Limit of $\sqrt{2^\sqrt{2^\sqrt{2^\sqrt{\ldots}}}}$? How do I compute the limit of the sequence:
$$\sqrt{2^\sqrt{2^\sqrt{2^\sqrt{\ldots}}}}$$
I tried:
$$x = \sqrt{2^\sqrt{2^\sqrt{2^\sqrt{\ldots}}}}$$
$$x^2 = 2^\sqrt{2^\sqrt{2^\sqrt{\ldots}}}$$
$$x^2 = 2^x$$
But this equation has more than one solution. How do I interpret this?
By trying, it seems that the sequence converges to 2.

The question arose by rewriting $2$
\begin{align*}
2 &=\sqrt{4}\\
2 &=\sqrt{2^\sqrt{4}}\\
2 &=\sqrt{2^\sqrt{2^\sqrt{4}}}\\
  & \vdots\\
2 &= \sqrt{2^\sqrt{2^\sqrt{2^\sqrt{\ldots}}}}
\end{align*}
But is this equal to the sequence in the beginning?
 A: Writing an expression with $\cdots$ in it doesn't inherently mean something. It's often a suggestive informal way to define the limit of a sequence, but if there's any confusion about what sequence it's the limit of, it just needs clarification.
In this case, you can define a sequence by the recurrence relation $$x_{n+1} = \sqrt{2^{x_n}}.$$ This doesn't uniquely specify the sequence, because we still need to add an initial condition $x_0$. I would say that the limit of any sequence you get by specifying a value of $x_0$ has some right to being referred to as $$\sqrt{2^{\sqrt{2^{\sqrt{2^{\dots}}}}}}.$$ (Because the starting value $x_0$ is "hidden inside the $\cdots$", we can't really say what it is from the expression, which is where the ambiguity comes from.)
If you set $x_0 = 2$, then we'll have $x_n = 2$ for all $n$, and the limit is (trivially) $2$. Also, if we set $x_0 = 4$, then we'll have $x_n = 4$ for all $n$, and the limit is (trivially) $4$.
More interesting is the fact that for any $x_0 \in (-\infty, 4)$, the limit will be $2$, and for any $x_0 \in (4,\infty)$, the limit is $\infty$ (that is, the sequence diverges). So $2$ is an attractor and $4$ is a repeller, and I think that if you had to give some single answer to what $$\sqrt{2^{\sqrt{2^{\sqrt{2^{\dots}}}}}}$$ is, it would be $2$.
A: We have
$$u_{n+1}^2=2^{u_n} $$
or
$$u_{n+1}=2^{\frac {u_n}{2}} $$
let $$f (x)=2^{\frac {x}{2} }$$
$f $ is increasing at $[0,+\infty) $.
$u_1=\sqrt {2}<u_2 \implies (u_n) $ increases.
by induction we prove that
$$\forall n\geq 1\;\, u_n<2$$
thus $(u_n) $ converges to the fixed point $x=2=f (2) $.
A: If we define a sequence $x_n$ by the recurrence formula
$$x_{n+1}=\sqrt{2^{x_n}}$$
then the existence and value (if any) of the limit $\lim_{n\to\infty}x_n$ will depend on the choice of starting value $x_0$.  If we write $x_n=2(1+u_n)$, the recurrence becomes
$$u_{n+1}=2^{u_n}-1$$
The function $f(u)=2^u-1$ clearly satisfies $f(0)=0$ and $f(1)=1$.  It's easy to show that the iterates $f^{(n)}(u)=f(f^{(n-1)}(u))$ tend to $\infty$ if $u\gt1$ and $0$ if $u\lt1$.  All this corresponds to
$$x_n\to
\begin{cases}
\infty&\text{if }x_0\gt4\\
4&\text{if }x_0=4\\
2&\text{if }x_0\lt4\\
\end{cases}$$
