Should we say $0.\bar9$ *can* equal $1$ also in the hyperreals?

Question

Consider the sequence $R_n$ of repunits, defined as $\displaystyle\frac{10^n-1}{9}$. We have $$\frac{R_{n+1}}{R_n}=\frac{9}{9} \frac{10^{n+1}-1}{10^n -1}=\frac{10^{n+1}-1}{10^n -1}=\frac{\overbrace{99\cdots99}^{n \ \text{nines}}}{\underbrace{99\cdots99}_{n-1 \ \text{nines}}}=\frac{100\cdots009-10}{\underbrace{99\cdots9}_{n-1 \ \text{nines}}}=10.\overline{00\cdots09},$$ (where the first sequence of zeros contains $n$ of them and the second $n-1$), and thus $$\frac{R_2}{R_1}=\frac{99}{9}=11\cdot\frac99=10.\bar9. $$Let $n.d_0d_1d_2\cdots;\cdots d_{H-1}d_{H}d_{H+1}\cdots$ be the decimal representation of a hyperreal number between $n$ and $n+1$. To the left of the semicolon one finds the standard decimal part, the non-standard one to the right of it. Since we do not want $\frac99<1$, isn't the equality above an argument for the representation $n.999\cdots;\cdots999\cdots$, perhaps more succintly expressable as $n.\overset{\infty}{\bar9}$, of $n+1$? Then again, how does that combine with the non-existence of $0.000\cdots;\cdots999\cdots$ (all the digits after the semicolon are nines)? Do we need to deny the existence of $0.999\cdots;\cdots000\cdots$?