Prove $\text{lim} \ b^n=0$ for $0 < \vert b \vert< 1$ My work starts with a supposition of $N$, so that for $n > N$ we have $\vert b \vert ^n < \epsilon$.
Since $0 < \vert b \vert < 1$, we see the logarithm with base $\vert b \vert$ is a decrescent function meaning it will invert the inequality once taken.
$$\vert b \vert ^n < \epsilon $$
$$n > \text{log}_{\vert b \vert}\epsilon $$
Done the scrapwork, lets start the formal proof.
Let $\epsilon > 0$. Let $N = \text{log}_{\vert b \vert}\epsilon$. If $n > N$ and $0 < \vert b \vert< 1$, then
$$\vert b \vert ^n < \vert b \vert ^{\text{log}_{\vert b \vert}\epsilon} = \epsilon$$
Hence by definition, $\text{lim} \ b^n=0$ 
Have I done everything correctly? I am using in an implicit manner that $\vert b^n \vert = \vert b \vert ^n$, is everything fine with this? (Something is really pinching me up!)
 A: If $0<|b|<1$ then $|b|=\frac{1}{1+A}$ for some $A>0$.
But $(1+A)^n\ge 1+nA$ for each $n\in \Bbb N$, so you have $$|b|^n=\frac{1}{(1+A)^n}\le\frac{1}{1+nA}$$
Now taking $n$ large enough you are going to reach $\frac{1}{1+nA}<\varepsilon$, for each $\varepsilon>0$. Hence $|b|^n$ too.
A: My answer can be over killing, but it has different view point.
If $0<|b|<1$, note that
$\sum_{n=0}^{\infty}|b|^n = \frac{1}{1-|b|} <\infty$.
Thus, $\lim_{n\rightarrow \infty}|b|^n=0$ so
$|\lim_{n\rightarrow \infty} b^n|\leq |\lim_{n\rightarrow \infty}|b|^n| =0$ $\implies \lim_{n\rightarrow \infty}b^n=0$.
A: You can make janmarqz's result
more explicit.
Since
$|b^n| = |b|^n$,
we can assume that
$0 < b < 1$.
Then
$b = \dfrac1{1+a}$
with $a > 0$,
so that,
from Bernoulli's inequality,
$(1+a)^n \ge 1+an
$.
Therefore,
since
$a = \dfrac1{b}-1$,
$\begin{array}\\
b^n
&=\dfrac1{(1+a)^n}\\
&\le \dfrac1{1+na}\\
&\lt \dfrac1{na}\\
&=\dfrac1{n(\frac1{b}-1)}\\
\end{array}
$
To make
$b^n < \epsilon$,
it is enough if
$\epsilon > \dfrac1{n(\frac1{b}-1)}$
or
$n > \dfrac1{\epsilon(\frac1{b}-1)}$.
This is,
of course,
far worse than using the log,
but it is 
completely elementary.
This is also not original with me.
I first saw this in
"What is Mathematics?"
by Courant and Robbins,
which I highly recommend
and is available in paperback
for less than $20.
