Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

$$ \lim_{n\rightarrow\infty}|x_n| = 0$$

share|improve this question
1  
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
1  
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
1  
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

2 Answers 2

up vote 0 down vote accepted

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

share|improve this answer
    
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.

share|improve this answer
    
@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
1  
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
1  
Its increasing already!!! :) –  alexandreC Dec 15 '12 at 21:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.