Reversing the digits of an infinite decimal

Question

Let $x$ be a real number in $[0,1)$, with decimal expansion $$ x = 0.d_1 d_2 d_3 \cdots d_i \cdots \;. $$ If the decimal expansion is finite, ending at $d_i$, then extend with zeros: $d_k = 0$ for all $k > i$. Define a sequence $x_k^R$ by digit reversals, as follows: \begin{eqnarray} x_1^R & = & 0.d_1 \\ x_2^R & = & 0.d_2 d_1 \\ x_3^R & = & 0.d_3 d_2 d_1 \\ x_4^R & = & 0.d_4 d_3 d_2 d_1 \\ & \cdots &\\ x_k^R & = & 0.d_k d_{k-1} \cdots d_3 d_2 d_1\\ & \cdots & \end{eqnarray} Finally, define $x^R = \lim_{k\to\infty} x_k^R$, when that limit exists.

Q. For which $x$ does the limit exist? In particular, must $x$ be rational for the limit $x^R$ to exist? If not, what are some irrationals with limits?

If the decimal expansion of $x$ is finite, then the extension by zeros leads to $\lim_{k\to\infty} x_k^R = 0$.

Answer 1

Suppose that $\lim_{k \to \infty} x^R_k = x^R$ exists. Then, for $k$ large enough, the first digit of $x^R_k$ is equal to the first digit of $x^R$, say $a$.

Hence, $d_k = d_{k+1} = \dots = a$ and $x^R = 0.aaaa \dots = a/9$.

So, the number $x$ must have a decimal expansion where eventually only the digit $a$ appears.

Answer 2

The limit exists precisely when the sequence of digits is eventually constant. If it is eventually constantly $d$, the limit is $\frac{d}9$.