Show that if $0Show that if $0<b<1$ it follows that
$$\lim_{n\to\infty}b^n=0$$
I have no idea how to express $N$ in terms of $\varepsilon$. I tried using logarithms but I don't see how to find $N$ from this.
 A: Set $b=\dfrac{1}{1+x}$ , $x >0.$
Note: $ (1+x)^n \gt 1+nx.$
$0<b^n =\dfrac{1}{(1+x)^n} \lt \dfrac{1}{1+nx}\lt$
$(1/x)\dfrac{1}{n}.$
Take the limit.
A: Welcome to MSE. What you want is, given some $\epsilon > 0$, $b^N \leq \epsilon$. Note that, by properties of the logarithm, and since both $b$ and $\epsilon$ are positive,
$$ b^N \leq \epsilon \Leftrightarrow N \ln(b) \leq ln(\epsilon).$$ 
I'll let you write the conclusion from here (there is a little thing to be careful of). 
A: From $0<b<1$, we get
$$0< (n+1)b^n < 1+b+\cdots+b^n = \frac{1-b^{n+1}}{1-b} < \frac{1}{1-b}.$$ 
Hence, 
$$0< b^n < \frac{1}{n+1} \frac{1}{1-b}.$$
And as $n$ goes to infinity, the limit of $b^n$ approaches to zero. 
A: By definition, if you choose $\varepsilon > 0$, then you can find a natural number $N$ such that $\lvert b^n\rvert  < \varepsilon$ when $n > N$. Since $0<b<1$, then you have $b^n < \varepsilon$, so $n > \log_b \varepsilon$, because $\log_b$ is a decreasing function. Therefore any natural number $N$ greater than $\log_b \varepsilon$ works, e.g. $N = \lfloor{\log_b \varepsilon \rfloor}+1.$
A: Recall that by the definition $\lim_{n\to\infty}a_n=L\in \mathbb{R}$ means that
$$\forall \epsilon>0 \quad \exists \bar n \quad \forall n>\bar n \quad |a_n-L|<\epsilon$$
In that case, note that for any $\epsilon>0$
$$|b^n-0|<\epsilon \iff b^n<\epsilon \iff n>\frac{ \log \epsilon}{\log b}$$
then by the definition $b^n\to 0$.
