How do we prove that $\lfloor0.999\cdots\rfloor = \lfloor 1 \rfloor$? Are the floor functions of $0.999\cdots$ and 1 equal?
It is true that $0.999\cdots=1$ but how does one justifies the integer part of $0.999\cdots$ being 1 , where it is not, or alternatively without using $0.999\cdots=1$ how can we show that  $\lfloor0.999\cdots\rfloor = \lfloor 1 \rfloor$ ?
 A: The floor function is a discontinuous function. This means that limits cannot be interchanged before and after the action of the function, i.e. $f(\lim\limits_{n\to\infty}x_n)\ne \lim\limits_{n\to\infty} f(x_n)$ generally. In particular,
$$\lim_{n\to\infty}\left\lfloor 0.\underbrace{99\cdots9}_n \right\rfloor=\lim_{n\to\infty}0=0 $$
$\hskip 3.2in$ but
$$\left\lfloor \lim_{n\to\infty} 0.\underbrace{99\cdots9}_n \right\rfloor=\lfloor1\rfloor=1. $$
Now the expression $\lfloor0.999\dots\rfloor$ denotes the latter, which is $1$, but intuitively and at face value one notices that the floor function evaluated at the partial sums always returns $0$, so we might be tempted to accept the first formula above as the real answer, but appearance $\ne$ reality in general.
A: The floor function $\lfloor x \rfloor$ is càdlàg (continue à droite, limite à gauche, i.e. right continuous with left limits).
Since there is no point between $0.999\ldots$ and $1$, and the real numbers are continuous, this right continuity implies  $\lfloor0.999\cdots\rfloor = \lfloor 1 \rfloor$. 
(Since the identity function is also càdlàg, you can use a similar argument to show $0.999\cdots =  1$.) 
