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.

Note that as we add more terms to the continued fraction, it oscillates between $1$ and slightly higher than $2$.
$$
\begin{align}
n&=1&
3&=3&
3-2&=1\\\\
n&=2&
3-\cfrac23&=\frac73&
3-\cfrac{2}{3-2}&=1\\\\
n&=3&
3-\cfrac{2}{3-\cfrac23}&=\frac{15}7&
3-\cfrac2{3-\cfrac2{3-2}}&=1\\\\
n&=4&
3-\cfrac2{3-\cfrac2{3-\cfrac23}}&=\frac{31}{15}&
3-\cfrac2{3-\cfrac2{3-\cfrac2{3-2}}}&=1\\\\
&&
3-\cfrac2{\cfrac{2^n-1}{2^{n-1}-1}}&=\frac{2^{n+1}-1}{2^n-1}&
3-\cfrac21&=1
\end{align}
$$
Therefore, the limit of the continued fractions with $2n-1$ twos and threes is
$$
\lim_{n\to\infty}\frac{2^{n+1}-1}{2^n-1}=2
$$
and the limit of the continued fractions with $2n$ twos and threes is
$$
\lim_{n\to\infty}1=1
$$
Therefore, one value represents the limit of an odd number of twos and threes and the other value represents the limit of an even number of twos and threes.
