$A^{\infty}$

Let $A$ be the matrix in the question.

Since the column totals of $A$ are all equal to $1$ , the quantity $x(k)+y(k)+z(k)$ is constant, and from the given values when $k=0$ , is always $7$ .

Also, all entries of $A$ , and the given initial values, are positive; so $x(k),y(k),z(k)$ are always positive.

These two facts mean everything is well enough behaved in the limit.

Say that, in the limit, the vector tends to $\bold{v}$ . Then $A\bold{v}=\bold{v}$ so $\bold{v}$ is an eigenvector of $A$ corresponding to the eigenvalue $1$ .

This eigenvector is parallel to $8,5,4)^T$ , and (from above) its components sum to $7$ .

Scaling, we find $\lim_{k\to \infty} x(k)=\frac{8}{8+5+4}\times7=\frac{56}{17}\approx \boxed{3.294}$