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Suppose I have these two inequalities:

$$-\epsilon < l < \epsilon$$ $$ -\epsilon + 1 < l < \epsilon +1 $$

where $\epsilon$ and $l$ can be any number and $\epsilon \gt 0$.

How can I show that these two inequality is not true for all $\epsilon$?

I found that $\epsilon$ between 0 to 0.5 will make this inequality false but I was wondering if there is another approach that does not involve trial and error.

Thank you in advance for any help provided.

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Hint: study the lengths of both intervals. Also, look for their starting point. What is the distance between them? –  Sigur Jul 21 '13 at 15:46
1  
(An implicit assumption seems to be $\epsilon>0$.) When is $\epsilon<-\epsilon+1$? –  egreg Jul 21 '13 at 15:47
    
@egreg Yes, assuming $\epsilon > 0$. Sorry, I missed that. –  mauna Jul 21 '13 at 19:20

3 Answers 3

up vote 1 down vote accepted

You're implicitly assuming $\epsilon>0$, probably this question arises in connection with uniqueness of limits.

The numbers $l$ satisfying $-\epsilon<l<\epsilon$ form an interval; also those satisfying $-\epsilon+1<l<\epsilon+1$ do. The center of the second interval is at the right of the first interval's center. So we can look at the supremum of the first interval and the infimum of the second one: if $$ \epsilon<-\epsilon+1 $$ what can you say?

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$\epsilon \lt 0.5$. This means that the two intervals do not overlap when $\epsilon$ is less than 0.5. When the two intervals don't overlap, then there is no $l$ that can satisfy the inequality. Right? –  mauna Jul 21 '13 at 19:19
    
@mauna Exactly. –  egreg Jul 21 '13 at 19:36

The first equation reads $$|l|<\epsilon,$$the second one is $$|l-1|< \epsilon.$$ By triangular inequality $$1=|l-(l-1)|\le|l|+|l-1|< \epsilon+\epsilon,$$ which is impossible if $\epsilon\le 1/2$

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+1 for the beautiful solution. But, please, note that it was tagged as homework. Now there is no work for him/her. lol –  Sigur Jul 21 '13 at 15:59
    
@Sigur They still have to prove that for $\epsilon>1/2$ such an $l$ exists=) –  TZakrevskiy Jul 21 '13 at 16:01
    
Well, this could be a nice exercise. But since he/she only need to prove that the inequalities are false, you gave him/her a counter example... :) –  Sigur Jul 21 '13 at 16:03

You could consider modulus, I will use x instead of l, so the first inequality means mod x < e and the second is mod x-1 < e, you can plot the functions mod x and mod x-1 and then go from there to deduce what values of e make the inequality false etc, its important to see especially where the intersections occur between the two 'mod' graphs

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Why change the variable from $l$ to $x$? –  Sigur Jul 21 '13 at 16:00
    
its a dummy variable so it doesnt matter really, i prefer x :p –  WhizKid Jul 21 '13 at 17:37

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