Sequence of Powers of Number less than One/Necessary Condition/Proof 3

Proof
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.

By the Archimedean Principle, there exists a natural number $M$ such that:
 * $M > \dfrac 1 {\paren {1 - \size x} \epsilon}$

By the Well-Ordering Principle, there exists a smallest natural number $m$ such that:
 * $\exists N \in \N: m > M \size x^N$

Note that:
 * $m - 1 \le M \size x^{N + 1}$

By elementary algebra, it follows that:
 * $1 - \dfrac 1 m = \dfrac {m - 1} m \le \size x < 1 - \dfrac 1 {M \epsilon}$

Hence, $m < M \epsilon$.

Therefore:
 * $\size x^N < \epsilon$

By Absolute Value of Product:
 * $\forall n \in \N: n \ge N \implies \size {x^n} = \size x^n \le \size x^N < \epsilon$

Hence the result, by the definition of a limit.