$a_{n+1} = \sqrt{2 + a_n}$ Specific Theorem Needed I know this question has been asked many times, but I need a specific part of it.
When we get to the limit part, I had it written like this:
$\lim_{n \to +\infty} a_{n+1} = \sqrt{2+\lim_{n \to +\infty} a_n}$
What is the reason I am allowed to put the limit under the square root? Why am I just allowed to put it there? I know it's true, but I am not sure why. I need to justify that in order to receive credit. My instructor said it is a "special word." I have no idea what the word is.
 A: IF the limit exists, it will be a fixed point of the function $\sqrt{2+x}$, in other words a solution to the equation
$$x=\sqrt{2+x}$$
Here's why:
$$\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\sqrt{2+a_n}$$
Given two functions $f$ and $g$, as long as $f$ is continuous and $\lim_{x\to x_0}g(x)$ exists, then
$$\lim_{x\to x_0}f(g(x))=f\left(\lim_{x\to x_0}g(x)\right)$$
This can be shown fairly routinely with the $\epsilon ,\delta$ definition of the limit.
Since $\sqrt{2+x}$ is continuous on its domain,
$$\lim_{n\to\infty}\sqrt{2+a_n}=\sqrt{2+\lim_{n\to\infty}a_n}=\lim_{n\to\infty}a_{n+1}$$
Since $x:=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}a_n$, the initial statement follows.
A: We have a few claims:
Claim: The sequence $\{ a_{n}\}_{n = 1}^\infty$ is bounded above by $2$.
Proof. The proof can be done by induction. I'll just write the induction step
$k + 1$.
$$ a_{k + 2}^2 = 2 +  a_{k + 1} < 2 + 2 \implies a_{k + 2} < 2 $$
The rest is as goes with the formal writing.
Lemma: $\{ a_n\} = \left\{ 2 \cos \frac{\pi}{2^{n + 1}} \right\} $
Proof. Similar by induction.  Assume $$ a_k = 2\cos \left( \frac{\pi}{2^{k + 1}} \right) $$
Then $$ a_{k +1}^2 = 2 + 2 \cos\left( \frac{\pi}{2^{k + 1}} \right) = 2^2  \cos^2 \left( \frac{\pi}{2^{k + 2}} \right)  $$
$$ \implies a_{k + 1} = 2 \cos \left( \frac{\pi}{2^{k + 2}} \right) $$
Hence inductively we are done.
Now, we have a specific closed form. Then we have that $\lim\limits_{n \to \infty} a_n$ exists, as $\cos \left( \dfrac{\pi}{2^{n + 1}} \right) \to \cos 0 = 1 $, as $\cos$ is a continuous function on $\left[0, \dfrac{\pi}{2} \right]$.
Thus finally we have that
$$ \lim_{n \to \infty} a_n = \lim_{n \to \infty} 2\cos \left( \frac{\pi}{2^{n + 1}} \right) = 2 $$

Edit: I have tried to find a closed-form.
A: As mentioned before in comments and answers, once the convergence if the sequence $a_n$ has been established, finding the limit is a matter if using the continuity of the function $f(x)=\sqrt{x+2}$.
Here is yer another proof of convergence of $a_n$ when the initial value $a_0$ is taken in $[0,\infty)$.

*

*If $0\leq a_0\leq 2$, then $a_n\leq a_{n+1}\leq 2$ for all $n\in\mathbb{Z}_+$. This can be shown by induction. For instance, for $n=0$,
$$a_0\leq\sqrt{2a_0}\leq a_1=\sqrt{2+a_0}\leq \sqrt{4}=2$$
and so on...


*If $2\leq a_0$, then $2\leq a_{n+1}\leq a_n$ for all $n\in\mathbb{Z}_+$. Again, this can be argue by indiction. For instance, for $n=0$
$$ 2=\sqrt{2+2}\leq a_1=\sqrt{2+a_0}\leq\sqrt{2a_0}\leq a_0$$
ans so on.
Being $a_n$ a bounded monotone sequence, it converges to a finite limit, say $a_*$. Since $a_{n+1}=f(a_n)$ and $f$ is continuous,
$$a_*=f(a_*)$$
whence $a_*=2$.
