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

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

$$3^{rd}$$case

if $$x_0>3$$ then I showed $$x_n (decreasing sequence) and $$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, $$L=\lim_{n\to \infty}x_{n+1}=2+\sqrt{L-2}\iff L=2\text{ or }L=3$$

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 $$L = 3$$.
• You can apply it, but before your edit you wrote in each case that '$x_{n-1} < 3$' is an increasing sequence and so $x_{n-1} \rightarrow 3$' which is false, you only know it converges to something less than or equal to $3$. You've fixed that in your edit, and so now the solution looks better. You still need to mention you are using continuity of $\sqrt{.}$ though to get the equation involving $L$. Commented Dec 6, 2020 at 17:10