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.
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

$$b_n = b_1^{2^{n-1}} = 3^{2^{n-1}}
\quad\implies\quad
a_n = \frac{b_n + 1}{b_n - 1} = \frac{3^{2^{n-1}} + 1}{3^{2^{n-1}} - 1}
\to 1 \quad\text{ as }\quad n \to \infty
$$
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 $x$.