As the limit of $n$ goes to infinity, prove that $x^n = 0$ if $\operatorname{abs}(x)<1$. 
As the limit of $n$ goes to infinity, prove that $x^n = 0$ if $\operatorname{abs}(x)<1$.

So I want to prove it this by observing that $\operatorname{abs}(x) < 1$ which means $1/(\operatorname{abs}(x)) > 1$.  How would I proceed from here?

So you want to focus on $1/|x|$ being larger than $1$. Let $a=1/|x|$.  Because $a>1$, we have $a=1+h$ with $h>0$.  There exists a positive integer $k$ such that $kh>1$. By Bernoulli's inequality, $a^k=(1+h)^k\geq 1+hk>2$.  Then for each positive integer $m$, $a^{km}>2^m>m$.  This implies that $\lim\limits_{m\to\infty}a^{km}=+\infty$, which implies that $\lim\limits_{n\to\infty}a^n=+\infty$ because the sequence is monotone increasing. This in turn implies that $\lim\limits_{n\to\infty}\dfrac{1}{a^n}=\lim\limits_{n\to\infty}|x|^n = 0$.
