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 $|x|=\dfrac{1}{1+t}$. 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 $\dfrac{1}{nt}\lt \epsilon$.
