Sequence of Powers of Number less than One/Necessary Condition/Proof 4

Proof
Define:
 * $\displaystyle L = \inf_{n \mathop \in \N} \left\vert{x}\right\vert^n$

By the Continuum Property, such an $L$ exists in $\R$.

Clearly, $L \ge 0$.

Suppose that $L > 0$.

Then, by the definition of the infimum, we can choose $n \in \N$ such that $\left\vert{x}\right\vert^n < L \left\vert {x}\right\vert^{-1}$.

But then $\left\vert{x}\right\vert^{n+1} < L$, which contradicts the definition of $L$.

Therefore, $L = 0$.

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

By the definition of the infimum, there exists an $N \in \N$ such that $\left\vert{x}\right\vert^N < \epsilon$.

It follows that:
 * $\forall n \in \N: n \ge N \implies \left\vert{x^n}\right\vert = \left\vert{x}\right\vert^n \le \left\vert{x}\right\vert^N < \epsilon$

where either Absolute Value Function is Completely Multiplicative is applied.

Hence the result, by the definition of a limit.