<s>[INST] 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???? 
 [/INST] Enforcing $u_n=1/a_n$ gives 
$$2a_{n+1}a_n = 4a_n + 3a_{n+1}\Longleftrightarrow 4u_{n+1}+ 3u_n =2\Longleftrightarrow u_{n+1}=- \frac34u_n +\frac12$$
Now set $x_n = u_n-\frac27$ then you can easily check that we have $x_{n}= -\frac34x_{n-1}$ which is a geometric sequence 
$$x_{n}= -\frac34x_{n-1}\implies x_n =\left(-\frac{3}{4}\right)^{n-1}x_1 \\\implies \color{red}{\frac{1}{a_n}= u_n =\frac27+\left(-\frac{3}{4}\right)^{n-1}(u_1 -\frac27)\to \frac27}$$

Alternatively, a more general formula is obtained  using this How to find the closed form of $u_{n+1}=a_nu_n+b_n~~~\text{where $a_1$, $a_n$ and $b_n$ are given.}$ you get that 
  $$ u_n = \left(\sum_{p=1}^{n-1}\frac{ \frac12}{\prod_{k=1}^{p}- \frac34}+u_1\right)\prod_{k=1}^{n-1}- \frac34 = \left(-\frac{3}{4}\right)^{n-1}\left(1+\frac12\sum_{p=1}^{n-1}\left(-\frac{4}{3}\right)^{p-1} \right)\\=\color{red}{\frac{1}{a_n}= u_n =\frac27+\frac57\left(-\frac{3}{4}\right)^{n-1} \to \frac27}$$

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