Trouble proving $\lim_{n\to\infty} x^n=0$ when $|x|<1$.

Question

Intuitively speaking this is very clear, however when doing a formal proof I get stuck; this is what I have so far.

Let $\epsilon>0$, now, if $x<\epsilon$ the result follows immediately since $x^n$ is decreasing.

So the non trivial case comes when $x\in(\epsilon,1)$, for this it is clear, intuitively, that there should exist $N\in\mathbb{N}$ such that we get $x^N<\epsilon$. This last part is where I'm having trouble with, don't know how to argue that this $N$ indeed exists.

I've tried also doing this by contradiction, this would mean that for every $N\in\mathbb{N}$ don't get $x^N<\epsilon$ i.e. $x^N\geq\epsilon$, and I get stuck once again.

Any insights or hints would be appreciated.

Answer 1

Firstly, we go to show that $\lim_{n\rightarrow\infty}x^{n}=0$ if $x\in(0,1)$. Let $x\in(0,1)$. Define $y=\frac{1}{x}-1$. Note that $y>0$. By Binomial Theorem, \begin{eqnarray*} & & (1+y)^{n}\\ & = & 1+ny+\frac{n(n-1)}{2!}y^{2}+\cdots\\ & \geq & ny. \end{eqnarray*} Therefore, \begin{eqnarray*} & & x^{n}\\ & = & \frac{1}{(1+y)^{n}}\\ & \leq & \frac{1}{ny}. \end{eqnarray*} That is, $0\leq x^{n}\leq\frac{1}{ny}$. Since $\frac{1}{ny}\rightarrow0$ as $n\rightarrow\infty$, by sandwich rule, $\lim_{n\rightarrow\infty}x^{n}=0$.

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If $x\in(-1,0)$, we have that $-|x|^{n}\leq x^{n}\leq|x|^{n}$. Since $\lim_{n\rightarrow\infty}|x|^{n}=0$ by the above result, we have $\lim_{n\rightarrow\infty}-|x|^{n}=-\lim_{n\rightarrow\infty}|x|^{n}=0$ too. By sandwich rule again, $\lim_{n\rightarrow\infty}x^{n}=0$.

Answer 2

If $x>1$ then you can write it as $1+y$ with $y>0.$ Next you observe that $$(1+y)^N > 1 + N\cdot y$$ for every $N>1;$ this last formula could either be proved by induction or by using Newton binomial formula. In conclusion we have that $$ lim \ x^N = lim \ (1+y)^N = lim \ 1+N\cdot y = +\infty$$ which means $$ lim \ x^N = +\infty$$ for every $x>1.$ Now to rigorously prove your limit, you have to replace $x$ with $1/x.$ To be extra rigorous, the sequence $1+N\cdot y$ goes to infinity basically because the real numbers are an archemidian field.