What is ⌊0.9 recurring ⌋? [duplicate]

$$\lim_{n\to\infty}\left\lfloor\sum_{k=1}^n\frac9{10^k}\right\rfloor=0\\ \left\lfloor\lim_{n\to\infty}\sum_{k=1}^n\frac9{10^k}\right\rfloor=1\\$$

You have to look at the $.9$ recurring as a sum... then you'll know the answer.

$$\bar{.9} = \sum_{i=1}^{\infty} \frac{9}{10^{i}}$$

So, $$\lfloor \bar{.9}\rfloor = \left\lfloor \sum_{i=1}^{\infty}\frac{9}{10^{i}}\right\rfloor=\lfloor 1 \rfloor = 1.$$ You cannot split up the floor function over a sum, i.e. $\lfloor a+b\rfloor \neq \lfloor a\rfloor + \lfloor b\rfloor$.

As Gregory Grant says, $\lfloor0.9\overline{9}\rfloor = 1$; your phenomenon illustrates the jump discontinuity in the floor function at each integer.

$\lfloor x\rfloor$ is defined as the unique integer $n$ such that $n\leq x<n+1$. Because $0.999...=1$ we have $1\leq 0.999...$ and obviously $0.999...<2$, so $\lfloor 0.999...\rfloor=1$.

The question embeds an alternate question of what is the division rule that can produce $0.\bar{9} = 0.999999$ (recurring)?

A normal (long) division rule produces an integer (complete division) and a remainder (aka "floor"). The number $0.\bar{9} = 0.999999$ (recurring) does not exist as the result of a classic division, hence the confusion.