is pretty slow, with n iterations to get accuracy to 1/n.
The formal Taylor series coefficients of $1/g(\frac{1}{x}) = \sqrt{z^2/(1-2z^2)}$ are:
$x + x^3 +
\frac{3        x^5    }{2     } +
\frac{5        x^7    }{2     } +
\frac{35       x^9    }{8     } +
\frac{63       x^{11} }{8     } +
\frac{231      x^{13} }{16    } +
\frac{429      x^{15} }{16    } +
\frac{6435     x^{17} }{128   } +
\frac{12155    x^{19} }{128   } +
\frac{46189    x^{21} }{256   } ...$
Empirically, for the $2^n$th iteration starting with $f_n$ in the limit above, the Taylor coefficients are accurate to approximately $O 2^{-n}$, so there is pretty good numerical computation evidence for the conjecture, but I have no idea how to prove it.
EDIT
I (https://math.stackexchange.com/users/39261/mick) will place a bounty for the following problem :
Basicly the inverse :
Suppose we are given $g(x)=\sqrt{x^2-2}$ and we are asked to find $f_n$ such that :
$$\lim_{n \to \infty} \;\;\; g(x)=f_n^{on}(x)$$
How do we solve such problems ??
EDIT
EDIT 2
As Sheldon's comment says $f_n(x)=\sqrt{x^2-\frac{2}{n}}$ is also a solution but I want to find the $f_n$ from the OP : $x - \frac{1}{nx}$ or another $f_n$ that is real-meromorphic on the entire complex plane. 
EDIT 2
 [/INST] Pick an $x > \sqrt{2}$ and $n > \dfrac{1}{x^2-2}$ and define a sequence by $x_0 = x$, $x_{k+1} = f_n(x_{k})$ for $k \ge 0$. 
Rearrange $x_{k+1} = f_n(x_k) = x_k - \dfrac{1}{nx_k}$, to get $x_k(x_k-x_{k+1}) = \dfrac{1}{n}$. 
Clearly, $x_0 \ge \sqrt{x^2}$. Now, suppose that $x_{m-1} \ge \sqrt{x^2-\dfrac{2(m-1)}{n}}$ for some $1 \le m \le n$. 
Then, $x_{m-1} \ge \sqrt{x^2-2} > \dfrac{1}{\sqrt{n}}$, and so, $x_m = x_{m-1} - \dfrac{1}{nx_{m-1}} > 0$. Furthermore, we have: 
$x^2-x_m^2 = \displaystyle\sum_{k = 0}^{m-1}(x_k^2-x_{k+1}^2) = \sum_{k = 0}^{m-1}\left[(2x_k-(x_k-x_{k+1}))(x_k-x_{k+1})\right]$ $= \displaystyle\sum_{k = 0}^{m-1}\left[2x_k(x_k-x_{k+1}) - (x_k-x_{k+1})^2\right] = \sum_{k = 0}^{m-1}\left[\dfrac{2}{n} - \dfrac{1}{n^2x_k^2}\right] = \dfrac{2m}{n} - \dfrac{1}{n^2}\sum_{k = 0}^{m-1}\dfrac{1}{x_k^2} \le \dfrac{2m}{n}$. 
Hence, $x_m^2 \ge x^2 - \dfrac{2m}{n}$. Then, since $x_m > 0$, we have $x_m \ge \sqrt{x^2-\dfrac{2m}{n}}$. 
So by induction, we have $x_n \ge \sqrt{x^2-2}$. (Note: all of that was necessary to ensure that $x_n > 0$ , otherwise, $x^2-x_n^2 \le 2$ doesn't imply that $x_n \ge \sqrt{x^2-2}$.)
Then, since $x^2-x_n^2 = 2 - \dfrac{1}{n^2}\displaystyle\sum_{k = 0}^{n-1}\dfrac{1}{x_k^2} \ge 2 - \dfrac{1}{nx_n^2} \ge 2 - \dfrac{1}{n(x^2-2)}$, we have $x_n^2 \le x^2-2 + \dfrac{1}{n(x^2-2)}$, i.e. $x_n \le \sqrt{x^2-2 + \dfrac{1}{n(x^2-2)}}$.
Therefore, $\sqrt{x^2-2} \le f_n^n(x) \le \sqrt{x^2-2 + \dfrac{1}{n(x^2-2)}}$ for all $n > \dfrac{1}{x^2-2}$. 
Taking the limit as $n \to \infty$ and squeezing yields $g(x) = \displaystyle\lim_{n \to \infty}f_n^n(x) = \sqrt{x^2-2}$, as desired. 
Note that the proof that $g(x) = \sqrt{x^2-2}$ for $x < -\sqrt{2}$ is similar.
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