Limit of $a_{n+1}= (2-a_n)a_n$ Let $(a_n)_{n \in\ \mathbb{N}}$ be a recursive sequence with $a_0= \frac{1}{2}$ and $a_{n+1}= (2-a_n)a_n ~~n \in\mathbb{N}$
(a) Show that $0 < a_n<1$ for all $n \in\mathbb{N}$
(b) Show that the sequence $(a_n)_{n \in\ \mathbb{N}}$ converge and calculate their limit.
I've already proofed (a) and (b), there's only one last step missing, which is to determine the limit:
$a=\lim\limits_{n \rightarrow \infty}a_{n+1}=\lim\limits_{n \rightarrow \infty}((2-a_n)a_n)=(2-a)a \Leftrightarrow 0=a^2-a=a(a-1)$ 
I think it must be $1$ because the sequence is raising.
 A: If
$a_{n+1}= (2-a_n)a_n$,
$\begin{align}
1-a_{n+1}
&=1- (2-a_n)a_n\\
&=1- 2a_n+a_n^2\\
&=(1-a_n)^2\\
\end{align}
$
so
$1-a_n = (1-a_0)^{2^n}$.
Since $0 < a_0 < 1$,
$1-a_n \to 0$
so
$a_n \to 1$.
A: $a_{n+1}= (2-a_n)a_n$ is a first order non linear difference equation of the form $a_{n+1}=f(a_{n})$ where $f(u)=-u^2+2u$
as you proved it should have two equilibrium points(points that satisfy f(x)=x).
now evaluate the absolute value of the derivative at each of this two points.
$f^{'}(0)=2$ and $f{'}(1)=0$. 
$x_{1}=0 $is said to be unstable (repeller or a source since the absolute value of the derivative evaluated at the point $x_{1}=0$ is greater than one ) we may start our sequence near zero but we never stay close to zero as a matter of fact we never converge to zero.
$x_{2}=1$ is said to be locally asymptotically stable as $|f{'}(1)=0|<1$  if we start close we stay close and also converge. and because the sequence is monotonic it is said globally asymptotically stable you can pick $a_{0}$ to be anywhere in the positive reals, the sequence will eventually converge to 1.
another way to prove the convergence is by proving that the sequence is increasing and bounded then prove that $x_{2}=1$ is the least upper bound of the sequence:
observe that $\dfrac{a_{n+1}}{a_{n}}=2-a_{n}>1$ which implies that $a_{n+1}>a_{n}$when $0<a_{n}<1$.
finally to prove that $X_{2}=1$ is the least upper bound assume it is not for the sake of contradiction then that implies the seuquence will converge to a point different than $1$ and $0$ which implies that  point to be also an equilibrium ( satisfying $f(x)=x$ as f is continuous ) and that is a contradiction since the only equilibrium points are zero and one.
A: I would start by showing that it is increasing. That is, prove $0<a_n<1 \forall n \Rightarrow a_n < a_{n+1} \forall n$. Since it is increasing and bounded, you know it is convergent. But increasing doesn't directly imply the limit is 1. To show that, you have to show that 1 is the lowest upper bound of the sequence. The typical way to do this is to assume $\exists M < 1:M \geq a_n \forall n$ and show this is a contradiction.
