TITLE: Epsilon-Delta Proof for $x^n$ tends to 0
QUESTION [6 upvotes]: What is the epsilon proof that $x^n \rightarrow 0$ as $n \rightarrow \infty$ provided $|x| < 1 $? I only know it's true because I know the geometric series converges, which implies its terms must tend to 0, but never seen an epsilon proof of this simple fact.

REPLY [3 votes]: Let us assume the limit of the sequence $\{x_n\}$ is $L$. Here $x_n=x^n$. We can easily show that limit exists by using Ratio Test (given $|x|<1$) 
Since $L$ is the limit, we can always find an $N \in \mathbb{N}$ for every $\epsilon > 0\,$ s.t.
$$|x_n-L|<\epsilon \qquad \forall n\ge N$$
$\Rightarrow$$L-\epsilon <x^n<L+\epsilon$  and $L-\epsilon <x^{n+1}<L+\epsilon$
Also the above equaion imlies $(L-\epsilon)x <x^{n+1}<(L+\epsilon)x$
$\Rightarrow$ $L-\epsilon < (L+\epsilon)x$
$\Rightarrow$ $L<\epsilon \frac{1+x}{1-x}$
Now since $\epsilon$ can be arbitrarily small, you can easily show that $L=0$