Real analysis question about limits? If $0  < x < 1$, show that $x^n → 0$ as $n → ∞$ .
I'm thinking it has something to do with rational numbers (declaring $x = p/q$) and then using exponent laws to show that $x^n = p^n/q^n$, and since $q > p$ ... but I'm not sure where to go with this.  Any help is appreciated.  Thanks
 A: The sequence is monotone decreasing, since $x < 1$ and hence $$x^{n+1} = x \cdot x^{n} < x^n$$
and it is bounded below by $0$, since $x > 0$ and hence $$\underbrace{x \cdot x \cdots x}_{\text{ $n$ times }} > 0$$ for every $n$. 
Therefore, we know that the sequence converges to some limit $0 \le \gamma \le x$. 
We must have  $$\gamma = \lim_{n\rightarrow\infty} x^{n} = \lim_{n\rightarrow\infty} x^{n+1} = x \lim_{n\rightarrow\infty} x^n = x \gamma$$
And since $0 < x < 1$ we know that $\gamma = 0$. 
A: Note that since $x \in (0,1)$, we have $x=\dfrac1{1+M}$, where $M>0$. By binomial theorem, we now have
$$x^n = \dfrac1{(1+M)^n} < \dfrac1{1+nM}$$
Use this to conclude what you want.
A: Hints:
$$0<x<1\implies x>x^2>x^3>\ldots >x^n\;,\;\;\forall\,n\in\Bbb N$$
$$a_n>0\;\;\forall\,n\in\Bbb N\implies \lim_{n\to\infty}a_n=\infty\iff \lim_{n\to\infty}\frac1{a_n}=0$$
A: One way is to use the natural log : you want to show, for all $\epsilon>0$ there is an $N>0$ with $x^n < \epsilon$ for all $n>N$ . Then, from: $$ x^n <\epsilon$$, apply $ln$ on both sides (using the fact that $ln$ is monotone increasing, so that if $0<a<b$ , then $lna<lnb$), so that $x^n <\epsilon$ then $lnx^n <ln\epsilon$....
EDIT: Maybe a more fundamental fact is this: a monotone bounded sequence converges to its (least upper, or greatest-lower ) bound. We can show $x^n$ is monotone decreasing in $(0,1)$ ( as a function of $n$ , i.e., for fixed $x$): $x^{n+1}-x^n=x^n(x-1)$, which is negative in $(0,1)$ (since $x^n$ is positive in $(0,1)$ ).Using that $x^n>0$ in $(0,1)$  we have that $x^n$ is monotone and bounded, so it must converge to its least upper bound. Then you can use the $ln$ to show this GLB is $0$ 
