Show that $\lim x_n$ exists. 
Consider the sequence $x_n $ defined by $x_{n+1}=\dfrac{x_n+3}{3x_n+1}$ with $0<x_1<1$.
Show that $\lim x_n$ exists.

My try:
Let $x_1=a$ then $x_2=\dfrac{a+3}{3a+1},x_3=\dfrac{5a+3}{3a+5},x_4=\dfrac{7a+9}{9a+7},x_5=\dfrac{17a+15}{15a+17}$.
If I assume the limit to exist then it can be found out to be $1$,but how to show that the limit exists actually?
Please help.
 A: If
$$
x_{n+1}=\frac{x_n+3}{3x_n+1}
$$
then
$$
\frac{x_{n+1}-1}{x_{n+1}+1}=-\frac12\frac{x_n-1}{x_n+1}
$$
Therefore,
$$
\bbox[5px,border:2px solid #C0A000]{x_n=\frac{1+\left(-\frac12\right)^n\frac{x_0-1}{x_0+1}}{1-\left(-\frac12\right)^n\frac{x_0-1}{x_0+1}}}
$$
A: Let $f : x \in \mathbb{R}_{+}\mapsto \dfrac{x + 3}{3x+1} $.
$f$ is a homographic function, and since $1 \times 1 - 3 \times 3 < 0$, $f$ is decreasing.
$f(0) = 3$ and $f(3) = \dfrac{3}{5}$ therefore $ f([0,3]) \subset [0,3] $.
As $ x_1 \in [0,3] $, it gives : 
$\forall n \in \mathbb{N}^{*}, 0 \leq x_n \leq 3$.
Moreover,
$\forall n \in \mathbb{N}^{*}, x_{n+1} - x_n = 3\dfrac{x_n^2 + 1}{3x_n+1} > 0$.
$(x_n)$ is therefore increasing and has an upper bound : it converges, and lim $x_n$ exists.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Set $\ds{x_{n} \equiv {p_{n} \over q_{n}}}$ such that
  $\ds{x_{n + 1} = {x_{n} + 3 \over 3x_{n} + 1}}$  becomes
  $\ds{{p_{n + 1} \over q_{n + 1}} = {p_{n} + 3q_{n} \over 3p_{n} + q_{n}}}$. Then, choose $\ds{p_{n + 1} = p_{n} + 3q_{n}\quad\mbox{and}\quad q_{n + 1} = 3p_{n} + q_{n}}$ such that the original sequence is identically satisfied. 

The above setting is equivalent to ( $\ds{\sigma_{i}}$ with
$\ds{i = x,y,z}$ is a
Pauli Matrix )
\begin{align}
{p_{n + 1} \choose q_{n + 1}} & =
\pars{\begin{array}{cc}
\ds{1} & \ds{3}
\\
\ds{3} & \ds{1}
\end{array}}
{p_{n} \choose q_{n}} = \pars{1 + 3\sigma_{x}}{p_{n} \choose q_{n}}
\end{align}

For simplicity, scalars as added quantities to matrices are implicitly multiplied by the identity matrix.

Then,
\begin{align}
{p_{n + 1} \choose q_{n + 1}} & =
\pars{1 + 3\sigma_{x}}^{\, 2}{p_{n - 1} \choose q_{n - 1}} =
\pars{1 + 3\sigma_{x}}^{\, 3}{p_{n - 2} \choose q_{n - 2}} = \cdots =
\pars{1 + 3\sigma_{x}}^{\, n}{p_{1} \choose q_{1}}
\end{align}

Note that
$\ds{\pars{\totald[2]{}{k} - 1}\exp\pars{k\sigma_{x}} = 0\,;\qquad
\left.\exp\pars{k\sigma_{x}}\right\vert_{\ k\ =\ 0} = 1\,,\quad
\left.\totald{\exp\pars{k\sigma_{x}}}{k}\right\vert_{\ k\ =\ 0} = \sigma_{x}}$

because $\ds{\sigma_{x}^{2} = 1}$ $\ds{\implies \exp\pars{k\sigma_{x}} = \cosh\pars{k} + \sinh\pars{k}\sigma_{x}}$
such that
\begin{align}
\pars{1 + 3\sigma_{x}}^{n} & =
n!\bracks{z^{n}}\exp\pars{z\bracks{1 + 3\sigma_{x}}} =
n!\bracks{z^{n}}\exp\pars{z}\exp\pars{3z\sigma_{x}}
\\[5mm] & =
{1 \over 2}\,n!\bracks{z^{n}}\bracks{\pars{\expo{4z} + \expo{-2z}} +
\pars{\expo{4z} - \expo{-2z}}\sigma_{x}}
\\[5mm] & =
{1 \over 2}\braces{\bracks{4^{n} + \pars{-2}^{n}} +
\bracks{4^{n} - \pars{-2}^{n}}\sigma_{x}}
\\[5mm] & =
2^{n - 1}\pars{\begin{array}{cc}
\ds{2^{n} + \pars{-1}^{n}} & \ds{2^{n} - \pars{-1}^{n}}
\\
\ds{2^{n} - \pars{-1}^{n}} & \ds{2^{n} + \pars{-1}^{n}}
\end{array}}
\end{align}
and
\begin{align}
x_{n + 1} & = {p_{n + 1} \over q_{n + 1}} =
{\bracks{2^{n} + \pars{-1}^{n}}p_{1} + \bracks{2^{n} - \pars{-1}^{n}}q_{1} \over \bracks{2^{n} - \pars{-1}^{n}}p_{1} + \bracks{3^{n} + \pars{-1}^{n}}q_{1}}
\\[5mm]
\implies & \bbx{x_{n + 1} =
{\bracks{2^{n} + \pars{-1}^{n}}x_{1} + 2^{n} - \pars{-1}^{n} \over
\bracks{2^{n} - \pars{-1}^{n}}x_{1} + 2^{n} + \pars{-1}^{n}}}
\end{align}
