Two limit-based proofs that 0.999... = 1:

=====First=====

0.999... is clearly 1 minus the limit of 10^-n as n grows without bound.

|10| > 1.

The limit of any z^-n as n grows without bound where |z| > 1 is by definition 0.

Therefore, 0.999... is 1 - 0, which is 1.

=====Second=====

0.999... is clearly 9 times the limit of the series 10^-1 + ... + 10^-n as n grows without bound.

Let g = 10^-1 + ... + 10^-n

Then g = 10*(10^-2 + ... + 10^-n) + 10^-n

Then g = 10(g - 10^-1) + 10^-n

Then g = 10g - 1 + 10^-n

Then -9g = -1 + 10^-n

Then 9g = 1 - 10^-n

Then g = (1 - 10^-n)/9

Therefore, 0.999... is 9 times the limit of (1 - 10^-n)/9 as n grows without bound.

|10| > 1.

The limit of any z^-n as n grows without bound, where |z| > 1, is by definition 0.

Therefore, 0.999... is 9 times 1/9.

Therefore, 0.999... = 1.