Let $x_n$ denote a sequence given by a recurrence relation: $$ x_{n+1} = x_n(a-x_n)\\ 0 < x_1 <a\\ n\in\Bbb N $$ Show that: $$ \begin{align} a > 1 &\implies \lim_{n\to\infty}x_n = a-1 \\ 0<a\le 1 &\implies \lim_{n\to\infty}x_n = 0 \end{align} $$
This one is bugging me for a reasonable amount of time already. So far I've done the following. I was trying to translate the recurrence in terms of $x_n$ into a simpler one. I've used a similar method for simpler recurrences of the similar kind.
Consider the recurrence relation: $$ \begin{align*} x_{n+1} &= ax_n - x_n^2 \\ - x_{n+1} &= x_n^2 - ax_n \\ a - 1 - x_{n+1} &= x_n^2 - ax_n + a - 1 \\ a - 1 - x_{n+1} &= (x_n + 1)(x_n - 1) - a(x_n - 1) \\ a - 1 - x_{n+1} &= (x_n - 1)(x_n + 1 - a) \\ x_{n+1} + 1 - a &= (1 - x_n)(x_n + 1 - a) \tag 1 \end{align*} $$
Here I was trying to introduce a new sequence. Namely: $$ y_n = x_n + 1 - a $$
Then for the case of $a > 1$ if I show that $y_n$ is bounded and monotonic I could prove that the limit exists. Then I need to show $\lim y_n = 0$. From which it would follow immediately that $\lim x_n = a -1$. However the $(1 - x_n)$ multiplier in $(1)$ does not allow me to do what I intend. At this point I got stuck.
Is it possible to use my reasoning above in order to show those two limits exist and prove what's in the problem statement? If not what would be the proper way to do it?
Thank you!
If the sequence converges, it will converge to a fixed point of the function $g(x)=x(a-x)$. Hence, the only possible limits are $0$ and $a-1$. This is, however, a very rich and well studied case in discrete dynamical systems.
Depending on the value of $a$ you can get all sorts of periodic or erratic behavior of the orbits (orbit = sequence). So, in general, the sequence is not convergent and you can get sequences which are approximately cyclic.
In particular, it is surely not true that the sequence converges to $a-1$ for $a>1$. In the picture, the x-axis corresponds to the parameter $a$ and the y-axis represents the orbit. You can see that, up to a certain value of $a$ (around 3), the orbit is contant, i.e. the sequence converges (to $a-1$), but then you see orbits with different periods ($2,4,8, \cdots, 3, \cdots$). When you have non-trivial (non.constant) orbits, the sequence does not converge.