We consider a Markov Chain $\left(X_n\right)_{n \in \mathbb{N}}$ on $S=\{1,2,3,4,5\}$ with transition matrix
Q = $\begin{pmatrix} 0 & 1/2 & 1/2 & 0 & 1/3\\ 1 & 0 & 1/2 & 0 & 0 \\ 0 & 1/4 & 0 & 1 & 0\\ 0 & 1/4 & 0 & 0 & 1/3 \\ 0 & 0 & 0 & 0 & 1/3 \end{pmatrix}$
We have the two classes $\{5\}$(which is transient) and $\{1,2,3,4\}$(which is recurrent) It is clear that under $P_1(X_n+i) = \mathbb{P}(X_n=i \lvert X_0=1)$ the chain is irreducible, since it will always move around the four states. To compute the equilibrium measure I solved $Q\mu = \mu$ since under $P_1$ the chain is irreducible it is unique and I obtained: $\mu = \left(\frac{3}{10},\frac{2}{5},\frac{1}{5}, \frac{1}{10}\right)$
Now I am asked to computed the following limits: $\lim_{n \rightarrow \infty}P_1(X_n=3)$ and $\lim_{n \rightarrow \infty}P_5(X_n=3)$.
How can I compute these? I think $\lim_{n \rightarrow \infty}P_1(X_n=3)$ = $\mu_3 = \frac{1}{5}$
But what about $\lim_{n \rightarrow \infty}P_5(X_n=3)$?
You are correct that $\lim_{n\to\infty}\mathbb P_1(X_n=3)$ is given by the equilibrium probability $1/5$.
Now observe that for $n\geq1$, we have that $$\mathbb P_5(X_n=3)=\frac{\mathbb P_1(X_{n-1}=3)+\mathbb P_4(X_{n-1}=3)+\mathbb P_5(X_{n-1}=3)}{3}$$ So $$\lim_{n\to\infty}\mathbb P_5(X_n=3)=\lim_{n\to\infty}\frac{\mathbb P_1(X_n=3)+\mathbb P_4(X_n=3)}{2}=\frac{1}{5}.$$