TITLE: Reversing the digits of an infinite decimal
QUESTION [11 upvotes]: 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 $\lim_{k\to\infty} x_k^R = 0$.

REPLY [15 votes]: The limit exists precisely when the sequence of digits is eventually constant. If it is eventually constantly $d$, the limit is $\frac{d}9$.

REPLY [14 votes]: 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.