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Given $a\in\mathbb{R}$ and $0, 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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    @Anonymous In that case Andre's answer will work fine...2012-04-05

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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 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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    Sorry, typo, should be for n>N. Very informally, if we know that |a^n-L|<0.001 if n>N, it follows immediately that |a^{n+1}-L|<0.001 if n>N.2012-04-05