Consider the function $f(x) = \sqrt{2 + x}$ for $x \geq -2$ and the iteration $x_{n+1} = f(x_n) ; n \geq 0$ for $x_0 = 1$. What are the possible limits of the iterations ?
$\sqrt{2 + \sqrt{2 +\sqrt{2 + ...}}}$
-1
2
1
I think $x_1 = \sqrt3$, $x_2 = \sqrt {2 + \sqrt {3}}$, $x_3 = \sqrt{2 + \sqrt {2 + \sqrt {3}}}$, ..........Similarly like this, we get a sequence $\{ x_n \}$.
Observe that $\{x_n\}$ is an increasing sequence.This can be shown by induction. Suppose $x_n \ge x_{n-1}$. Then $x_{n+1}= \sqrt{2+x_n} \ge \sqrt{2+x_{n-1}}=x_n$ Moreover it is bounded above by $2$. Assume that that $x_n \le 2$ . Then $x_{n+1} = \sqrt{2+x_n} \le \sqrt{2+2}=2$. Hence $x_n$ is convergent. Now let is converge to $l$. Then $l^2-l-2=0 \implies (l-2)(l+1)=0$. But since every term is positive, $l=2$.
See option (1) is nothing but the equation $l^2-l-2=0$. Because if you let $l= \sqrt{2 + \sqrt{2 +\sqrt{2 + ...}}}$, then $l =\sqrt{2+l}$ which is the above equation
One should note, that all iterates are positive.
If the iteration converges, $x_n \to x$, then the limit satisfies (by continuity) $$ f(x) =x, $$ which is in your case $$ x = \sqrt{2+x}. $$ Can you conclude?
