Let $(x_n)$ be a sequence of real numbers.
I am wondering we can formally prove that:
$$ \lim_{n\rightarrow\infty}|x_n| < a$$
for all $a > 0$, implies
$$ \lim_{n\rightarrow\infty}|x_n| = 0$$
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1$\begingroup$ Clearly $\lim_{n\to\infty} |x_n|\geq 0$, so what non-negative number satisfies being smaller than every other postive number? $\endgroup$ – Stefan Hansen Dec 14 '12 at 18:57
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$\begingroup$ Note that denote $b=lim_{n\to\infty}|x_n|$, then $0\leq b<a$, for all $a>0$, so by contradiction, $b=0.$ $\endgroup$ – ougao Dec 14 '12 at 18:58
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1$\begingroup$ Instead of $lim_{n\rightarrow\infty}|x_n|<a$, could you possibly want something like "eventually, $|x_n|$ is always less than $a$ (but we're not assuming the limit exists yet)"? In that case, it's still true that $lim_{n\rightarrow\infty}|x_n|=0$, but it's less trivial. $\endgroup$ – Lopsy Dec 14 '12 at 19:05
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1$\begingroup$ What's more, $$ \limsup_{n\to\infty}|x_n|<a\quad\text{for all }a>0\qquad\Longrightarrow\qquad\lim_{n\to\infty}|x_n|=0 $$ $\endgroup$ – robjohn♦ Dec 14 '12 at 19:14
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$\begingroup$ robjohn's comment is a rephrasing of (a very slightly weaker version of) Lopsy's comment. $\endgroup$ – Jonas Meyer Dec 14 '12 at 19:22
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Let $L=\lim_{n\rightarrow\infty}|x_n|$. Then $0\le L<a $ for all $a>0$. Suppose $L>0$. Then $\frac{L}{2}>0$ and so using $a=\frac{L}{2}$ gives $$L<a=\frac{L}{2}\Rightarrow L<0$$ which is a contradiction
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$\begingroup$ No need to divide by $2$, since strict inequality is already assumed: $L<L$. $\endgroup$ – Jonas Meyer Dec 14 '12 at 19:01
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$\begingroup$ @JonasMeyer Sure you don't. But because when dealing with limits and a positive number pops up, we usually devide it by $2$ and use it as $\epsilon$. Also this covers the case $L\le a$ $\endgroup$ – Nameless Dec 14 '12 at 19:04
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Suppose $\varepsilon > 0$ is the limit under consideration. Choosing $a=\frac{\varepsilon}{2}$ gives us a contradiction from your first equation. Therefore $\varepsilon$ has to be smaller or equal to zero.
Now, from your first equation, and because there is a $\delta$ such that for $n> n_0$ all the elements of $|x_n|$ belong to $]0, \delta[$, we have to have the limit of $|x_n|$ belonging to $[0, \delta]$, i.e. it has to be non-negative.
The only non negative number which is smaller or equal to zero is zero.
q.e.d.
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$\begingroup$ @Stefan Hansen, thank you for your edit! I have to learn how to LaTeX in here! // I dont actually think "Nameless"'s answer, accepted as the correct one, is fully correct, as it explicitely assumes the limit has to be non-negative,( but I am not allowed to comment in there!) :S:S $\endgroup$ – alexandreC Dec 15 '12 at 13:28
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1$\begingroup$ No problem :) I guess you'll have to earn some more reputation in order to for you to comment on his answer. I'm sure you can do that $\endgroup$ – Stefan Hansen Dec 15 '12 at 13:41
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