Show convergence of recursive sequence and find limit value

We already proved (in Show limit for recursive sequence by induction) that $a_n\geq {1\over 3}$ so $$a_{n+1}-a_n=a_n{1-3a_n\over 3a_n+3}\leq 0$$ so the sequence is decreasing and bounded so it is convergent. Say $a$ is it limit, then \begin{eqnarray*} a&=&\lim _{n\to \infty} a_{n+1} \\ &=& \lim _{n\to \infty} {4a_n\over 3a_n+3}\\ &=& {\lim _{n\to \infty}4a_n\over \lim _{n\to \infty}(3a_n+3)}\\ &=& {4a\over 3a+3} \end{eqnarray*} So we have to solve the equation $$a={4a\over 3a+3}\Longrightarrow 3a^2-a=0 \Longrightarrow a=0\;\; {\rm or}\;\; a=1/3$$ Since all members of sequence are $\geq {1/3}$ we have $$a ={1\over 3}$$

All the sequences fulfilling a recurrence relation of the form $a_{n+1}=\frac{A a_n + B}{C a_n + D}$ admit simple closed forms. The trick is to notice that by setting $\frac{p_n}{q_n}=a_n$, $p_0=a_0, q_0=1$ and $v_n=(p_n,q_n)^T$ we have $$ v_{n+1} = \begin{pmatrix}A & B \\ C & D\end{pmatrix} v_n $$ hence $a_n$ is completely fixed by the powers of the involved matrix $M$. By the Hamilton-Cayley theorem the sequences $\{p_n\}_{n\geq 0}$, $\{q_n\}_{n\geq 0}$ have the same characteristic polynomial of $M$. If the eigenvalues of $M$ are distinct, it follows that $$ a_n = \frac{a \lambda_1^n + b \lambda_2^n}{c \lambda_1^n + d \lambda_2^n}$$ where $\lambda_1,\lambda_2$ are the eigenvalues of $M$ and $a,b,c,d$ are suitable constants depending on the initial values. In your case the involved matrix is $$ M = \begin{pmatrix}4 & 0 \\ 3 & 3\end{pmatrix} $$ which is lower triangular. It follows that $a_n = \frac{a4^n+b 3^n}{c 4^n+d 3^n}$ and by interpolating through the initial values we get $$ a_n = \frac{3\cdot 4^n}{9\cdot 4^n-8\cdot 3^n} = \frac{1}{3}\left[1+\frac{1}{\frac{9}{8}\left(\frac{4}{3}\right)^n-1}\right]$$ so it is pretty obvious that $a_n\to\frac{1}{3}$ as $n\to +\infty$.