Are you sure it's not
$\displaystyle \displaystyle a_1 = \sqrt{2}$ and $\displaystyle \displaystyle a_{n + 1} = \sqrt{2 + a_n}$?
If so, this is a relatively easy limit to evaluate...
It should be clear that the sequence goes $\displaystyle \displaystyle \left\{\sqrt{2}, \sqrt{2 + \sqrt{2}}, \sqrt{2 + \sqrt{2 + \sqrt{2}}}, \dots \right\}$.
So you want to see if $\displaystyle \displaystyle \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots} }}}}$ can be evaluated (has a limit...). Call this limit $\displaystyle \displaystyle L$.
If $\displaystyle \displaystyle L = \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots} }}}}$ then
$\displaystyle \displaystyle L^2 = 2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots} }}}}$
$\displaystyle \displaystyle L^2 = 2 + L$
$\displaystyle \displaystyle L^2 - L - 2 = 0$
$\displaystyle \displaystyle (L-2)(L+1) = 0$
$\displaystyle \displaystyle L - 2 = 0$ or $\displaystyle \displaystyle L+1 = 0$
$\displaystyle \displaystyle L = 2$ or $\displaystyle \displaystyle L = -1$.
It should be clear that $\displaystyle \displaystyle L = -1$ is an extraneous solution which came from the original squaring of the equation, so the limit is $\displaystyle \displaystyle L = 2$.
The difference equation can be written as...
$\displaystyle \displaystyle \Delta_{n}= a_{n+1}-a_{n} = \sqrt{2+a_{n}} -a_{n} = f(a_{n})$ (1)
The function $\displaystyle f(*)$ is illustrated here...
There is only one 'attractive fixed point' in $\displaystyle x_{0}=2$ but that's only a necessary condition of convergence. In this case however is $\displaystyle |f(x)|\le |x_{0}-x|$ [red line...] so that any $\displaystyle a_{0}>-2$ will generate a sequence monotonically convergent at $\displaystyle x_{0}$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$