Compute limit if it exist of $a_1 = 2$, $a_{n+1} = \frac{1}{2} (a_n + \frac{1}{a_n})$ $$a_1 =  2$$
$$a_{n+1} = \frac{1}{2} (a_n + \frac{1}{a_n})$$
I tried to use induction but it looks as if the sequence isn't monotonic. 
 A: Let $\displaystyle\;b_n = \frac{a_n + 1}{a_n - 1}$, we have 
$$
b_{n+1} = \frac{a_{n+1}+1}{a_{n+1}-1} = \frac{a_n + \frac{1}{a_n} + 2}{a_n + \frac{1}{a_n} - 2} = \left(\frac{a_n + 1}{a_n - 1}\right)^2 = b_n^2$$
Together with
$\displaystyle\;b_1 = \frac{a_1-1}{a_1+1} = \frac{2+1}{2-1} = 3$, we get
$$b_n = b_1^{2^{n-1}} = 3^{2^{n-1}}
\quad\implies\quad
a_n = \frac{b_n + 1}{b_n - 1} = \frac{3^{2^{n-1}} + 1}{3^{2^{n-1}} - 1}
\to 1 \quad\text{ as }\quad n \to \infty
$$
A: Hint: This sequence represents a common method of approximating a quantity. What method and what quantity is it?

 Hint 2: A sequence $a_n = f^n(y)$, where $f^n$ means $f$ iterated $n$ times, often has its limit $x$ satisfy $x=f(x)$. 

A: we have $a_1=2$
by induction $2\geq a_n>1$.
$$a_{n+1}=a_n-\frac{a_n^2-1}{2a_n}$$
$$a_{n+1}-1=(a_n-1)(1-\frac{a_n+1}{2a_n})$$
$$=\frac{1}{2}(a_n-1)(1-\frac{1}{a_n})$$
$\implies$
$$|a_{n+1}-1| \leq \frac{1}{2}|a_n-1|$$
$....\leq (\frac{1}{2})^n|a_1-1|$
this proves $(a_n)$ tends to $1$.
A: Using AM-GM inequality we have $a_n \ge 1$ for all $n \ge 1 \implies 0 < \dfrac{1}{a_n} \le a_n \implies a_{n+1} \le a_n$. Thus $\{a_n\}$ is decreasing and bounded below by $1$, hence converges to $L$ that satisfies: $ L = \dfrac{L+\dfrac{1}{L}}{2} \implies L = 1$.
A: If you JUST want to find the limit the easiest thing to do is solve
$$L=\frac{1}{2}\left(L+\frac{1}{L}\right)$$ you get $L=\pm 1$, and since $a_n$ is positive, $L=1$.
If you want to show the limit exists, it is decreasing.
$$a_{n+1}=\frac{1}{2}\left(a_n+\frac{1}{a_n}\right)\leq a_n$$
is equivalent to $1\leq a_n^2$ and this is true because $$1\leq \frac{1}{2}\left(x+\frac{1}{x}\right)$$ for all $x$.
