$x_{k+1}=(x_k+c/x_k)/2$. Prove that {$x_k$} converges and find its limit. Fix any $c>0$.
Let $x_1$ be any positive number and define $x_{k+1}=(x_k+c/x_k)/2$.
a)Prove that {$x_k$} converges and find its limit.
b)Use this sequence to calculate $\sqrt5$, accurate to six decimal places.
I tried to solve $x_{k+1}=(x_k+c/x_k)/2$ in terms of $x_1$ but it all went wrong. How can I solve it when there are two terms of $x_k$ in an equation?
Plus, any hint to question b)?
 A: If $\lim_{k \rightarrow \infty }x_{k}=a$, then $$a=(a+c/a)/2$$
$$2a=a+c/a$$
$$a^2=c$$
$$a=\sqrt{c}$$
But we must still prove that {$x_k$} converges
A: Notice for all $k \ge 1$,
$$\frac{x_{k+1}-\sqrt{c}}{x_{k+1}+\sqrt{c}}
= \frac{\frac12(x_k + \frac{c}{x_k}) - \sqrt{c}}{\frac12(x_k+\frac{c}{x_k}) + \sqrt{c}}
= \left(\frac{x_k-\sqrt{c}}{x_k+\sqrt{c}}\right)^2
$$
Using induction, it is easy to see for all $n \ge 1$, we have
$$\frac{x_n - \sqrt{c}}{x_n + \sqrt{c}} = \alpha^{2^{n-1}}
\iff x_n = \sqrt{c}\left(\frac{1 + \alpha^{2^{n-1}}}{1 - \alpha^{2^{n-1}}}\right)
\quad\text{ where }\quad \alpha =  \frac{x_1-\sqrt{c}}{x_1+\sqrt{c}}$$
When $x_1 > 0$, we have $|\alpha| < 1$ and hence $\alpha^{2^{n-1}}$ converges to $0$ as $n \to \infty$.
As a result, $x_n$ converges to $\sqrt{c}\left(\frac{1+0}{1-0}\right) = \sqrt{c}$.
For the special case $c = 5$, we know $\sqrt{5} \approx 2.236$. 
If we start from $x_1 = 2$, we have 
$$\alpha = \frac{2-\sqrt{5}}{2 + \sqrt{5}} \approx -0.0557$$
Since $\alpha$ is relatively small, even for $n = 1$, we have
$$|x_n - \sqrt{c} | = 2\sqrt{c}\left|
\frac{
\alpha^{2^{n-1}}
}{
1 - \alpha^{2^{n-1}}}
\right|\approx 2\sqrt{c}|\alpha|^{2^{n-1}}$$
For $n = 4$, the RHS is about $4\times 10^{-10}$. This implies
$x_4 = \frac{51841}{23184} \approx 2.23606797791580$
is an approximation of $\sqrt{5} \approx  2.23606797749979$
accurate to around $9$ decimal places.
A: Define $f(x)=\dfrac{x+\dfrac{c}{x}}{2}-x$ and notice that $f$ is strictly decreasing. 
Also $f(x)=0\iff x=\sqrt{c}$. If $x>\sqrt{c}$, then $f(x)<0$ and if $x<\sqrt{c}$, then $f(x)>0$. 
Now, if $x_1>\sqrt{c}$, then $f(x_1)<0\implies \sqrt{c}<x_2<x_1$. Using induction we can prove that $\sqrt{c}<x_{k+1}<x_k$. Thus, our sequence will converge (to $\sqrt{c}$).
Similarly,if $x_1<\sqrt{c}$, then $\sqrt{c}>x_{k+1}>x_k$ and our sequence will converge (to $\sqrt{c}$).
Obviously, if $x_1=\sqrt{c}$, then our sequence is constant.
