Find the limit if it exists of $S_{n+1} = \frac{1}{2}(S_n +\frac{A}{S_n})$ 
Suppose that $S_0$ and A are positive numbers, let $$S_{n+1} = \frac{1}{2}\left(S_n +\frac{A}{S_n}\right)$$ with $n \geq 0 $.
(a)Show that $S_{n+1} \geq \sqrt{A} $ if $n \geq 0$ 
(b)Show that $S_{n+1} \leq S_n $ , if $n \geq 1$
(c) Show that $s= \lim\limits_{n \rightarrow \infty} S_n$ exists
(d) find s

(a) Show that $S_{n+1} \geq \sqrt{A} $ if $n \geq 0$
Given $$P_n:  S_{n+1} = \frac{1}{2}\left(S_n +\frac{A}{S_n}\right) \geq \sqrt{A}$$
$$P_0: S_{1} = \frac{1}{2}\left(S_0 +\frac{A}{S_0}\right) \geq \sqrt{A}  $$
We assume that $P_n$ is true
$$P_{n+1}: S_{n+2}= \frac{1}{2}\left(S_{n+1} +\frac{A}{S_{n+1}}\right)$$
by assumption
$$S_{n+2}= \frac{1}{2}\left(S_{n+1}\left(1 +\frac{A}{(S_{n+1})^2}\right)\right) \geq \frac{1}{2}\left(\sqrt{A}\left(1 +\frac{A}{(\sqrt{A})^2}\right)\right)$$
$$ S_{n+2}= \frac{1}{2}\left(S_{n+1} +\frac{A}{S_{n+1}}\right) \geq \sqrt{A}  $$
It follows that $S_{n+1} \geq \sqrt{A} $

(b) Show that $S_{n+1} \leq S_n $ , if $n \geq 1$
$$S_{n+1} \leq S_n$$
$$\frac{1}{2}\left(S_n +\frac{A}{S_n}\right) \leq S_n $$
Dividing by $S_n$
$$\frac{1}{2}\left(1 +\frac{A}{S_n^2}\right) \leq 1 $$
$$\frac{A}{2S_n^2} \leq \frac{1}{2}$$
$$A \leq S_n^2$$
$$S_n \geq \sqrt{A}$$
As  $S_{n+1} \leq S_n$ yields a true statement, it follows $S_{n+1} \leq S_n$ is true.

(c) Show that $s= \lim\limits_{n \rightarrow \infty} S_n$ exists
Since $S_{n+1} \leq S_n$, the sequence is non-increasing, 
using the non-increasing theorem stating that 

if $\{S_n\}$ is non-increasing then $$\lim\limits_{n \rightarrow
> \infty} S_n = \inf\{S_n\} $$


(d) find s

Is the argumentation in (a) and (b) appropriate? Also, I have to admit I m getting less confident in my argumentation (c) and (d). How to proceed in (c) and (d)? 
Much appreciated for your input or help. 

a) $\frac{1}{2}(S_{n+1}(1 +\frac{A}{(S_{n+1})^2}) \geq \frac{1}{2}(\sqrt{A}(1 +\frac{A}{(\sqrt{A})^2})$ is not necessarily true for $S_{n+1} \ge \sqrt{A}$ because $\frac{A}{(S_{n+1})^2} \le \frac{A}{(\sqrt{A})^2}$. You can prove $\frac{1}{2}(S_{n+1} +\frac{A}{S_{n+1}}) \ge \sqrt{A}$ simply transforming it into $(S_{n+1} - \sqrt{A})^2 \ge 0$
d) To find $s$ just replace $S_{n+1}$ and $S_{n}$ in the recurrence formula with $s$ and solve the resulting equation
