Epsilon-Delta Proof for $x^n$ tends to 0 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. 

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 $L=0$
