Proving limit of an iterative sequence Let
$a_1=\sqrt{2}$, and $a_2=\sqrt{a_1+2}$, $a_{n+1}=\sqrt{a_n+2}$ I need to show that the limit of this sequence is 2, but I can I do this, I can not use epsilon criterion here, since it is iterative. I have already shown that this sequence is bounded by 2 with induction, but I also need to prove that the exact limit 2.
edit: thank you everyone for the answers! such great approaches that I am having a difficulty to chose the accepted answer
 A: The sequence is monotonically increasing between 1 and 2, so it has a limit. At the limit, $$l = \sqrt{l +2},$$ so solving the quadratic you get $l=2.$
A: First, you can readily see that, if the limit exists, it must be 2. Assume that the limit is $L$. Then, since $a_{n+1}=\sqrt{a_n+2}$, and thus $\lim a_{n+1} = \lim a_n =L$, we must have that $L = \sqrt{L+2}$. From here you conclude that $L=2$. Regarding the existence of limit,

*

*$a_n \ge \sqrt{2}$ and $a_n <2$ (the sequence is bounded)


*$
a_{n+1}-a_n =\sqrt{a_n+2}-a_n = \frac{(a_n+1)(2-a_n)}{\sqrt{a_n+2}+a_n}>0
$ (the sequence is monotonous)
and, therefore, the sequence in convergent.
A: $$a_{n+1}=\sqrt{a_n+2}$$
$$2=\sqrt{2+2}$$
so
$$|a_{n+1}-2|=|\frac{a_n-2}{\sqrt{a_n+2}+2}|$$
$$\le \frac{|a_n-2|}{2}$$
$$\cdots$$
$$\le \frac{|a_1-2|}{2^n}$$
$$\implies \lim_{n\to+\infty} a_n=2$$
A: First observe that $a_n>0$ for all $n$. Now prove that the sequence is strictly increasing, we have$$\begin{align*}a_{n+1}>a_n & \iff \sqrt{a_n+2}>a_n \\
& \underset{a_n>0\forall n}{\iff}a_n+2>a_n^2 \\
& \iff a_n^2-a_n-2<0
\end{align*}$$but the latter is true since the function $f(x)=x^2-x-2$ is negative on $(0,2)$ because its principal coefficient is positive and has roots $x_1=-1$ and $x_2=2$. Therefore, the sequence is increasing. 
Since it is increasing and bounded, it has a limit $\ell$, which is nonnegative because every term in the sequence is. Now$$\ell =\lim \limits _{n\to \infty}a_n=\lim \limits _{n\to \infty}a_{n+1}=\lim \limits _{n\to \infty}\sqrt{a_n+2}=\sqrt{\ell +2}$$hence $\ell =\sqrt{\ell +2}$, so $\ell ^2=\ell +2$, this quadratic has roots $-1$ and $2$, since $\ell \geq 0$, we get $\ell =2$, as we wanted.
A: Alternate way,
$$\begin{align}\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}_{ n ~ \text{terms},~n≥1}=2\cos \frac{\pi}{2^{n+1}}\end{align}$$
