I have the following question:
7. If $x_1=3$ and $x_{n+1}=\sqrt{2+x_n}$, $n\ge1$, then $\displaystyle\lim_{n\to\infty} x_n$ is
(1) $-1$
(2) 2
(3) $\sqrt 5$
(4) 3
And I found this solution for it:
7. (2) $x_{n+1} = \sqrt{2+x_n}$
or $\displaystyle \lim_{n\to\infty} x_{n+1} = \sqrt{2+\lim_{n\to\infty} x_n}$
or $t=\sqrt{2+t}$ $\displaystyle \left( \because \lim_{n\to\infty} x_{n+1} = \lim_{n\to\infty} x_n = t \right)$
or $t^2-t-2=0$
or $(t-2)(t+1)=0$
or $t=2$ $(\because x_n>0 \ \forall \ n,t>0)$
But how can we say that these two limits are equal when $n$ approaches to infinity as we can’t compare infinity?
First of all, we need to proof that this sequence is convergent. $$X_1 = 3 \space \text{and} \space X_{n+1} = \sqrt{2+X_n}, n \ge 1.$$ We are going to use Monotone Convergence Theorem. If you see Elementary Real Analysis, Bartle, 4th ed, page 71. It says
A sequence is convergence if and only if monotone and bounded.
First we are going to proof the sequence is monotone by induction. It is clear that $$X_2 = \sqrt{2+3} < \sqrt{9}=X_1.$$ So we have some intuition, this sequence is decreasing. If $X_{n+1} < X_n$, then $$X_{n+2} = \sqrt{2+x_{n+1}} < \sqrt{2+X_{n}} = X_{n+1}.$$ So that the sequence is decreasing.
We all know that this sequence is bounded by $0$.
We have proven that the sequence is monotone and bounded. So the sequence is convergent.
Let $a$ is the limit of $X_n$. That is $$a = \lim_{n \to \infty} X_n.$$ So that $$\lim_{n\to \infty} X_{n+1} = a.$$
We have $$a = \sqrt{2+a} \Longleftrightarrow a^2-a-2 = 0 \Longleftrightarrow (a-2)(a+1) = 0.$$ Finally $$a = 2 \vee a = -1.$$ Because of the sequence is bounded by $0$, so it must be positive sequence. We choose $2$ as a limit of the sequence.
If you are worry about the step of comparing at infinity, see $$X_n = \frac{1}{n}, \space \text{so that} \space X_{n+1} = \frac{1}{n+1}.$$ The limit $$\lim_{n \to \infty} \frac{1}{n} = 0.$$ And so does $$\lim_{n \to \infty} \frac{1}{n+1} = 0 = \lim_{n \to \infty}\frac{1}{n}.$$