Study the convergence of $x_n$, $ x_{n+1}=\frac{1}{2}\big(\frac{x_n+3}{ x_n}\big)$, with $x_0=1.$ How can we prove that the limit of the sequence $x_n$ defined by:
$ x_{n+1}=\frac{1}{2}\big(\frac{x_n+3}{ x_n}\big)$, with $x_0=1$ exists? I have tried to prove that it is Cauchy , but I failed? Can I get some help, and thanks in advance.
 A: As @Michael Hardy noted, letting $n\to\infty$ in the recurrence relation suggests that $\lim_{n\to\infty} x_n = \frac32$.
We can prove inductively that $x_n \in \left[\frac54,2\right]$ for all $n \ge 1$. Indeed, we have $x_1 = 2$. Furthermore, if $x_n \in \left[\frac54,2\right]$ then also
$$\frac54 = \frac12 + \frac32\cdot \frac12\le \underbrace{\frac12 + \frac32 \frac1{x_n}}_{=x_{n+1}} \le \frac12 + \frac32\cdot \frac45 = \frac{11}{10} \le 2$$
so $x_{n+1} \in \left[\frac54,2\right]$.
Now for all $n \ge 1$ have
$$\left|\frac32-x_{n+1}\right| = \left|\frac32-\frac12-\frac32\frac1{x_n}\right| = \left|1-\frac32\frac1{x_n}\right| = \frac1{x_n}\left|x_n - \frac32\right| \le \frac45\left|x_n - \frac32\right|$$
so iterating this yields
$$\left|x_n - \frac32\right| \le \left(\frac45\right)^{n-1}\left|x_1 - \frac32\right| = \frac12\left(\frac45\right)^{n-1} \xrightarrow{n\to\infty} 0$$
We conclude that $\lim_{n\to\infty} x_n = \frac32$.
A: Hint : Show that $x_n$'s (for $n\geq 1$) lie in the interval $[\frac 54,2]$ and the function $f(x)=\frac 12+\frac 3{2x}$ on $[\frac 54,2]$ gives a contraction mapping.
A: Just rewriting$$x_{n+1}=\frac{1}{2}\left(\frac{x_n+3}{ x_n}\right)\implies x_{n+1}=x_n-\frac{2 x_n^2-x_n-3}{2 x_n}$$ and recognize the Newton formula for finding the zero's of function $$f(x)=2x^2-x-3$$
A: I think with fixed-point theorem : let's define
$$ f(x) := \frac{x+3}{2x},$$
such that $x_{n+1}=f(x_n)$.
The interval $[1;2]$ is stable by $f$, so for all $n$, $1 \leqslant x_n \leqslant 2$.
Moreover $f$ is continuous on $[1;2]$ and $3/2$ is the unique fixed point in $[1;2]$, so $(x_n)$ converges to $3/2$.
