0.9 = 1

$\dfrac {1}{9} = 0.11111.........$

$1 =9\times (0.11111......) = 0.9999.......$

Hence, $0.9999.......=1$

$ANSWER:\boxed {TRUE}$

If $x=0.9999...$ ,

$10x=9.9999...$

Subtracting the two equations, $9x=9$ , $x=1$

1 ÷ 3 = 0.333 (recurring). 0.333 (recurring) x 3 = 1 or 0.999 (recurring). Can be proved other ways too.

$\begin{aligned} 0.999\ldots &= \frac{9}{10} + \frac{9}{100} + \cdots \\ &= \sum_{i=1}^\infty \frac{9}{10^i} \\ &= \frac{9}{10}\sum_{i=0}^\infty \frac{1}{10^i} && \color{#3D99F6}\text{The sum converges since }-1<\frac{1}{10}<1 \\ &= \frac{9}{10} \times \frac{1}{1-\frac{1}{10}} \\ &= \frac{9}{10} \times \frac{10}{9} = \boxed{1} \end{aligned}$

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$\begin{aligned} 0.999\ldots &= \lim_{n\to\infty} \frac{\overbrace{99\cdots9}^{n\text{ }9\text{s}}}{1\underbrace{00\cdots0}_{n\text{ }0\text{s}}} \\ &= \lim_{n\to\infty} \frac{10^n-1}{10^n} \\ &= \lim_{n\to\infty} \left(1-\frac{1}{10^n}\right) && \color{#3D99F6} \lim_{n\to\infty}\frac{1}{10^n} = 0 \\ &= 1 - 0 = \boxed{1} \end{aligned}$