How do I compute the limit of this sequence $x_{n+1}=\frac{(x_n)^2+b}{2}$, $x_o=0$ and $b\in [0,1]$ I need to study the limit behavior of the sequence and discuss all possible situations with the parameters given. The sequence is  $x_{n+1}=\frac{(x_n)^2+b}{2}$, $x_o=0$ and $b\in [0,1]$. This can also be written as $\frac{(x_n)^2}{2}+\frac{b}{2}$.
I studied the first few terms to get an idea of the behavior of the sequence, and have a feeling that when $b=0, x_{n+1}=0$ and thus $\lim\ x_n=0$ and when $0<b\leq1, \lim\ x_n=b/2$ because as n gets larger $\frac{(x_n)^2}{2}$ tends to $0$ and $\lim\ (0+\frac{b}{2})=(b/2)$.
Those were my thoughts on the problem, I'm not sure if they are right, but I got stuck trying to prove them.
Proof of when $b=0, x_{n+1}=0$ for all n, which will imply that $\lim\ x_n=0$:
This can be proved by induction (but I'm not sure how to do it properly):$$x_0=0=0$$$$x_{0+1}=x_1=\frac{0^2+0}{2}=0$$
$x_{n+1}=>x_{n+2}$
$$x_{n+2}=x_{(n+1)+1}=\frac{(x_{n+1})^2+b}{2}=\frac{(x_{n+1})^2}{2}=\frac{0}{2}=0$$
I really don't know if that is how you do the induction. Any help?
Proof of $0<b\leq1$
When $b=1$, $x_{n+1}=\frac{(x_n)^2+1}{2}=\frac{(x_n)^2}{2}+\frac{1}{2}$. I can't really explain this besides saying that $\frac{(x_n)^2}{2}$ seems to be tending to $0$. Is this correct, and can I have a hint at how to solve it? I'm thinking I may be able to split it into 2 limits,  saying $\lim \frac{(x_n)^2}{2}+\lim\ (b/2)=0+(b/2)$.
Thanks for any help!
 A: We deal with positive $b$. First we show by induction that the sequence $\{x_n\}$ is increasing, that is, that $^x_n\lt x_{n+1}$ for all $n$. This is easy to verify for $n=1$. 
Suppose it is true for $n=k$. We show it is true for $n=k+1$. So we need to show that $x_{k+1}\lt x^{k+2}$. 
From the induction hypothesis $x_k\lt x_{k+1}$ we conclude that 
$$\frac{x_k^2+b}{2}\lt \frac{x_{k+1}^2+b}{2}.\tag{$1$}$$
Since $x_{k+1}=\frac{x_k^2+b}{2}$ and $x_{k+2}=\frac{x_{k+1}^2+b}{2}$, the Inequality $(1)$  says precisely that $x_{k+1}\lt x_{k+2}$.
An easy induction shows that the sequence $\{x_n\}$ is bounded above by $1$, and now we can conclude convergence.
The value is the easy part. We have $\lim_{n\to\infty} x_{n}=\lim_{n\to \infty} x_{n+1}$. So if $L$ is their common limit, we have
$$L=\frac{L^2+b}{2}.$$
Solve this quadratic for $L$. The root less than $1$ gives $L=1-\sqrt{1-b}$.  
A: *

*$\{x_n\}$ is monotone increasing: 
$x_2\geq x_1.$ Suppose $x_{k+1}\geq x_k$ whenever $1\leq k\leq m~(m\in\mathbb N)$. Consequently, $x_n\geq 0~\forall~ n\leq m.$
$x_{k+2}-x_{k+1}=\dfrac{(x_{k+1})^2-1}{2}-\dfrac{(x_{k})^2-1}{2}=\dfrac{(x_{k+1})^2-(x_k)^2}{2}\geq 0$ whence $\{x_n\}$ is monotone increasing.

*$\{x_n\}$ is bounded above:
$x_1=\dfrac{b}{2}\leq b.$ Let $x_k\leq b$ for some $m\in\mathbb N.$
Then $x_{k+1}=\dfrac{(x_k)^2+b}{2}\leq b$ (Since $b\in [0,1]$). Thus by induction $b$ is an upper bound of $\{x_n\}.$
By $1$ and $2,$ $\{x_n\}$ converges to some limit $l$ (say). Taking limit as $n\to\infty$ on both sides of $x_{n+1}=\dfrac{(x_n)^2+b}{2}$ we get $l=\dfrac{l^2+b}{2}$ i.e. $l^2-2l+b=0.$ Conseqently $l=1-\sqrt{1-b}$ (Recall $x_n\leq b\implies \lim x_n\leq b\leq 1$).
