Prove convergence of a sequence and find it's limit if $x_{0} = 2$ I'm studying for my end of semester exam. And I'm having trouble understanding how to prove convergence of a sequence and finding it's limit.
$$x_{n} = \dfrac{1}{2}\cdot\left(x_{n-1} + \dfrac{1}{x_{n-1}}\right) $$
I need to prove that this sequence converges and find it's limit if $x_{0} = 2$. $$$$ I had an idea to prove it by Cauchy's criteria, but I'm not sure how to do it here.
 A: Here is a more fundamental way to prove:
First, we use the AM-GM to find the lower bound:
$$x_{n} = \dfrac{1}{2}\cdot\left(x_{n-1} + \dfrac{1}{x_{n-1}}\right) \ge \frac{1}{2}\cdot2\sqrt{x_{n-1}\cdot\frac{1}{x_{n-1}}} = 1$$
Now, we will show that it is decreasing:
$$
x_n - x_{n-1} = \frac{1}{2}\left(-x_{n-1} + \frac{1}{x_{n-1}}\right) = \frac{1-x_{n-1}^2}{2x_{n-1}} \le \frac{1-1}{2x_{n-1}^2} = 0 \\ x_n \le x_{n-1}$$
Hence, it is a decreasing sequence and lower bounded with $1$ which proves the convergence.
Now, let the limit of $x_n$ be $L$. Then, limiting the given recurrence we get
$$L = \frac{1}{2}\left(L + \frac{1}{L}\right) \implies 2L^2 = L^2 + 1 \implies L^2 = 1$$ Since $L$ must be positive, we have $L = 1$.
A: Hint
Prove by induction that the sequence $\{x_n\}$ is decreasing and bounded below by one. Therefore it converges.
To one which is the only positive root of the equation
$$x=f(x) = \frac{1}{2}\left(x + \frac{1}{x}\right)$$ where $f$ is continuous.
A: This is an iteration formula. (More specifically, it is Heron's Formula for approximating square roots). Iteration works by shrinking the distance between each successive $x_{n-1}, x_n$ to as close to $0$ as possible, so you can find it's limit by setting $x_{n}-x_{n-1}=0\implies x_n=x_{n-1}$. In this case, the limit is $\lambda=1$
To prove this is convergent, show that it is both decreasing (i.e. $x_{n}<x_{n-1}$, or simply that $\frac12(x+\frac 1x)<x$ when $x\in(1,2]$) and that $x_t>1$ for all $t$. With a decreasing sequence that is bounded below, we have convergence.
A: Update: Just realized an almost identical question  was asked a while ago (no wonder it looks familiar to me) and there is a way to get the closed form solution of the sequence.
I just copied and pasted my solution here and made a few changes (but I'll keep the $a_n$s from that problem):
$$\frac{a_{n+1}+1}{a_{n+1}-1} = \frac{a_n+\frac{1}{a_n}+2}{a_n+\frac{1}{a_n}-2}=\left(\frac{a_n+1}{a_n-1}\right)^2 \\\implies \frac{a_n+1}{a_n-1}=\left(\frac{a_0+1}{a_0-1}\right)^{2^n}=3^{2^n} \\ \implies 
a_n =\frac{3^{2^n}+1}{3^{2^n}-1} = 1+\frac{2}{3^{2^n}-1} \to 1$$

Original answer:
Via AM-GM $x_{n} = \dfrac{1}{2}\cdot\left(x_{n-1} + \dfrac{1}{x_{n-1}}\right) \ge 1$
Then $$x_n-1 = \frac 12 \left(x_{n-1}+\frac{1}{x_{n-1}} - 2\right) = \frac{(x_{n-1}-1)^2}{2x_{n-1}} \le \frac{(x_{n-1}-1)^2}{2}\\
\implies |x_n-1| = x_n-1 \le \frac{1}{2^n} (x_0-1)^{2^n} \to 0$$
Therefore $x_n \to 1$, as $n \to +\infty$.
