Showing the sequence is monotone, bounded, and finding the limit The problem I am having is figuring out the way show the following sequence is monotone:
let $x_1 = \frac{3}{2}$ and $x_{n+1} = {x_n}^2-2x_n+2$, show that the sequence $x_n$ is monotone and bounded and find the limit.
I have found the first three terms, and found that the sequence is decreasing, I have followed an example in my text that is the opposite however my text is vague and I'm not sure how they found where the sequence is bounded I am led to believe by math that it is bounded by 1.
any hints or suggestions on how to approach the problem would bennefit me greatly
thanks
 A: Let $f(x) = x^2 - 2x + 2$. Then it is easy to show that $f([1,2]) \subset [1,2]$. Since $x_1 \in [1,2]$, this implies that
$$x_n = f(x_{n-1}) = (f \circ f)(x_{n-2}) = ... = (f\circ ... \circ f)(x_1) \in [1,2]$$
so $\{x_n\}$ stays inside $[1,2]$ and, in particular, bounded. Now when $1 \le x \le 2$ we have
$$f(x) - x = x^2 - 3x + 1 \le 2x - 3x + 1 \le 0$$
so that
$$x_{n+1} = f(x_n) \le x_n$$
and $\{x_n\}$ is decreasing. Hence
$$x = \lim_{n \to \infty} x_n$$
exists. Since $f$ is a continuous function,
$$x = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} f(x_n) = f \left( \lim_{n \to \infty} x_n \right) = f(x)$$
or, in other words,
$$ 0 = f(x) - x = x^2 - 3x + 1 = (x-1)(x-2).$$
Therefore, $x \in \{1,2\}$. But since we showed $\{x_n\}$ is decreasing we must have $x \le x_1 = \frac{3}{2}$ and we conclude $x = 1$.
A: If your second equation is supposed to be $x_{n+1} = x_n^2-2x_n+2$, it might be useful to note that this is equivalent to $x_{n+1}-1 = (x_n-1)^2$. Thus, you might want to study the sequence $y_n = x_n-1$.
A: My turn... request check for typos:
We first rewrite the sequence as (Abel's bright idea: complete the square),
$$x_{n+1}={x_n}^2-2x_n+2=(x_n-1)^2+1.$$
$\textbf{Boundedness:}$ We prove that $x_n$ is bounded on the half open interval $1 \leq x_n<2$ using proof by induction on $n$.
For a basis, let $n=1$. Then $x_1=\frac{3}{2}$, and since $1 \leq \frac{3}{2}<2$, the basis holds.
For the inductive hypothesis we assume that the sequence is bounded for $n=k$, which is to say that
$$1 \leq x_k<2.$$
Our burden is to show that the sequence holds for $n=k+1$.
\begin{align*}
& 1 \leq x_k<2 \\
\Rightarrow & 0 \leq x_k-1<1 \\
\Rightarrow & 0 \leq (x_k-1)^2 <1 \\
\Rightarrow & 1 \leq (x_k-1)^2+1 <2 \\
\Rightarrow & 1 \leq x_{k+1} <2.
\end{align*}
Thus we have shown that the sequence is bounded for all $n \in \mathbb{N}$.
$\textbf{Monotonicity:}$ We now prove that the sequence is monotone decreasing on the half open interval $[1,2)$ by showing that for all $n$, $x_{n+1}-x_n \leq 0$.
\begin{align*}
x_{n+1}-x_n &= (x_n-1)^2+1-x_n \\
&= {x_n}^2 -3x_n+2 \\
&=(x_n-2)(x_n-1).             
\end{align*}
Now since $1 \leq x_n <2$, $(x_n-2)(x_n-1) \leq 0$, and we conclude that the sequence is monotone decreasing.
$\textbf{Limit:}$ We can now safely assume that the limit exists.
Suppose that 
$$\lim_{n \rightarrow \infty} (x_n)= L.$$
Then
\begin{align*}
& x_{n+1} = {x_n}^2-2x_n+2 \\
\Rightarrow & \lim_{n \rightarrow \infty}x_{n+1} = \lim_{n \rightarrow \infty}\left({x_n}^2-2x_n+2  \right) \\
\Rightarrow &L=L^2-2L+2 \\
\Rightarrow &L^2-3L+2=0 \\
\Rightarrow & (L-1)(L-2)=0 \\
\Rightarrow & L=1 \vee L=2.
\end{align*}
Now since the sequence is monotone decreasing on the interval, $L \neq 2$, and so we conclude that
$$\lim_{n \rightarrow \infty}(x_n) = 1.$$
