how to solve $3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$ $$A = 3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$$
My answer is:
$$\begin{align}
&A = 3 - \frac {2}{A}\\
\implies &\frac {A^2-3A+2}{A}=0\\
\implies &A^2-3A+2=0\\
\implies &(A-1)\cdot(A-2)=0\\
\implies &A=1\;\text{ or }\; A=2
\end{align}$$
I should note that I'm not sure if the above answer is true. Because I expected just one answer for A (A is a numeric expression), but I found two, $1$ and $2$. This seems to be a paradox.

This continued fraction is the limit of 
the sequence $a_n=3-2/a_{n-1}$. Computing the first few terms shows that $2$ is the correct limit; if our initial term $a_1=3$ were different then the limit could be $1$.
