Why do the solutions of this recurrence have a limit independent of the initial value? For $0<a_0<1$ the limit $\lim\limits_{n\rightarrow \infty}\sqrt{1-a_n}= \frac{\sqrt{5}-1}{2}$ for the sequence $a_{n+1}=\sqrt{1-a_n}$.
But the problem is that the choice of $a_0$ does not matter.
The question is why?
 A: Since one iterates the function $v:x\mapsto\sqrt{1-x}$ which is not contractive, an idea is to use monotonicity instead of contraction, hence to consider a power of $v$ instead of $v$.
Consider $u=v\circ v$, hence  $u(x)=\sqrt{1-\sqrt{1-x}}$ for every $x$ in $[0,1]$. Then $a_{n+2}=u(a_n)$ for every $n\geqslant0$, hence the two sequences $(b_n)$ and $(c_n)$ defined by $b_n=a_{2n}$ and $c_n=a_{2n+1}$ are both such that $b_{n+1}=u(b_n)$ and $c_{n+1}=u(c_n)$.
Note that $u([0,1])=[0,1]$, $u(0)=0$, $u(1)=1$, $u(z)=z$ for $z=\frac12(\sqrt5-1)$, $u$ is increasing on $[0,1]$, $u(x)>x$ if $0<x<z$ and $u(x)<x$ if $z<x<1$.
Thus, any sequence $(x_n)$ defined by $x_{n+1}=u(x_n)$ and $x_0$ in $[0,1]$ has the following behaviour. 


*

*If $x_0=0$ or $x_0=z$ or $x_0=1$, $(x_n)$  is constant, that is $x_n=x_0$ for every $n\geqslant0$. 

*If $0<x_0<z$, $(x_n)$  is increasing, $0<x_n<z$ for every $n\geqslant0$ and $x_n\to z$ when $n\to\infty$.

*If $z<x_0<1$, $(x_n)$ is decreasing, $z<x_n<1$ for every $n\geqslant0$ and $x_n\to z$ when $n\to\infty$. 


One sees that, for every $x_0$ in $(0,1)$, $x_n\to z$. Applying this to $(b_n)$ and to $(c_n)$, which start from $b_0=a_0$ and $c_0=a_1=\sqrt{1-a_0}$ respectively, and noting that $a_1$ is in $(0,1)$ for every $a_0$ in $(0,1)$, one sees that, for every $a_0$ in $(0,1)$, both $b_n\to z$ and $c_n\to z$, hence $a_n\to z$ when $n\to\infty$.
A: We are considering a question equivalent to $f(x) = \sqrt{1 - x}$ for $x \in (0,1)$. It is not so hard to show that this function has a fixed point, and exactly one fixed point in the interval $(0,1)$. (In fact, this fixed point is our limit).
In your question, you seem to know that the limit exists and are able to calculate it (take the even and odd terms separately, for example). This means that you know that no matter what starting value, the iterative function $f(x)$ will converge. But there is only one fixed point in the interval of its range, and thus it must converge to that point.
This is the same reasoning that goes to show that $\cos(x) = x$, or rather $x_{n+1} = \cos(x_n)$, will always have the same limit regardless of starting point, more or less. One shows a limit exists, and then the limit must be the fixed point (for cosine, this number has a name, the Dottie Number).
