= Solution
The self-loop at zero makes the irreducible chain <aperiodic Markov chain>[aperiodic]. The convergence theorem for irreducible, aperiodic, positive recurrent countable-state chains gives
$$
\mathbb P(X_n=i)\longrightarrow\pi_i.
$$
Because $X$ and $Y$ are <independent random variables>[independent],
$$
\mathbb P(X_n=0,Y_n=1)
=\mathbb P(X_n=0)\mathbb P(Y_n=1)
\longrightarrow\pi_0\pi_1.
$$
Therefore
$$
\boxed{\lim_{n\to\infty}\mathbb P(X_n=0,Y_n=1)
=\frac12\cdot\frac14=\frac18}.
$$
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