How determine limits of bounded series? Let a>2 .

$\phi=a- \frac{1}{a- \frac{1}{a-\frac{1}{a-\frac{1}{....}}}}$
$\phi_0=a$ and $\phi_{n+1}=a-\frac{1}{\phi_n}$
I would like to determine by recurrence that   $1 \leq \phi_n \leq a$

there is no limit to apply here. All I have  is that:    $a -\frac{1}{a} 
\neq 0$
Any hint please?
 A: Look at the function $f(x)=a-\frac{1}{x}$, because of $\phi_{n+1}=f(\phi_{n})$. Observe that $f'(x)=\frac{1}{x^2}>0$ (i.e. increasing) and $\color{red}{\phi_{1}}=a-\frac{1}{a}\color{red}{<}a=\color{red}{\phi_{o}}$, thus
$$\color{red}{\phi_{2}}=f(\phi_{1})\color{red}{\leq} f(\phi_{0})=\color{red}{\phi_{1}}$$
and by induction
$$\phi_{n+1}\leq \phi_{n} \Rightarrow 
\color{red}{\phi_{n+2}}=f(\phi_{n+1})\color{red}{\leq} f(\phi_{n})=\color{red}{\phi_{n+1}} \tag{1}$$
the sequence is decreasing.
Also 
$$1\leq x\leq a \Rightarrow 1\geq\frac{1}{x}\geq\frac{1}{a}\geq0 \Rightarrow \\
-1\leq-\frac{1}{x}\leq-\frac{1}{a}\leq0 \Rightarrow\\
1\leq a-1\leq a-\frac{1}{x}\leq a-\frac{1}{a}<a$$
or 
$$x\in [1,a] \Rightarrow f(x)\in[1,a] \tag{2}$$
Because $\phi_0\in[1,a]$ then by induction 
$$\phi_n\in[1,a] \Rightarrow \phi_{n+1}=f(\phi_n)\in[1,a]$$
The sequence is monotonic and bounded, thus the limit exists. Now, you can find its value $L=\lim\limits_{n\to\infty}\phi_n$ from
$$L=a-\frac{1}{L} \iff L^2-aL+1=0$$
