how can one solve for $x$, $x =\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}$ 
Possible Duplicate:
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$ 

how can one solve for $x$, $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}}}}$
we know, if $x=\sqrt[]{2+\sqrt{2}}$, then, $x^2=2+\sqrt{2}$
now, if $x=\sqrt[]{2+\sqrt{2}}$, then, $(x-\sqrt{2})(x+\sqrt{2})=\sqrt{2}$
 A: put $x = \sqrt{2+x} 
 \implies x^2 = 2+x \implies x^2 -x-2=0$
Now just solve the quadratic equation.
A: $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\ldots }}}}}}$
$x^2 =2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\ldots }}}}}$
$x^2 =2+x$
$x^2-x-2=0$
$x^2-2x+x-2=0$
$(x-2)(x+1)=0$
we have, $x=2$
A: This is to address the convergence issue. Assume that $x_0$ is given, and 
with $c>0$ and $0<\alpha<1$,
let me define 
$$
x_{n+1}=(c+x_n)^\alpha\quad\textrm{for }n=1,2,\ldots,\quad
\textrm{and}\quad x=\lim_{n\to\infty}x_n,
$$
if the latter limit exists. So the problem is about the fixed point iteration 
$x_{n+1}=\phi(x_n)$ with $\phi(x)=(c+x)^\alpha$. It is well known that a fixed point $x$ of this iteration is attracting if $|\phi'(x)|<1$. There is only one fixed point $x_*>0$ with $x_*=\phi(x_*)$. We check
$$
\phi'(x_*)=\frac{\alpha(c+x_*)^\alpha}{c+x_*}=\frac{\alpha x_*}{c+x_*}<1,
$$
so $x_*$ is an attracting fixed point. From here it is easy to see that any iteration with initial point $x_0\geq-c$ will converge to $x_*$.
A more direct way to see the convergence is to note that for $-c\leq x<x_*$, we have $(c+x)^\alpha<x_*$ and $(c+x)^\alpha>x$, and for $x>x_*$, we have $(c+x)^\alpha>x_*$ and $(c+x)^\alpha<x$. Hence the sequence is increasing and bounded from above if $-c\leq x_0<x_*$, and is decreasing and bounded from below if $x_0>x_*$.
A: Define a sequence $\{a_n\}_{n \geq 1}$ such that $a_1 = \sqrt 2$ and $a_{n+1} = \sqrt{2 + a_n}$. It should be clear that $x$ is the limit of this sequence as $n$ goes to infinity, i.e:
$$x = \lim_{n \rightarrow \infty} a_n $$
To prove that this limit exists, it is sufficient to show that the sequence is bounded above, and monotonically increasing. Both of these facts may be proved by induction. 
The fact that $a_n$ is bounded above follows since $a_1 < 2$ and $a_{n+1} = \sqrt{2+a_n}$ which is less than $2$ if $a_n$ is, because then $a_{n+1} < \sqrt{2+2} = 2$.
To prove that $a_n$ is monotonically increasing note that $a_1 > a_2$ and $a_{n+1} > \sqrt{2+a_{n-1}} = a_n$, assuming of course that $a_n > a_{n-1}$.
Every monotonically increasing sequence that is bounded from above converges, so the other solutions are justified in concluding that $x=2$.
Edit: As pointed out by did, the argument above only applies when $a_1 < 2$ as there is no reason to choose $a_1 = \sqrt{2}$ specifically, as $a_1$ occurs in the "$\dots$" portion of the nested radicals.
If $a_1>2$ we may show that the sequence $\{a_n\}$ is bounded below by 2 and monotonically decreasing, by a similar argument. Finally, it is simple to show that $x$ converges if $a_1 = 2$. Thus, $x$ converges to $2$ regardless of the starting value $a_1$.
A: Well, $x = (2 + x)^.5 $ from the expression of x. Implying $x^2 = 2 + x$. And now you can solve the quadratic for x, giving x = -1 and x = 2. X can't be negative, so x = 2
A: In general, the function $f(x)=\displaystyle\bigg(\small\sqrt{\normalsize x+\small\sqrt{\normalsize x+\sqrt{ x+\sqrt{x}}}}\;\normalsize\bigg)$ is:
$0.5(1+\sqrt[]{1+4x})$
Ref: Math-Integration of nested square roots of x
