**Lemma** 1. If $|x|<1$ then as $n \to \infty$, $x^{n} \to 0$. 2. If $x>1$ then $x^{n} \to \infty$ as $n \to \infty$. **Proof** 1. WLOG suppose $0<x<1$. We write $x = \frac{1}{1+h}$, so $h=x^{-1}-1>0$. Then using [[Bernoulli's Inequality]], $x^{n}= \left( \frac{1}{1+h} \right)^{n} \leq \frac{1}{ 1+nh }$ Given $\epsilon>0$, by [[Archimedean Property of Real Numbers|AP]] $\exists N \in \mathbb{N} \left( N> \frac{1}{h}\left( \frac{1}{\epsilon} -1 \right) \right)$ so $\frac{1}{1+Nh}<\epsilon$ then for any $n \geq N$ we have $|x^{n}-0| \leq \frac{1}{1+nh} \leq \frac{1}{1+ Nh} < \epsilon. \quad \square$ 2. If $x>1$. Write $x=1+h$. Then $x^{n}= (1+h)^{n} \geq 1+nh $Given $R \in \mathbb{R}$, pick $N$ such that $1+Nh>R$. Then for $n \geq N$, $x^{n}\geq 1+nh \geq 1+Nh >r.$