# Simple Limit Proof

Given $a\in\mathbb{R}$ and $0<a<1$, let $X_n=a^n$, $\forall n\in\mathbb{N}$.

Prove that $\lim \limits_{n\to \infty}X_n=0$ using limit definition or limits arithmetics(including the squeeze theorem if needed).

Thanks a lot.

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What have you tried? Where are you getting stuck? Is this homework? If it is, please tag it as homework. –  Matthew Conroy Apr 5 '12 at 21:46
Hey, no it's not homework, it's something my professor assumed was obvious but he didn't prove it. –  Anonymous Apr 5 '12 at 21:46
Do you know of logarithms? –  anon Apr 5 '12 at 21:50
@anon No, we haven't gotten to logarithms yet. –  Anonymous Apr 5 '12 at 21:52
What "limit definition" do you have? The one with $\epsilon$ ...? –  Thomas Apr 5 '12 at 21:54
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Note that $\dfrac{1}{a}>1$. Let $\dfrac{1}{a}=1+k$.

By induction, or by using the Binomial Theorem, we can show that $(1+k)^n \ge 1+nk$. It follows that $$0<a^n=\frac{1}{(1+k)^n}\le\frac{1}{1+kn}.$$ Now it should not be hard to use the $\epsilon$-$N$ definition, or Squeezing, to get the result.

Remark: One could use fancier tools. The sequence $(a^n)$ is decreasing. It is bounded below by $0$. So the sequence has a limit. Let $L$ be the limit. Then $$L=\lim_{n\to\infty} a^n=\lim_{n\to\infty}a^{n+1}=a\lim_{n\to\infty} a^n=aL.$$ so $L(1-a)=0$ and therefore $L=0$.

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why didn't you write $\frac{1}{(1+k)^n} \le \frac{1}{1+kn}$? Thanks a lot. –  Anonymous Apr 5 '12 at 22:08
@Anonymous: Thanks, fixed. –  André Nicolas Apr 5 '12 at 22:10
why does $lim_{n\to\infty} a^n=\lim_{n\to\infty}a^{n+1}$? –  Anonymous Apr 5 '12 at 22:14
Anonymous also wishes to note that your first inequality is Bernoulli's inequality. –  anon Apr 5 '12 at 22:21
Definition of limit. If $|a^n-L|<\epsilon$ whenever $n >N$, then $|a^{n+1}-L|<\epsilon$ for $n>\epsilon$. –  André Nicolas Apr 5 '12 at 22:25
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