(rectified) proof by induction - Fibonacci Sequence Define the sequence $ (a_n) $ by induction, putting $ a_1 = a_2 = 1 $ and $ a_ {n + 2} = a_ {n + 1} + a_n $, $ \forall n \in \mathbb{N} $ , thus obtaining the Fibonacci sequence $ (1,1,2,3,5,8,13,...)$.
Write $ x_n = \frac{a_n}{a_ {n + 1}} $ and prove that $ \lim x_n = c $, where $ c $ is the only positive number such that $ \frac{1}{c + 1} = c $ (ie $ c $ is the root of the equation $ c^{2} +c-1 = 0$).
Could someone help me continue my induction prove?
$$(a_n) = \begin{cases}
 1 & \mbox{for}\quad a_1\\
 1 & \mbox{for}\quad a_2 \hspace{1cm} (\forall n \in \mathbb{N})\\
 a_{n+1}-a_{n+2} & \mbox{for}\quad a_n
\end{cases}$$
Suppose $ \lim x_n = c $, then $ \lim x_{n + 1} = c $, since $ x_ {n + 1} $ is a subsequence of $ x_n $.
We can write
\begin{align*} 
a_{n + 2} = a_{n + 1} + a_{n} \Rightarrow \frac{a_{n + 2}}{a_{n + 1}} &= \frac{a_{n}+a_{n + 1}}{a_{n + 1}}\\
&= 1 +\frac{a_{n}}{a_{n + 1}}\\
\end{align*}
We know that
\begin{align*}
x_n=\frac{a_n}{a_{n + 1}}\
\end{align*}
follow that
$$x_{n+1}=\frac{a_{n+1}}{a_{n+2}}$$
therefore
$$ \frac{1}{x_{n + 1}}=x_n+1\Rightarrow x_{n+1}=\frac{1}{x_n+1}.$$
We have
$$\lim x_n=\lim x_{n+1}=c$$
Hence
\begin{align*}
\lim x_n=c &\Rightarrow \frac{1}{c+1}=c\\
&\Rightarrow 1=c^2+c \\
&\Rightarrow 0=c^2+c-1.
\end{align*}
Thus
\begin{align*}
c&=\frac{-1+\sqrt{1^2-4(1)(-1)}}{2\cdot(1)}\\
&= \frac{-1+\sqrt{5}}{2}
\end{align*}
Consequently
$$x_n \longrightarrow c=\frac{-1+\sqrt{5}}{2}$$

There are several mistakes/typos in your proof, and I suggest you go over your proof much more carefully. Once you have reached the equation
$$
\frac{1}{x_{n+1}} = 1+x_n
$$
you can simply apply the limit as $n \to \infty$ from both sides as the limits being finite.
