Find the limit of the sequence $(x_n)$, where $x_1=1$ and $x_{n+1}=(2+x_n)^{1/2}$ I have been given this hint in the problem: In order to determine the limit, it may be helpful to use the fact that if $y_n \rightarrow y$ and $y_n > 0$ for every $n$,then $\sqrt{y_n} \rightarrow \sqrt{y}$. You can use this without proving it. You can also use without proof the fact that if $0\le a \le b,$ then $\sqrt{a} \le \sqrt{b}$.
 A: You have: $|x_n - x_{n-1}| = \left|\sqrt{x_{n-1}+2}-\sqrt{x_{n-2}+2}\right|=\dfrac{|x_{n-1} - x_{n-2}|}{\sqrt{x_{n-1}+2}+\sqrt{x_{n-2}+2}}\le \dfrac{|x_{n-1}-x_{n-2}|}{2\sqrt{2}}$. This shows it's a Cauchy sequence, hence converges to $L$ that also satisfies: $L = \sqrt{L+2} \implies L = 2$.
A: hint
To be sure the sequence is well-defined, Prove by induction that
$$(\forall n\ge 1)\;\;  x_n\ge 0$$
then use
$$|x_{n+1}-2|=\frac{|x_n-2|}{\sqrt{2+x_n}+2}$$
$$\le \frac 12|x_n-2|$$
$$\le \frac{1}{2^n}|x_1-2|$$
A: My attempt :
In the first, is easy to showing $x_n$ is incremental, just pose $f(x) =\sqrt{2+x}$ and use recurrence for showing $x_n<x_{n+1}$
And  you can see that $x_2=x_3=x_4=.......=2$
So $x_n\leq 2$ and for showing that also you can use recurrence
For $n=1$ $x_1\leq2$$\Rightarrow $$1\leq 2$ that is correct
Suppose $x_n\leq 2$
Let's show $x_{n+1}\leq 2$
We know $x_n\leq 2$
So $f(x_n) \leq f(2)$
so $x_{n+1}\leq 2$ because f is incremental
we see $x_n\leq 2$ and $x_n$ is incremental so $x_n $ is convergent and $\lim x_n=l$
So. $l=\sqrt{2+l} $$\Rightarrow $$ l^2-l-2=0$$\Rightarrow l=2$ or $l=-1 $but this is impossible because $l\geq 0$
So finally
$\lim x_n=2. $
