What is ⌊0.9 recurring ⌋? For a ceiling and floor function, the number is taken to 0 decimal places. Does this process mean that 0.9 recurring inside a floor function would go to 0? Or would the mathematician take 0.9 recurring to be equal to 1, thus making the answer 1?

And if 0.9 recurring does equal 1, does that mean (by definition) that ⌊1⌋ = 0?
$$\lim_{n\to\infty}\left\lfloor\sum_{k=1}^n\frac9{10^k}\right\rfloor=0\\
\left\lfloor\lim_{n\to\infty}\sum_{k=1}^n\frac9{10^k}\right\rfloor=1\\$$
Because $0.\bar{9} = 0.999999$ (recurring) is equal to $1$, the floor function of it is equal to the floor of $1$, and that is $1$.