calculus - Recursive sequence problem...

Consider the recursive sequence
\begin{equation*}
\begin{split}
a_{n+1} = \frac{5}{6-a_n} \quad \textit{with} \quad a_1 = 4.
\end{split}

\end{equation*}
Prove that the sequence $(a_n)$ converges and find its limit, by working out the following steps.
1. First assume that the limit $L = \lim_{n\to \infty} a_n$ exists and find its possible values.
Let $L = \lim_{n\to\infty}a_n$. Then we get
\begin{equation*}
\begin{split}
L &= \lim_{n\to\infty}a_{n+1} \\
&= \lim_{n\to\infty} \frac{5}{6-a_n} \\
&= \frac{5}{6-L}.
\end{split}
\end{equation*}
So we have $L^2-6L+5 = 0 \Longleftrightarrow (L-1)(L-5) = 0$. So the possible values of $L$ are $L = 1$ or $L = 5$.
2. Starting with the initial value $a_1 = 4$, write down the first five entries in the sequence $(a_n)$. Can you see any pattern?
We have

\begin{equation*}
\begin{split}
a_2 &= \frac{5}{6-a_1} = \frac{5}{6-4} = \frac{5}{2} = 2.5 \\
a_3 &= \frac{5}{6-a_2} = \frac{5}{6-5/2} = \frac{10}{7} \approx 1.42857 \\
a_4 &= \frac{5}{6-a_3} = \frac{5}{6-10/7} = \frac{35}{32} = 1.09375 \\
a_5 &= \frac{5}{6-a_4} = \frac{5}{6-35/32} = \frac{160}{157} \approx 1.01091 \\
a_6 &= \frac{5}{6-a_5} = \frac{5}{6-160/157} = \frac{785}{782} \approx 1.00384.
\end{split}
\end{equation*}
3. Find the real valued function $f(x)$ defining the sequence, i.e. $a_{n+1} = f(a_n)$.
This is the question I'm having trouble with. Please help!

Answer

Hints. For $3.$ $f(x)=\frac{5}{6-x}$ with $f'(x)=\frac{5}{(6-x)^2}>0$, i.e. the function is ascending. Now $a_2=f(a_1)=\frac{5}{2}<4=a_1$. Then $f(a_2)\leq f(a_1) \Rightarrow a_3\leq a_2$ and by induction
$$a_n\leq a_{n-1} \Rightarrow f(a_n)\leq f(a_{n-1}) \Rightarrow a_{n+1}\leq a_n$$
or the sequence is descending.

Let's show that it is also bounded. From $$x\in[1,5] \Rightarrow 1\leq x \leq 5 \Rightarrow 5\geq 6-x \geq 1 \Rightarrow 1\leq \frac{5}{6-x}\leq 5$$
or
$$x\in[1,5] \Rightarrow f(x)\in[1,5]$$
Now, $a_1\in[1,5] \Rightarrow a_2=f(a_1)\in[1,5]$ and, again by induction, $a_n\in[1,5]$. So, the sequence is bounded in monotone.