Finding a limit of recursively given sequence Given a sequence $\{a_n\}_{n=1}^\infty$:
$$a_1=2$$
$$\forall n\in \Bbb{N}:a_{n+1}=4-\frac{3}{a_n}$$ find $\lim_{n\to\infty}a_n$.
Writing out first few terms i found out $a_n\to 3$ as $n\to\infty$. But i would like it to be in more precise way (the best would be the $\epsilon$-definition). So I attempted to get a direct formula for $a_n$ not depending on any other $a_i$'s but also couldn't find a suitable way to do that. Could you please give me some hints, how to go about this problem in more mathematical way?
 A: For the main question "how to find the limit", please see other answers.
As a supplement, here is a trick to obtain a direct formula for this particular $a_n$.
If $L_a$ is the limit of sequence $a_n$, it will satisfy
$$L_a = 4 - \frac{3}{L_a} \iff L_a^2 - 4L_a + 3 = (L_a-1)(L_a-3) = 0$$
In general, if you have a sequence $b_n$ whose limit $L_b$ satisfy some polynomial equation with roots $\lambda_1, \lambda_2, \ldots, \lambda_n$, you can construct auxillary sequences $c_n$ of the form $\frac{(b_n - \lambda_{i_1})(b_n-\lambda_{i_2})\cdots(b_n - \lambda_{i_p})}{(b_n - \lambda_{j_1})(b_n-\lambda_{j_2})\cdots(b_n - \lambda_{j_q})}$ and see whether any of them is easier to solve or estimate the bounds.
For the sequence at hand, let 
$\displaystyle\;c_n = \frac{a_n - 3}{a_n - 1} \iff a_n = \frac{c_n-3}{c_n-1}\;$, we have
$$c_{n+1} = \frac{ c_n - 3}{c_n - 1}
= \frac{1 - \frac{3}{a_n}}{3 -\frac{3}{a_n}} = \frac13\frac{a_n-3}{a_n-1} = \frac13 c_n$$
Together with $c_1 = \frac{2-1}{2-3} = -1$, we obtain
following explicit form for $c_n$ and hence the one for $c_n$.
$$c_n = -\frac{1}{3^{n-1}}
\quad\implies\quad a_n = \frac{-\frac{1}{3^{n-1}} - 3}{-\frac{1}{3^{n-1}} - 1} = \frac{3^n+1}{3^{n-1}+1}
$$
A: One shows that the sequence is strictly increasing and bounded from above. From this conclude that the limit exists. For the limit $a$, we must have $a=4-\frac 3a$. Conclude that $a=3$ (and not $a=1$).
A: Technical point:
To show that $a_n$ is strictly increasing and bounded by $3$ note that we have
I:
$$3>x>0\implies (x-3)(x-1)<0\implies x^2-4x+3<0\implies x^2<4x-3\implies x<4-\frac 3x$$ 
II:  $$0<x<3\implies \frac 3x>1\implies 4-\frac 3x<3$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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$\ds{a_{n + 1} = 4 - {3 \over a_{n}} = {4a_{n} - 3 \over a_{n}}}$. Lets $\ds{a_{n} = {x_{n} \over y_{n}}}$ such that
  $\ds{{x_{n + 1} \over y_{n + 1}} = {4x_{n} - 3y_{n} \over x_{n}}}$

Set $\ds{x_{n + 1} = 4x_{n} - 3y_{n}}$ and $\ds{y_{n + 1} = x_{n}}$ which can be written as $\ds{\pars{~\mbox{with}\ x_{1} = y_{1} = 1~}}$ 
\begin{align}
{x_{n + 1} \choose y_{n + 1}} & =
\pars{\begin{array}{rr}\ds{4} & \ds{-3} \\ \ds{1} & \ds{0}\end{array}}
{x_{n} \choose y_{n}} =
\pars{\begin{array}{rr}\ds{4} & \ds{-3} \\ \ds{1} & \ds{0}\end{array}}^{2}
{x_{n - 1} \choose y_{n - 1}} =
\pars{\begin{array}{rr}\ds{4} & \ds{-3} \\ \ds{1} & \ds{0}\end{array}}^{3}
{x_{n - 2} \choose y_{n - 2}}
\\[5mm] & = \cdots =
\pars{\begin{array}\ds{4} & \ds{-3} \\ \ds{1} & \ds{0}\end{array}}^{n}
{x_{1} \choose y_{1}} =
\pars{\begin{array}\ds{4} & \ds{-3} \\ \ds{1} & \ds{0}\end{array}}^{n}
{1 \choose 1}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
\lambda^{n}\,\mathbf{u}\,\mathbf{u}^{\mathsf{T}}{1 \choose 1}
\end{align}

$\ds{\lambda}$ is the eigenvalue, with the largest magnitude, of the above mentioned matrix and $\ds{\mathbf{u}}$ is the correspondent eingenvector.

Then,
$$
\lim_{n \to \infty}a_{n + 1} =
{\mathbf{v}_{+}^{\mathsf{T}}\mathbf{u}\,\mathbf{u}^{\mathsf{T}}{1 \choose 1} \over
\mathbf{v}_{-}^{\mathsf{T}}\mathbf{u}\,\mathbf{u}^{\mathsf{T}}{1 \choose 1}}\,,\qquad
\mathbf{v}_{+} \equiv {1 \choose 0}\,,\quad
\mathbf{v}_{-} \equiv {0 \choose 1}
$$

It turns out that $\ds{\mathbf{u} \propto {3 \choose 1}}$ such that
  
  $\ds{\mathbf{u}\mathbf{u}^{\mathsf{T}}{1 \choose 1} =
\pars{3 \quad 1}{3 \choose 1}{1 \choose 1} =
\pars{\begin{array}{cc}
\ds{9} & \ds{3}
\\
\ds{3} & \ds{1}
\end{array}}{1 \choose 1} =
{\color{red}{12} \choose \color{red}{4}}}$ which yields

$$
\lim_{n \to \infty}a_{n + 1} =
{\mathbf{v}_{+}^{\mathsf{T}}\mathbf{u}\,\mathbf{u}^{\mathsf{T}}{1 \choose 1} \over
\mathbf{v}_{-}^{\mathsf{T}}\mathbf{u}\,\mathbf{u}^{\mathsf{T}}{1 \choose 1}} =
{12 \over 4} = \bbx{\large 3}
$$
