Find the values of $x_0 \in \mathbb{R}$ for which the recurrent sequence $x_{n+1} = 2^{\frac{x_n}{2}}$ has $2$ as its limit. I am given the recurrent sequence $(x_n)_{n \ge 0}$ with $x_0 \in \mathbb{R}$ and:
$$x_{n+1} = 2^{\frac{x_n}{2}}$$
and I am asked to find the interval of values for which this sequence has $\lim\limits_{n \to \infty}x_n=2$. Also, if it is helpful (it didn't help me) I am given the following options:
A. $x \in \{2\}$
B. $x \in [-2, 2]$
C. $x \in (-\infty, 2]$
D. $x \in [2, 4)$
E. Other answer
I tried to take the logarithm of both sides, ending up with:
$$\ln(x_{n+1}) = \dfrac{x_n}{2} \ln (2)$$
$$2 \ln(x_{n+1}) = x_n \ln(2)$$
But I got nowhere with it. How should I approach something like this?
 A: As a suggestion, consider the function $f(x)=2^{x/2}$. Notice that $0<f'(2)<1$ and $f'(4)>1$. This will imply that $2$ is an attractor and $4$ is a repellent.
A: The equation $2^{\frac{x}{2}}=x$ has the two solutions $x=2$ and $x=4$. Those are the only solutions since the derivative of $2^{\frac{x}{2}}$ is always increasing. Since $2^{\frac{x}{2}}$ and $x$ are continuous we get
$$2^{\frac{x}{2}}>x\Longleftrightarrow \begin{cases}x>4\\x<2\end{cases}$$

For any $x_n<2$ we get (since $2^{\frac{x}{2}}<2\Leftrightarrow \sqrt{2^x}<2\Leftrightarrow x<2$)
$$x_n<x_{n+1}<2$$
Thus for $x_0<2$ the sequence is increasing and bounded and therefore convergent. And as "Math1000" pointed out the limit $x$ must satisfy $x=2^{\frac{x}{2}}$ but since $x_n<2$ $\forall n\in\mathbb{N}$ we get $x=2$.
For $2<x_0<4$ the sequence is decreasing and bounded by the same logic ($2^{\frac{x}{2}}<x\Leftrightarrow 2<x<4$) so the limit is $2$ here aswell.
For $x_0>4$ the sequence is increasing but has to diverge since the only possible limits are $2$ and $4$.
For $x=2$ or $x=4$ the sequence is constant. The sequence converges to $2$ for $x_0\in(-\infty,4)$ so the correct answers are A,B, C and D.
Edit: Here are the first few terms of the sequence for $x_0\in[-2,4.1]$:

(left click to open gif)
