Find closed formula and limit for $a_1 =1$, $2a_{n+1}a_n = 4a_n + 3a_{n+1}$ Tui  a sequence $(a_n)$ defined for all natural numbers given by
$$a_1 =1, 2a_{n+1}a_n = 4a_n + 3a_{n+1}, \forall n \geq 1$$

Find the closed formula for the sequence and hence find the limit. 
Here, what I have done:
$$2a_{n+1}a_n = 4a_n + 3a_{n+1}
\implies a_{n+1} = \frac{4a_n} {2a_n - 3} 
\implies a_{n+1} = \frac{\frac{4a_n} {a_n} } {\frac{2a_n}{a_n} - \frac{3} {a_n} }
\implies  a_{n+1} = \frac{4} {2 - \frac{3} {a_n} } 
\implies \frac{1 } {a_{n+1}} = \frac{2 - \frac{3} {a_n} } {4}
\implies \frac{1 } {a_{n+1}} =\frac{1 } {2 } - \frac{3} {4a_n}$$
Then go to where????
hint

If $a_n=0$ then $a_{n+1}=0$.

assume that for $n&gt;0 \;\;a_n\ne 0$.

dividing by $2a_na_{n+1} $, we get

$$1=\frac {2}{a_{n+1}}+\frac {3}{2a_n} $$

Put $b_n=2/a_n $.
then

$$b_{n+1}=1-\frac {3}{4}b_n $$
Now look for $p$ such that

$$b_{n+1}-p=-3/4(b_n-p) $$
thus
$$p=4/7$$

$$b_n=4/7+(-3/4)^{n-1}(b_1-4/7) $$
$$=4/7+10/7 (-3/4)^{n-1} $$


  $$a_n=\frac {7}{2+5 (-3/4)^{n-1}} $$
  $$\lim_{n\to+\infty}=7/2$$
If you have proved that the sequence has a limit $l$, then $l$ can be calculated by the following:

$$\lim_{n\to \infty}2a_{n+1}a_n = \lim_{n\to \infty}(4a_n + 3a_{n+1}) $$

and hence

$$2l^2=4l+3l$$