Find the limit of the bounded decreasing sequence 
Find the limit of the bounded decreasing sequence defined by
$a_1=3 \\a_{n+1}= \frac{1}{4-a_n}+1$

Can anyone teach me how to do this question? Thanks.
The answer is $\frac{5-\sqrt5}{2}$.
 A: The recurrence relation is $$a_{n+1}=\frac{a_n-5}{a_n-4},$$ and for this kind of recurrence we have a trick to calculate the general formula.
The equation $$\frac{x-5}{x-4}=x$$ is called the characteristic equation of the recurrence relation and let us denote its roots (called characteristic roots) by $\alpha$ and $\beta$. Then after some (ugly) calculations we can get $$\frac{a_{n+1}-\alpha}{a_{n+1}-\beta}=k\frac{a_n-\alpha}{a_n-\beta},$$ where $k$ is some constant. Now this is a geometric progression and can be easily solved.
A: Use the fact that $\lim_{n \to \infty} a_n = \lim_{n \to \infty} a_{n+1}$. 
Write $\lim a_n = k$ and you have  by simple limit arithmetic properties that $k  = \dfrac{1}{4-k} + 1$.
From here you can find $k$.
A: We'll prove that $a_n\leq3$.
Indeed, $a_1\leq3$ and $a_{n+1}=\frac{1}{4-a_n}+1\leq\frac{1}{4-3}+1=2<3$.
In another hand $a_1>\frac{5-\sqrt5}{2}$ and
$$a_{n+1}-\frac{5-\sqrt{5}}{2}=\frac{1}{4-a_n}-\frac{3-\sqrt5}{2}=\frac{1}{4-a_n}-\frac{2}{3+\sqrt5}=$$
$$= \frac{1}{4-a_n}-\frac{1}{4-\frac{5-\sqrt5}{2}}=\frac{a_n-\frac{5-\sqrt5}{2}}{(4-a_n)\left(\frac{3}{2}+\frac{\sqrt5}{2}\right)},$$
which gives $a_n>\frac{5-\sqrt{5}}{2}$ by induction and
$$a_n-\frac{5-\sqrt{5}}{2}<\left(\frac{2}{3+\sqrt5}\right)^1\left(a_{n-1}-\frac{5-\sqrt{5}}{2}\right)<$$
$$<\left(\frac{2}{3+\sqrt5}\right)^2\left(a_{n-2}-\frac{5-\sqrt{5}}{2}\right)<...<\left(\frac{2}{3+\sqrt5}\right)^{n-1}\left(a_1-\frac{5-\sqrt{5}}{2}\right)$$
and since $\frac{2}{3+\sqrt5}<1$ we obtain
$$\lim_{n\rightarrow+\infty}a_n=\frac{5-\sqrt5}{2}$$
by the limit definition.
