Proof of Convergence: Babylonian Method $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$ a) Let $a>0$ and the sequence $x_n$ fulfills $x_1>0$ and $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$ for $n \in \mathbb N$. Show that $x_n \rightarrow \sqrt a$ when $n\rightarrow \infty$.
I have done it in two ways, but I guess I'm not allowed to use the first one and the second one is incomplete. Can someone please help me?

*

*We already know $x_n \rightarrow \sqrt a$, so we do another step of the iteration and see that $x_{n+1} = \sqrt a$.


*Using limit, $x_n \rightarrow x, x_{n+1} \rightarrow x$ (this is the part I think it's incomplete, don't I have to show $x_{n+1} \rightarrow x$, how?), we have that
$$x = \frac x 2 (1 + \frac a {x^2}) \Rightarrow 1 = a/x^2 \Rightarrow x = \sqrt a$$
b) Let the sequence $x_n$ be defined as $x_{n+1}= 1 + \frac 1 {x_n} (n \in \mathbb N), x_1=1$. Show that it converges and calculate its limit.
"Tip: Show that sequences $x_{2n}$ and $x_{2n+1}$ monotone convergent to the limit."
I didn't understand the tip, how can this help me? Does it make a difference if the number is odd or even?
Thanks in advance!

For a):
The proof of convergence can be deduced from the question/answer LFT theory found in
Iterative Convergence Formulation for Linear Fractional Transformation with Rational Coefficients
Proof when $x_1^2 > a$
Note: If both $a$ and $x_1$ are rational numbers, then this solution is obtained  without recourse to the real number system.
Let $S$ represent $a$ and $K$ denote $x_1$.
We have our LFT theory for
$F(x) = \frac{S + Kx}{K + x}$
as espoused in the above link.
Considering Proposition 2 & 3, we have a sequence
$\{F^1, F^2, F^3, ..., F^n, ...\}$ of LFTs
with the corresponding decreasing sequence of Ks
$\{K, K_2, K_3, ..., K_n, ...\}$ and $(K_n)^2$ converges to $S$.
Might as well set $K_1$ to $K$ now.
Now with a little thought, you can see the following holds:
$K_2 = (S + K_1^2)/2K_1$
$K_4 = (S + K_2^2)/2K_2$
$K_8 = (S + K_4^2)/2K_4$
$K_{16} = (S + K_8^2)/2K_8$
etc.
But this is exactly the Babylonian Method.
So the method can  actually be described as calculating numbers that are a subsequence of our convergent sequence
$\{K_1, K_2, K_3, ..., K_n, ...\}$
So we have shown that the squares of the numbers produced by the Babylonian Method converge to $S$.
Proof when $x_1^2 < a$
Let $K$ denote $(a + x_1^2)/2x_1$. We know by the LFT theory that the square of this number is greater than $a$. So, the proof given above can now be applied, by simply 'throwing out' the first number $x_1$ of the sequence.
