Limit of the geometric sequence $\{r^n\}$, with $|r| < 1$, is $0$? 
Prove that the $\lim_{n\to \infty} r^n = 0$ for $|r|\lt 1$.

I can't think of a sequence to compare this to that'll work.  L'Hopital's rule doesn't apply.  I know there's some simple way of doing this, but it just isn't coming to me. :(

If $r=0$ it's trivial, so we can skip this case. Now assume $r\neq0$. 
$$\lim_{n\to\infty}r^n=0 \Longleftrightarrow(\forall \epsilon \in \mathbb{R}^+)(\exists \delta\in\mathbb{R})(\forall n_{> \delta})(|r^n| < \epsilon)$$
Now we can note, that both sides of above inequality are positive. We can use logarithms.
$$|r^n|<\epsilon \overset{|r|<1}{\Longleftrightarrow} n > \log_{|r|}\epsilon $$
Because $\log_{|r|}\epsilon$ is const, $\delta = \log_{|r|}\epsilon$ satisfy thesis.
$\mathscr{Q.E.D.}$
