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. :(

For fun,
I'll try to do
a proof by contradiction.
Suppose
$\lim_{n \to \infty} r^n
\ne 0
$.
Then,
since $S =(r^n)_{n=1}^{\infty}$
is a decreasing sequence
(since $0 < r < 1$),
$s = \lim S$
exists
and $s > 0$.
(Only $\lim\inf$
is needed
for what follows.)
Then,
for any $c > 0$,
there is an
$n(c)$
such that
$r^{n(c)}
< s+c
$.
Now,
consider
$r^{n(c)+1}$.
$r^{n(c)+1}
=r^{n(c)}r
<(s+c)r
$.
But this can be made
less than $s$
if
$(s+c)r
< s
$
or
$sr+cr < s$
or
$cr < s-sr = s(1-r)$
or
$c < \frac{s(1-r)}{r}$.
But,
by assumption,
$r^{n(c)+1}
\ge s
$.
Therefore
the assumption that
$\lim r^n > 0$
leads to a contradiction,
so
$\lim r^n = 0$.
