Real Analysis question about proving limits using boundedness and monotony I am trying to get a hang of solving problems that ask to "prove the limit". One of the questions on our previous homework was:

The solution starts with 
.
How do we know that sn is bounded below and not above? I believe the next step is to plug in the formula for Sn+1 into Sn^2 - x >= 0 but I am not sure why. I think I will understand the rest of the solution if you can help me understand this. Thank you.
 A: $s_n$ is bounded below because if you examine the formula
$$
s_{n+1} = \frac{s_n^2 + x}{2s_n}
$$
then the numerator is positive. So as long as you know that $s_n$ is nonzero and nonnegative at each finite $n$ (nonzero so the formula is well-defined), $s_{n+1}$ will in fact be nonnegative, hence bounded below by $0$.
As to why $s_n$ is nonzero: if at some $n$ we had $s_n = 0$, but $s_m \neq 0$ for all $m < n$ (i.e. we find the earliest $n$ where we get $0$), then
$$
0 = \frac{s_{n-1}^2 + x}{2s_{n-1}}
$$
and therefore $s_{n-1}^2 + x = 0$. But $x>0$ and $s_{n-1}^2 \geq 0$, so this is impossible.
As to why $s_n$ is nonnegative: a similar argument, and proceeds by induction. You've got the base case for free.
Finally, showing why it's not bounded above: that's the rest of the proof!
A: First by induction, it is easy to prove that $s_n>0$ for all n. Then
$$s_{n+1}=\dfrac{s^2_n+x} {2s_n}\geqslant \dfrac{2s_n\sqrt{x}}{2s_n}=\sqrt{x}$$
So $s_n$ is bound below. Now use induction to prove for all $n, s_{n}\leqslant s_{n-1}.$
For $n=2$
\begin{align}
s_2-s_1&=\dfrac{s^2_1+x}{2s_1}-s_1
\\
&=\dfrac{x-s^2_{1}}{2s_{1}}
\\
&\leqslant 0
\end{align}
Assume it be true for $k<n$, then 
\begin{align}
s_{n+1}-s_n&=\dfrac{s^2_n+x} {2s_n}-\dfrac{s^2_{n-1}+x} {2s_{n-1}}
\\
&=\dfrac{s^2_ns_{n-1}-s_ns^2_{n-1}+x(s_{n-1}-s_{n})}{2s_ns_{n-1}}
\\
&=\dfrac{(s_{n-1}-s_{n})(x-s_{n-1}s_{n})}{2s_ns_{n-1}}
\\
&\leqslant 0 
\end{align}
The last step is true for $s_{n-1}-s_{n}\geqslant 0$ and $x-s_{n-1}s_{n}\leqslant 0$
Since $s_n$ is monotonic decreasing and bounded below, it has limt. By taking the limit on both side, we have
$$s=\dfrac{s^2+x} {2s} \quad\text{where}\quad s=\lim \limits_{n \to \infty}s_n$$
Thus $s=\sqrt{x}$
