Limit of a recursive sequence $a_{n+1}=\frac{1}{2}\left(a_n + \frac{5}{a_n}\right)$ A sequence is given by $a_{n+1}=\dfrac{1}{2}\left(a_n + \dfrac{5}{a_n}\right)$ for each integer $n \geq 0$.  Does the limit of $a_n$ as $n$ goes to infinity equal $\sqrt{5}$ for $a_0 =2$?  What if $a_0=3$?
Fix $k&gt;0$ and $t&gt;0$.  I shall prove that the sequence $\left\{a_n\right\}_{n\in\mathbb{Z}_{\geq 0}}$ defined by $a_0=t$ and
$$a_{n+1}=\frac{1}{2}\,\left(a_n+\frac{k^2}{a_n}\right)\text{ for every }n=0,1,2,3,\ldots\tag{*}$$ satisfies
$$\lim_{n\to\infty}\,a_n=k\,.$$  This problem is a particular case where $k:=\sqrt{5}$ and $t\in\{2,3\}$.

First of all, it is clear that $a_n&gt;0$ for all $n\in\mathbb{Z}_{\geq 0}$.  By the AM-GM Inequality, $$a_n=\frac{1}{2}\,\left(a_{n-1}+\frac{k^2}{a_{n-1}}\right)\geq \sqrt{a_{n-1}\left(\frac{k^2}{a_{n-1}}\right)}= k$$ for every $n=1,2,3,\ldots$.  Note from (*) that
$$a_{n+1}-a_n=\frac{1}{2}\left(\frac{k^2-a_n^2}{a_n}\right)\leq 0\text{ for }n=1,2,3,\ldots\,.$$
That is, the subsequence $\left\{a_n\right\}_{n\in\mathbb{Z}_{&gt;0}}$ is nonincreasing and is bounded from below by $k$.  Hence, $\lim\limits_{n\to\infty}\,a_n$ exists (and equals $\inf\big\{a_n\,\big|\,n=1,2,3,\ldots\big\}$).  If $L$ is this limit, then $L\geq k&gt;0$.  By (*), we have
$$L=\frac{1}{2}\,\left(L+\frac{k^2}{L}\right)\text{ or }L=k\,.$$
Disclaimer: I'm in a hurry so I'll just show, if the base value is acceptable, that the sequence converges but I don't think its hard to find the limit using subsequences.

$\textbf{Claim 1}: \forall y&gt;0, y+25/y\geq 10$.

$\textit{Proof:} 0\geq (y-5)^2=y^{2}-10y+25\Rightarrow y^{2}+25\geq 10y\Rightarrow y+y/25\geq 10$.

$\textbf{Theorem}: \text{The sequence} \hspace{.5pc} (x_{n}), x_{n}=\displaystyle \frac{1}{2}\left(x_{n}+\frac{5}{x_{n}} \right)$ converges.

$\textit{Proof:} \forall x&gt;0, \displaystyle \left(x+\frac{5}{x}\right)^2=x^{2}+\frac{25}{x^{2}}+10\underset{x^{2}&gt;0\wedge\text{Claim 1}}{\geq}10+10=20\Rightarrow x_{n}^{2}=\frac{1}{4}\left(x_{n}+\frac{5}{x_{n}}\right)^{2}\geq \frac{1}{4}\cdot 20=5\Rightarrow x_{n}^{2}\geq 5\forall n\Rightarrow x_{n}-x_{n+1}=x_{n}-\frac{1}{2}\left(x_{2}+\frac{5}{x_{n}}\right)=\frac{1}{2}x_{n}-\frac{5}{x_{n}}=\frac{x_{n}^{2}-5}{2x_{n}}\underset{x_{n}^{2}\geq 5}{\geq}0\Rightarrow (x_{n})$ is decreasing, bounded above by $x_{1}$ (if base value is appropriate) and bounded below by $\sqrt{5}$ (why?). Then $(x_{n})$ converges [1].

What this means is that we can find a sequence of rationals that approximate square roots. Can you generalize this idea to any $\sqrt{c}, c&gt;0$?

I didn't check out the source but just google monotone convergence theorem or trust wiki :)

[1] https://en.wikipedia.org/wiki/Monotone_convergence_theorem 

[2] Understanding Analysis; Stephen Abbott, 2nd ed.