>>11523262limits are only viable without infinity
or rather, convergence is.
divergence will remain expanding regardless of anything.
but if there an implicit notion of infinite allowable terms between any two integers, then limits of convergence are complete nonsense.
put it another way:
A: [ 0 , 1 , 2 , 3 , 4 , ... ] -> ?
mapped to
B: [ 0 . 9 , 9 , 9 , 9 , ... ] -> 1
has the same meaning.
A cannot reach infinity, much less get close to it.
B cannot reach 1, much less get close to it.
because infinity is insurmountable.
if there were a finite function there rather than infinite, then you could approach.
For example, with a limit of 7 decimals.
[ 0 . 9 0 0 0 0 0 0 ]
[ 0 . 9 9 0 0 0 0 0 ]
[ 0 . 9 9 9 0 0 0 0 ]
[ 0 . 9 9 9 9 0 0 0 ]
[ 0 . 9 9 9 9 9 0 0 ]
[ 0 . 9 9 9 9 9 9 0 ]
[ 0 . 9 9 9 9 9 9 9 ]
if you doubled that up to 14, at this most recent step we'd at least be halfway there. 14/2 = 7
if you expand it to ? though, well, what's ?/2?
?/2 doesn't exist.
can't divide infinity.
so you can't show that any n amount of 9's are "approaching" infinite 9's.
