I have the sequence $a_{1} =1 $ and $a_{n+1}=0.5(a_{n}+x/a_{n})$ with $x \in \mathbb{R} $ and $x>0$. Then it is given that $a_{n}^2 \ge x$ for all $n\ge2$ and $a_{n+1}\le a_{n}$.
How can I show with these properties that $\lim_{n \to \infty}a_{n}=\sqrt{x}$?
I have come so far: $a_{n} \ge a_{n+1} \ge \sqrt {x} $.
On
As $a$ is decreasing and positive, it converges to some $L\ge 0$. As the sequences decreases, $$ a_n \le a_2 \implies a_n = \frac12\left( a_{n-1} + \frac xa_{n-1} \right) \ge \frac12\left( a_2 + \frac xa_{2} \right) \ge \frac 12 \frac x{a_2} $$
Now use the continuity of the function $$ y\to \frac 12\left(y + \frac xy\right) $$ on $\left[\frac 12 \frac x{a_2}, \infty\right)$: you can take the limit $n\to\infty$ in the definition of the sequence and get
$$ L = \frac12\left( L + \frac xL \right) $$hence $L = \sqrt x$.
On
Notice that $$r_{n+1}=\frac{a_{n+1}-\sqrt x}{a_{n+1}+\sqrt x}=\frac{\frac12(a_{n}+\frac x{a_n})-\sqrt x}{\frac12(a_{n}+\frac x{a_n})+\sqrt x}=\frac{(a_n-\sqrt x)^2}{(a_n+\sqrt x)^2}=r_n^2.$$
On every iteration, the ratio is squared; as $|r_n|<1$, it converges to $0$, and $a_n$ converges (quickly) to $\sqrt x$, as
$$a_n=\sqrt x\frac{1+r_n}{1-r_n}.$$
Firstly you need to know if the sequence is increasing or decreasing and if it is bounded (this can be done easily by induction). Once you proved that is decreasing and bounded, by the monotone convergence theorem, $(a_n)$ converges. Let $\lim (a_n)=L$.
Remember that every $m$-tail of $(a_n)$ converges to the same limit as $(a_n)$, thus $\lim (a_n)=\lim (a_{n+1})=L$
Now because $$a_{n+1}=0.5(a_{n}+x/a_{n})$$ then $$\lim(a_{n+1})=0.5(\lim (a_n)+x/\lim (a_n))$$ $$\implies L=0.5(L+x/L) $$ $$\implies L=\sqrt x$$
Note that this second part could only be done once proved that $(a_n)$ indeed converges.