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 $\frac{5-\sqrt5}{2}$.

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.
