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I have to find a limit as $n\rightarrow\infty$ of 2 sequences:

$\lim\space (0,9999+\frac{1}{n})^n$

$\lim\space (1,00001-\frac{1}{n})^n$

Intutition tells me that as n goes to infinity $\frac{1}{n}$ becomes so small we can throw it out of the equation and it all comes down to finding limits of $0,9999^n$ and $1,00001^n$ which is trivial. But $\lim\space (1+\frac{1}{n})^n=e$ shows that this intuition may be wrong. So what should I do about these limits in order to prove them formally? Which theorems could be useful?

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    $\begingroup$ For a intuitive notion, please see this $\endgroup$ – cjferes Oct 30 '14 at 12:25
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Your intuition is right. To make it formal, find a lower and upper bounds: $$\left(0.9999+\frac 1n\right)^n\le0.99995^n$$ for sufficiently high $n$ and similarly for the other limit.

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  • $\begingroup$ Why is that an upper bound? $\endgroup$ – qiubit Oct 30 '14 at 12:50
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    $\begingroup$ @user36346 When $n$ is sufficiently large, we have $\frac1n\le0.00005$ and you only need the inequality to hold for sufficiently large $n$ as the limit doesn't depend on finitely many members. $\endgroup$ – user2345215 Oct 30 '14 at 12:54
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For $\;M\in\Bbb N\;$ such that $\;\frac1n=r<1-0.9999=0.0001\;\;\forall\, n>M$ , you get

$$0\le\left(0.9999+\frac1n\right)^n\le r^n\xrightarrow[n\to\infty]{}0$$

Try something similar with the second one.

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