# Prove that $\lim|x_n| < a$ for all $a > 0$ implies $\lim|x_n| = 0$

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? – Stefan Hansen Dec 14 '12 at 18:57
Note that denote $b=lim_{n\to\infty}|x_n|$, then $0\leq b<a$, for all $a>0$, so by contradiction, $b=0.$ – ougao Dec 14 '12 at 18:58
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. – Lopsy Dec 14 '12 at 19:05
What's more, $$\limsup_{n\to\infty}|x_n|<a\quad\text{for all }a>0\qquad\Longrightarrow\qquad\lim_{n\to\infty}|x_n|=0$$ – robjohn Dec 14 '12 at 19:14
robjohn's comment is a rephrasing of (a very slightly weaker version of) Lopsy's comment. – Jonas Meyer Dec 14 '12 at 19:22

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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No need to divide by $2$, since strict inequality is already assumed: $L<L$. – Jonas Meyer Dec 14 '12 at 19:01
@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 $L\le a$ – Nameless Dec 14 '12 at 19:04

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:S – alexandreC Dec 15 '12 at 13:28
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 – Stefan Hansen Dec 15 '12 at 13:41
Its increasing already!!! :) – alexandreC Dec 15 '12 at 21:50