Use the Monotone Convergence Theorem to prove that either $\lim_{n \to \infty}x_n=2$ or $3$ .

Question

Suppose that $x_0 \geq 2$ and $x_n = 2 + \sqrt{x_{n-1}-2}$ for $n \in \mathbb{N}$. Use the Monotone Convergence Theorem to prove that either $\lim_{n \to \infty}x_n=2$ or $3$ .

I divided this problem into $3$ cases but I don't know whether it is correct or not if there is anything wrong please tell me? I will explain what I did briefly?

$1^{st}$ case

if $x_0=2$ or $x_0=3$ it constant sequence so that either $\lim_{n \to \infty}x_n=2$ or $3$ .

$2^{nd}$case

if $2<x_0<3$ then I showed $x_n>x_{n-1}$ (increasing sequence) and $x_{n-1}<3$ using induction

$3^{rd}$case

if $x_0>3$ then I showed $x_n<x_{n-1}$ (decreasing sequence) and $x_{n-1}>3$ using induction

then we can say $\lim_{n \to \infty}x_n=2$ or $3$ (Monotone convergence theorem) since in either case $\{x_n\}$ is increasing or decreasing and bounded below or bounded above

In other word in either case, $$L=\lim_{n\to \infty}x_{n+1}=2+\sqrt{L-2}\iff L=2\text{ or }L=3$$

Answer 1

I don't think you can use monotone convergence directly to show that the limit is $2$ or $3$. As you have done though, you can show that it means for any $x_0 \geq 2$ the limit exists. Once you have this, you can write (using continuity of $\sqrt{.}$) $L = 2 + \sqrt{L-2}$ as you have done, and solve for $L$. So just to summarize:

1. Use monotone convegence to show the limit exists and is nonnegative for any $x_0$
2. Use continuity of $\sqrt{.}$ to get $L = 2 + \sqrt{L-2}$.
3. Conclude $L = 2$ or $L = 3$.