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???? 

So you got $\frac{1} {a_{n+1}} =\frac{1 } {2 } - \frac{3} {4a_n}, a_1=1.$ 
Denote: $b_n=\frac{1}{a_n}$, then the equation becomes $b_{n+1}=-\frac{3}{4}b_n+\frac12$ with $b_1=\frac{1}{a_1}=1$.
Now apply the general formula: $$\bbox[lightgreen]
{
If \  \ y_{n+1}=ay_n+b,y_1=c, \ then: y_n=\left(c+\frac{b}{a-1}\right)a^{n-1}-\frac{b}{a-1}.
\qquad
} $$
Thus: 
$$b_n=\left(1+\frac{\frac12}{-\frac34-1}\right)\left(-\frac34\right)^{n-1}-\frac{\frac12}{-\frac34-1}=\frac{5}{7}\left(-\frac34\right)^{n-1}+\frac27$$ and $$a_n=\frac{1}{b_n}=\frac{7}{2+5\left(-\frac34\right)^n}$$ 
And the limit: $$\lim_\limits{n\to+\infty} a_n=\frac72.$$
