Prove that $|x|<1$ implies that $\lim_{n \to \infty} x^n=0$ Show that $|x|<1$ implies that $\lim_{n \to \infty}x^n$ = 0. 
Can anyone help me define my $N$ (as used in the formal definition of a limit) in terms of $\varepsilon$? I define $x$ as $\dfrac{1}{1+p}$ where $p>0$ but I am not sure how also to define $N$ since both $P$ and $\varepsilon$ are variable. 
 A: Assuming you mean to show $\lim\limits_{n\to\infty} x^n=0$ here is a hint:
Without loss of generality assume $x\neq 0$, then there exists $h>0$ with $\dfrac{1}{|x|}=1+h$. Applying Bernoulli's inequality yields: $$\left|\frac{1}{x^n}\right|=\left(\frac{1}{|x|}\right)^n\geq 1+nh = 1+n\left(\frac{1}{|x|}-1\right)>n\cdot \frac{1-|x|}{|x|}.$$ Thus we have $$0\leq |x^n|<\frac{|x|}{1-|x|}\cdot \frac 1n$$
Now you should easily find such $N$ and $\varepsilon$ to use in the formal definition of a limit to prove $\lim\limits_{n\to\infty} x^n=0$.
Edit: although using the formal definition is not necessary as this would be a nice opportunity to apply the sandwich theorem.
A: Given your expression $\frac {1}{1+p}$ , consider $\frac {1}{(1+p)^N}$. We have that $1$ is fixed, and you can show that for fixed $p$ , $(1+p)^N$ grows without bound: $(1+p)^N = e^{Nln(1+p)}$. Let $N= \frac {K}{ln(1+p)}$. Then $(1+p)^n=e^{ \frac {K}{ln(1+p)}}(ln(1+p))=e^K$ goes to $\infty$ as
 $K \rightarrow \infty$, so that $\frac {1}{(1+p)^n} \rightarrow 0$
A: For fixed $p>0$ 
$$\forall \epsilon >0  \space \exists N >0 :n>N \Rightarrow  |x^n-0| <\epsilon \\ $$ 
$x=\frac{1}{1+p}$
$$\forall \epsilon >0  \space \exists N>0:n>N  \Rightarrow  |(\frac{1}{1+p})^n-0| <\epsilon \\ |(\frac{1}{1+p})^n-0| <\epsilon \to \\(\frac{1}{1+p})^n <\epsilon\\ (\frac{1}{1+p})^n <\frac{1}{1+np+\frac{n(n-1)}{2}p^2+...} <\frac{1}{1+np}<\epsilon\\$$ so 
$$\frac{1}{1+np}<\epsilon\\1+np> \frac{1}{\epsilon} \\ \to n>\frac{1}{p\epsilon}-\frac{1}{p} \\ \to N >n \geq \left \lfloor  \frac{1}{p\epsilon}-\frac{1}{p}\right \rfloor+1$$
for every $p$ such as $p_0$  you can say $N >n \geq \left \lfloor  \frac{1}{p_0\epsilon}-\frac{1}{p_0}\right \rfloor+1$
A: I am not sure why you need to use $x=1/(1+p)$. Using the definition of limit, $x^n\to 0$ for $n\to\infty$ iff $\forall \varepsilon>0, \exists N$ such that $n>N\Rightarrow |x^n - 0|<\epsilon$. Just use the definition:
$$
|x|^n<\varepsilon\quad\Rightarrow\quad \frac{1}{|x|^n}>\frac{1}{\varepsilon}
$$
Notice that $1/|x| >1$. Therefore, applying the increasing function $log_{\frac{1}{|x|}}$ to both sides, we get
$$
n>log_{\frac{1}{|x|}}\varepsilon^{-1} = \dfrac{\ln\varepsilon^{-1}}{\ln|x|^{-1}} = \dfrac{\ln\varepsilon}{\ln|x|}.
$$
Hence, if you set $N=ceil(-\ln(\varepsilon)/ln(|x|))$, you have $|x|^n<\varepsilon$, where $ceil(x)$ is the smallest integer greater than $x$.
A: I want to try an use only the definition of convergence and the ordered field properties of real numbers. Ignore the trivial $x=0$ case. Since $|x|<1$ we have that $\frac{1}{|x|}>1$. We know that we can find a $\delta\in(0,\infty)$ such that $\frac{1}{|x|}=1+\delta$ i.e. that $\frac{1}{|x|^n}=(1+\delta)^n$.
Because $\delta>0$, for any $K\in\Bbb N$ we can find an $M\in \Bbb N$ such that $M\delta>K$. So we have that 
$$\frac{1}{|x|^n}=(1+\delta)^n>1+\delta^n+n\delta>n\delta>K$$
as long as $n\geq M$ (prove this by induction). Rearranging we get that $\frac{1}{K}>|x|^n$. Now we have just shown that for any $K\in \Bbb N$ we can find an $M\in \Bbb N$ such that $\forall n\geq M$, $|x|^n<\frac{1}{K}$. Now for any $\epsilon>0$ there is a natural number $K$ such that $\frac{1}{K}<\epsilon$. Thus $|x|^n\rightarrow0$ as $n\rightarrow\infty$.
A: This is another approach. If $x = 0$ then clearly the limit is $0$. Let us first consider $x$ to be positive so that $0 < x < 1$. Then clearly as $n$ increases the sequence $x^{n}$ decreases and is bounded below (by $0$). Hence $L = \lim\limits_{n \to \infty}x^{n}$ exists. Further we have $$L = \lim_{n \to \infty}x^{n + 1} = \lim_{n \to \infty}x\cdot x^{n} = x\lim_{n \to \infty}x^{n} = xL$$ or $(1 - x)L = 0$. Since $(1 - x) \neq 0$ it follows that $L = 0$. Hence for $0 < x < 1$ we have $$\lim_{n \to \infty}x^{n} = 0$$ If $-1 < x < 0$ then we put $x = -y$ so that $0 < y < 1$ and $x^{n} = (-1)^{n}y^{n}$. Since $y^{n} \to 0$ it follows that $x^{n} \to 0$ as $n \to \infty$.
