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. 
 A: Let $|x|=\dfrac{1}{1+t}$. Since $|x|\lt 1$, we have $t\gt 0$.
By the Binomial Theorem, $(1+t)^n \ge 1+nt\gt nt$. Now it is easy to find $N$ such that if $n \gt N$, then $\dfrac{1}{nt}\lt \epsilon$.
A: 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 $L=0$
A: You don't need the binomial theorem.
$(1+t)^n \ge 1+nt$
is Bernoulli's inequality, which is easily proved by induction on $n$.
This proof is suitable for an introductory algebra class,
once induction has been presented.
I first saw this proof in "What is Mathematics?" by Courant and Robbins -
a wonderful book for learning math.
