If $x_{1} = \frac{1}{2}$ and $x_{n+1} = \sqrt{1-x_{n}}\;\;\forall\; n\; \geq 1\;,$ Then $\lim_{x\rightarrow \infty}x_{n}=$ 
If $\displaystyle x_{1} = \frac{1}{2}$ and $\displaystyle x_{n+1} = \sqrt{1-x_{n}}\;\;\forall\;  n\; \geq 1\;,$ Then $\lim_{x\rightarrow \infty}x_{n}=$

$\bf{My\; Try::}$ Although we know that when $x\rightarrow \infty\;,$ Then $x_{n+1}= x_{n} = l>0(\bf{say})$
Then we get $$l=\sqrt{1-l}\Rightarrow l^2=1-l\Rightarrow l^2+l-1=0$$
So we get $$l=\frac{-1+ \sqrt{5}}{2}\;,$$ bcz $l>0$
My question is how can we prove that the folowing sequence is Monotonic(either strictly increasing and decreasing function.)
and the sequence is bounded, Help required, Thanks
 A: This sequence is in fact not monotonic. You can see this by computing the first few terms:
\begin{align*}
a_1 &= 0.5\\
a_2 &= 0.7071\\
a_3 &= 0.5411\\
a_4 &= 0.6773
\end{align*}
So you cannot claim that the sequence is bounded and monotonic to ensure the existence of a limit. However, you can use the Banach Fixed Point Theorem.
First, you can show that $\frac12\leq x_n\leq\frac{\sqrt2}{2}$ for all $n\geq1$ using induction. The base case is given since $x_1=\frac12$. Now assume that $\frac12\leq x_n\leq\frac{\sqrt2}{2}$ for some $n\geq1$. Then $\frac{1-\sqrt2}{2}\leq 1-x_n\leq \frac12$, and taking the square root gives
$$\frac12<\frac{\sqrt{1-\sqrt2}}{\sqrt2}\leq x_{n+1}\leq\frac{\sqrt2}{2}.$$
Use $X=\big[\frac12,\frac{\sqrt2}{2}\big]$ as your metric space with the usual Euclidean metric and $T:X\to X$, $T(x)=\sqrt{1-x}$ as your contraction mapping. I'll leave it to you to check the conditions of the Banach Fixed Point Theorem. The key point is to find some $q\in[0,1)$ such that
$$\left|\sqrt{1-x}-\sqrt{1-y}\right|\leq q\left|x-y\right|$$
for all $x,y\in X$.
So the function $T$ has a fixed point (which is the $l$ that you have already found) and this fixed point is the limit of your sequence.
A: We have $0<x_n<1$ for all $n$ by induction.(That is, the positive number $x_n$ exists for all $n.)$
With $l=(-1+\sqrt 5)2$ we have $l^2=1-l.$ Using this, show that $$x_n<l\implies x_n<x_{n+2}<l$$ $$\text {and } \quad  x_n>l\implies x_n>x_{n+2}>l.$$  $$ \text {So } \quad x_1<x_3<x_5<...<l<...<x_6<x_4<x_2.$$ Now let $A=\lim_{n\to \infty}x_{2 n-1}$ and $B=\lim_{n\to \infty}x_{2 n}.$
To show $A=B,$ note that $$A=\lim_{n\to \infty}\sqrt {1-x_{2 n}}=\sqrt {1-B}$$ $$\text {and that }\quad  B=\lim_{n\to \infty}\sqrt {1-x_{2 n-1}}=\sqrt {1-A}.$$
So $A^2=1-B, $ and $ B^2=1-A,$ implying $A^2-B^2=A-B.$ Hence  $A\ne B \implies  A+B=1. $
But $A>x_1=1/2$ and $B\geq l.$ So $A+B>1/2+l>1.$ Therefore $A=B.$ 
A: Take $f(x) = \sqrt{1 -x}$. Then $f$ is strictly decreasing, which shows that $x_0 < x_1$, $x_1 > x_2$, $x_2 < x_3$, etc. The sequence is neither increasing nor decreasing. 
Though, the subsequences of odd and even indices are both monotonic.
Hint: Consider
$$f(x) = \sqrt{ 1 - \sqrt{ 1 -x}}$$
Prove that it is increasing and note that for all $n \ge 1$: $$\begin{cases} x_{2n + 2} = f(x_{2n}) \\ x_{2n + 1} = f(x_{2n - 1})\end{cases}$$
