Consider the following discrete-time dynamical system (or "infinite state machine", if you prefer)
$$x_{k+1} = \displaystyle\sqrt{2 x_k}$$
where $x_0 = \sqrt{2}$ is the initial condition. Let us introduce function $f (x) := \sqrt{2 x}$ so that the state-transition equation can be written in the form $x_{k+1} = f (x_k)$. You want to prove that
$$x_{\infty} := \displaystyle\lim_{k \rightarrow \infty} x_k = 2$$
Does $x_k$ converge to $2$? Here's a short numerical experiment in Haskell (using GHCi):
*Main> let f x = sqrt (2 * x)
*Main> let x0 = sqrt (2)
*Main> let xs = iterate f x0
*Main> take 10 xs
[1.4142135623730951,1.6817928305074292,1.8340080864093424,1.9152065613971472,1.9571441241754002,1.978456026387951,1.9891988469672663,1.9945921121709402,1.9972942257819404,1.9986466550053015]
Looks like $x_k$ does indeed converge to $2$. Let's now try to prove it. I introduce a new symbol
$$\tilde{x}_{k} := \ln (x_k)$$
which allows us to rewrite the state-transition equation $x_{k+1} = \displaystyle\sqrt{2 x_k}$ in the following form
$$\tilde{x}_{k+1} = \frac{1}{2} \tilde{x}_k + \ln(\sqrt{2})$$
which is a scalar linear dynamical system of the form $s_{k+1} = a s_k + b$, whose general solution is
$$s_ k = a^k s_0 + \displaystyle\sum_{i=0}^{k-1} a^i b$$
Therefore, we obtain the following general solution
$$\tilde{x}_k = \left(\frac{1}{2}\right)^k \tilde{x}_0 + \displaystyle\sum_{i=0}^{k-1} \left(\frac{1}{2}\right)^i \ln(\sqrt{2})$$
and, taking the limit
$$\tilde{x}_{\infty} := \displaystyle\lim_{k \rightarrow \infty} \tilde{x}_k = \displaystyle\lim_{k \rightarrow \infty} \sum_{i=0}^{k-1} \left(\frac{1}{2}\right)^i \ln(\sqrt{2}) = \left[\sum_{i=0}^{\infty} \left(\frac{1}{2}\right)^i\right] \ln(\sqrt{2}) = 2 \ln (\sqrt{2}) = \ln (2)$$
and, finally, we conclude that $x_{\infty} = \exp(\tilde{x}_{\infty}) = 2$.