Limit of the sequence $x_{n+1}=x_n(2-ax_n)$ For some real $a$ positive. Find the limit of the following recurrence relation $$x_{n+1}=x_n(2-ax_n)$$
For some real $a$ positive.

I don't know how to find a closed form of the given recurrence relation and since I don't have any initial value so I cannot check OEIS sequence for a possible solution.
 A: Allow complex values.  The solution is
$$
x_n = \frac{1-(1-ax_0)^{2^n}}{a} .
\tag1$$
So if $|1-ax_0|<1$, the limit is $1/a$.  If $|1-ax_0|>1$ then $|x_n| \to \infty$.  If $1-ax_0 = 1$ or $-1$, then $x_n \to 1/a$.
We could have complex value for $1-ax_0$ with $|1-ax_0|=1$ where the sequence does not converge.

How did I find $(1)$?  In general, iterating a quadratic polynomial is conjugate to the Mandelbrot iteration of $z^2+c$ for some $c$.  If $c=0$ or $-2$, there is a formula for the sequence, otherwise not.
So I tried $u_n = \alpha x_n +\beta$, derived the recurrence for $u_n$, then chose $\alpha, \beta$ to get $1$ as coefficient of $u_n^2$ and $0$ as coefficient of $u_n$.  The result was
$$
u_n = \frac{1-ax_n}{2},\qquad x_n = \frac{1-2u_n}{a}
$$
and the recurrence was
$$
u_{n+1} = u_n^2
$$
Since we got $c=0$, we get the formula
$$
u_n = u_0^{2^n}
$$
Retrace to get $(1)$.
A: Rewrite the condition as
$$x_{n+1} = x_n(2-ax_n) \\
ax_{n+1}= 2ax_n - a^2x_n^2 \\
ax_{n+1} -1 = -(ax_n - 1)^2$$
Now, set $y_n = ax_n - 1$, then
$$y_{n+1} = -y_n^2$$
I'm not sure if indeed no information given about the initial value since if $y_1 = 1$, then $\lim_{n\to\infty} y_n = -1$, but if $y_1 = 0$, then the limit is $0$. Anyway, you can find the limit with this observation.
