Determine $\lim_{n \to \infty} P_{ij}^n$ without using technology.

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Consider a Markov chain with transition matrix

$ \begin{matrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 & 0 & 0 & 0 \\ \frac{2}{3} & 0 & \frac{1}{3} & 0 & 0 & 0 & 0 \\ 0 & \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{4} & 0 & \frac{1}{2} & 0 & \frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{4} & 0 & \frac{1}{4} & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & \frac{1}{8} & \frac{1}{8} & \frac{1}{8} & \frac{1}{8} & \frac{1}{4} & \frac{1}{4} \end{matrix} $

Say the states space are $\{1,2,3,4,5,6,7 \}$.Identify the communication classes. Classify the states as recurrent or transient. For all $i$ and $j$, determine $\lim_{n \to \infty} P_{ij}^n$ without using technology.

I can identify the communication classes are $\{1,2,3\}$ and $\{4,5,6,7 \}$. The former is recurrent, and the latter is transient. But I don't know how to determine $\lim_{n \to \infty} P_{ij}^n$ without using technology.

My attempt:

The submatrix corresponding to the recurrent class is doubly-stochastic and thus the stationary distribution is uniform for these states. I also note that the state 1 and 3 has the property that $P_{11} =1/3=P_{33}$, thus it is necessarily aperiodic. Hence, this sub MC is ergodic. The stationary distribution is just the limiting distribution.

Also, I know that if $j$ is transient, then $\lim_{n \to \infty} P_{ij}^n=0$. But I don't know how to get when $j$ is recurrent and $i$ is transient.