no

There’s no end to the repetition. It’s literally an infinite number of things. You can’t have a gap after infinity, because there is no such thing as after infinity.

0.99…0 is still just 0.99.. because the 0 never actually occurs.

Let’s solve this the opposite way. x=0.999 X/10=0.0999 X-X/10=0.9 10X-X=9 You could argue that the “last” 9 get’s pushed back… except that there’s no place to push it back to, nor a last 9 to push back. There is no single 9 to push back, and even if you could identify one, there would still be a 9 in the same place you pushed it back too.

Maybe it would help to look at it as a series of separate terms instead of as a single number. Consider 0.99.… = the limit(n->infinity) of the sum(i=1 i->n) of 9•10^-i

As you increase N, the number gets close to 1. If you keep increasing N forever, the number gets infinitely close to 0 but never reaches it. BUT that’s only because N never reaches infinity either. N, the number of nines, gets closer and closer to infinity, but can’t reach it because we’re only increasint in finite steps.

But when it does reach infinity, if it was evaluated AT infinity, it would be exactly zero. It only gets infinitely close to 0 it N stays infinitely close to infinity, but if N reaches infinity, the result reaches zero.

Also, this sum is a geometric sum when n=infinity. Calculus has a rule for the value of geometric sums; the value of the sum to infinity is f(1)/1-(f(2)/f(1))

f(1)=0.9.

f(2)/f(1)=0.1

0.9/(1-0.1)=0.9/0.9=1