Finding the index of a sequence $x_n$ for it to be close enough to its limit Here is the question:
Define a sequence by $x_0=5$ and
$$
x_{n+1}=\frac{x_{n}^{2}+5 x_{n}+4}{x_{n}+6}.
$$
Let $m$ be the least positive integer such that 
$$
x_{m} \leq 4+\frac{1}{2^{20}},
$$
In which of the following intervals does $m$ lie?
$$
[9,26],[27,80],[81,242],[243,728],[729,+\infty].
$$
I noticed that these intervals involve powers of $3$ so I am thinking of using some approximations to $x_n$. This sequence seems to have no general formula, which makes me struggle to find a formula that could approximate $m$.
Is this the correct way or are there better solutions to this problem?
Here is the plot of the sequence.

 A: Write $x_n:=4+y_n$ $(n\geq0)$ with $y_0:=1$. Then you obtain the recursion
$$y_{n+1}={9y_n+y_n^2\over 10+y_n}=y_n\left({9\over10}+{y_n\over100+10 y_n}\right)\qquad(n\geq0)\ .\tag{1}$$
This shows that the $y_n$ are approximately multiplied with ${9\over10}$ at each step. Solving $$\left({9\over10}\right)^m=2^{-20}$$
for $m$ gives $m\approx131.576$, so that we conjecture the true $m_*$ to lie in the interval $[81,242]$. 
For a proof we write $(1)$ in the form
$$y_{n+1}=\lambda_n y_n\quad(n\geq0),\qquad\lambda_n:={9+y_n\over 10+y_n}\ .$$ 
As $0<y_n\leq1$ for all $n\geq0$ we have 
$${9\over10}\leq\lambda_n\leq {10\over11}\qquad(n\geq0)\ .$$
This implies $$\left({9\over10}\right)^n\leq y_n\leq\left({10\over11}\right)^n\qquad(n\geq0)\ .$$
Solving $\left({9\over10}\right)^m=2^{-20}$ and $\left({10\over11}\right)^m=2^{-20}$ then gives you a lower and an upper estimate for the true $m_*$, namely $132\leq m_*\leq146$.
A: Seems like I've got an approximation.
Let $y_n=x_n-4$, then 
$$
y_{n}=y_n\left(1-\frac{1}{y_{n-1}+10}\right)<\frac{9}{10}y_{n-1}<\cdots<\left(\frac{9}{10}\right)^ny_0=\left(\frac{9}{10}\right)^n.
$$
Let $\left(\frac{9}{10}\right)^n\leqslant \frac{1}{2^{20}}$, we get 
$$
n\leqslant \log_{9/10} \frac{1}{2^{20}}\approx 131.5762695792117.
$$
Seems like I should choose the interval $[81,242]$...? I still don't understand why the choices are divided by powers of 3.
