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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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    Clearly $\lim_{n\to\infty} |x_n|\geq 0$, so what non-negative number satisfies being smaller than every other postive number?2012-12-14
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    Note that denote $b=lim_{n\to\infty}|x_n|$, then $0\leq b, for all $a>0$, so by contradiction, $b=0.$2012-12-14
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    Instead of $lim_{n\rightarrow\infty}|x_n|, 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.2012-12-14
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    What's more, $$ \limsup_{n\to\infty}|x_n|0\qquad\Longrightarrow\qquad\lim_{n\to\infty}|x_n|=0 $$2012-12-14
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    robjohn's comment is a rephrasing of (a very slightly weaker version of) Lopsy's comment.2012-12-14

2 Answers 2

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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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    @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:S2012-12-15
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    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 that2012-12-15
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    Its increasing already!!! :)2012-12-15