0.999.../Proof by limit manipulation

Assumptions

• Definition from sequences
• Term-by-term operations on sequences
• The limit of a geometric sequence

Proof

$0.999\dots =\underset{n\to \mathrm{\infty }}{\lim }0.\underset{n}{\underbrace{99\dots 9}}=\underset{n\to \mathrm{\infty }}{\lim }\sum _{k=1}^n\frac{9}{10^k}=\underset{n\to \mathrm{\infty }}{\lim }(1-\frac{1}{10^n})=1-\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{10^n}=1.$