I. solution. In the steps of the ox, it alternately changes its first and second coordinates. With its odd-numbered steps, it modifies its first coordinate, always in the opposite direction and a quarter of the magnitude of the previous one. Therefore, the modification of the first coordinate at $k$, that is, after the $(2 k-1)$-th step, will be

$$x_{2 k-1}=1-\frac{1}{4}+\ldots+\left(-\frac{1}{4}\right)^{k-1}=\frac{1-\left(-\frac{1}{4}\right)^{k}}{1-\left(-\frac{1}{4}\right)}$$

and its value remains the same after the $2 k$-th step as well.

The second coordinate changes similarly, and its value after the $2 k$-th step is

$$y_{2 k}=\frac{1}{2}-\frac{1}{8}+\ldots+\frac{1}{2}\left(-\frac{1}{4}\right)^{k-1}=\frac{1}{2} x_{2 k-1}$$

and its value remains the same after the $(2 k+1)$-th step as well. Since the powers of 4 are different integers, the $k$-th one is greater than the $k$-th positive integer, that is, $4^{k}>k$. Therefore, $4^{k}$ grows beyond any bound as $k$ runs through the positive integers, and $4^{-k}$ approaches 0. Thus, the limit of $x_{n}$ is $4 / 5$, and the limit of $y_{n}$ is half of that, $2 / 5$.

Let $B_{n}$ denote the point $\left(x_{n} ; y_{n}\right)$ and $B$ the point $\left(\frac{4}{5} ; \frac{2}{5}\right)$. Since the distance between $B_{n}$ and $B$ is less than the sum $\left|x-x_{n}\right|+\left|y-y_{n}\right|$, and this sum approaches 0, the ox approaches the point $B\left(\frac{4}{5} ; \frac{2}{5}\right)$ more and more closely.

II. solution. Let $B_{n}$ still denote the position of the ox after the $n$-th step, and $B_{0}$ the origin.

!

The triangles $B_{n} B_{n+1} B_{n+2}$ for $n=0,1,2, \ldots$ are all similar, as they are right-angled and the ratio of the legs $B_{n} B_{n+1}$ and $B_{n+1} B_{n+2}$ is $2: 1$ regardless of $n$. Specifically, in the triangles $B_{0} B_{1} B_{2}$ and $B_{2} B_{3} B_{4}$, the angles at $B_{0}$ and $B_{2}$ are equal, and their sides are oppositely directed. Therefore, the hypotenuses of these triangles lie on the same line, and $B_{4}$ lies on the segment $B_{0} B_{2}$. Similarly, we find that $B_{5}$ lies on the segment $B_{1} B_{3}$, or the segment $B_{4} B_{5}$ can be obtained by a $16: 1$ reduction of the segment $B_{0} B_{1}$ from the point $B$. From this point onward, the entire path can be continued by applying this reduction to the broken line $B_{0} B_{1} B_{2} B_{3} B_{4}$, then to the result, and so on. In other words, if we apply a $16^{k}: 1$ reduction to the broken line $T_{0}=B_{0} B_{1} B_{2} B_{3} B_{4}$ from the center $B$, we get exactly the broken line $T_{k}=$ $B_{4 k} B_{4 k+1} B_{4 k+2} B_{4 k+3} B_{4 k+4}$. Since among the points of the broken line $T_{0}$, $B_{0}$ is the farthest from $B$, we get that the points of $T_{k}$ are at most $B_{0} B / 16^{k}$ away from $B$. Since this approaches 0 (which we can also verify as in the first solution, that $1 / 4^{k}$ approaches 0, although this can also be read directly from the latter statement), it follows that the ox gets closer and closer to the point $B$.

The equation of the line $B_{0} B_{2}$ is $y=x / 2$, and that of $B_{1} B_{3}$ is $y=2(1-x)$, so at their intersection, $x=4 / 5, y=2 / 5$, thus the coordinates of $B$ are $\left(\frac{4}{5} ; \frac{2}{5}\right)$.