Boundedness and convergence of $x_{n+1} = x_n ^2-x_n +1$ Suppose that $x_0 = \alpha \in \mathbb{R}$ and $x_{n+1} = x_n ^2-x_n +1$.
I am asked to study the boundedness of $(x_n)$ and then asked if $(x_n)$ converges. How can I show that $(x_n)$ is bounded?
I have noted that $$x_{n+1}-x_n = (x_n-1)^2\geq 0$$ so $x_n$ is increasing. 
Suppose that $(x_n)$ converges, then $l = l^2-l+1$ so $l = 1$. This means that the sequence converges only when $\alpha \leq 1$. And so $(x_n)$ is bounded above by $1$ if $\alpha\leq 1$ (since an increasing sequence converges to its supremum) otherwise the sequence is not bounded above.
Is my reasoning correct?

Case $\alpha >1$, show by induction that
$$x_n-1\geq  \alpha ^n(\alpha -1)$$
Proof:
$$x_{n+1}-1=x_n{}^2-x_n=x_n\left(x_n-1\right)\geq  \alpha  \alpha ^n(\alpha -1)=\alpha ^{n+1}(\alpha -1)$$
This shows $x_n$ diverges
Case $\alpha <0$, $x_1>1$ so we are back to the previous case
Case $0\leq  \alpha \leq  1$, $x_n$ is bounded so it converges
