$A$ is a Regular Transition Matrix $\Rightarrow$ $\lim\limits_{m \to \infty} A^m$ exists and rank 1 A is  a Regular Transition Matrix  $\Rightarrow$ $\lim\limits_{m \to \infty} A^m$ exists and rank 1
At the above proposition, what does "regular" mean?
 A: 
Def: $A$ is regular if some power of $A$ has all positive entries.

See also this question and here.
This comes up for regular Markov chains (i.e. Markov chains with a regular transition matrix). 
Concerning the statement you reference:
$$
A \;\,\text{regular} \;\;\implies\;\;  \text{rank}\left(\lim_{m\rightarrow\infty} A^m\right)=1
$$
this is because any regular transition matrix $A$ satisfies
$$
S:=\lim_{m\rightarrow\infty} A^m =
\begin{bmatrix}
s_1 & s_1 & \ldots & s_1 \\
s_2 & s_2 & \ldots & s_2 \\
\vdots & \vdots &\ddots & \vdots \\ 
s_n & s_n & \ldots & s_n
\end{bmatrix}
$$
where $s = (s_1, \ldots, s_n)$ is the steady state distribution of the Markov chain (with $||s||_1=1$).
Notice that the rank of $S$ is obviously 1, because there is only one linearly independent column. Intuitively, this means a regular Markov chain converges to a distribution $s$ after sufficient time passes, no matter what its initial distribution was and that this does not change as time passes further (i.e. $As=s$).
