Show that $(x_n)$ is decreasing and find its limit. Let $0<x_1<1$. For $n \in \mathbb{N}$, let $x_{n+1}=1- \sqrt{1-x_n}$. Show that $(x_n)$ is decreasing and find its limit.
I did:
$$x_{n+1} = 1- \sqrt{1-x_n}$$
$$x_{n+1} - x_n= 1- \sqrt{1-x_n} - x_n$$
We know that 
$$\sqrt{1-x_n}<1-x_n$$
and so,
$$x_{n+1} - x_n< 1- (1-x_n) - x_n< 1- 1+x_n- x_n<0$$
Therefore, $$x_{n+1} < x_n$$
Thus, $(x_n)$ is decreasing
How can I keep going to find the limit?
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*

*$\ds{\color{#0000ff}{\large x_{n + 1}} = 1 - \root{1 - x_{n}} = {x_{n} \over 1 + \root{1 - x_{n}}}
         \color{#0000ff}{\large < x_{n}}}$

*$\ds{\phi_{n} \equiv \root{1 - x_{n}}\quad\imp\quad \phi_{n + 1}
         = \phi_{n}^{1/2}}$:
    $$
    \phi_{n} = \phi_{n - 1}^{1/2} = \phi_{n - 2}^{1/4} = \phi_{n - 3}^{1/8}
    = \cdots = \phi_{1}^{1/2^{n - 1}}
    \quad\imp\quad\lim_{n \to \infty}\phi_{n} = 1
    $$
    Then, $\ds{\color{#0000ff}{\large\lim_{n \to \infty}x_{n} = 0}}$

A: First notice that if $x_n\gt0$, then $\sqrt{1-x_n}\lt1$, so $x_{n+1}=1-\sqrt{1-x_n}\gt0$. Furthermore,
$$
\begin{align}
x_{n+1}
&=1-\sqrt{1-x_n}\\
&=\frac{x_n}{1+\sqrt{1-x_n}}\\
&\lt x_n
\end{align}
$$
Therefore, $x_n$ is decreasing and bounded below. Thus, $x_\infty=\lim\limits_{n\to\infty}x_n$ exists and
$$
\begin{align}
x_\infty=1-\sqrt{1-x_\infty}
&\implies x_\infty(2-x_\infty)=x_\infty\\
&\implies x_\infty^2-x_\infty=0
\end{align}
$$
Since $x_\infty<1$, we must have $x_\infty=0$.
A: See $\{x_n\}$ is monotone decreasing and bounded below. Zero is a lower bound. So it is convergent and $L = \lim x_n$. As T. Bongers has written in his comment the limit should satisfy $L = 1 - \sqrt{1 - L}$
