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negation of limit definition

 
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Sep20-07, 02:22 PM   #1
 

negation of limit definition


1. The problem statement, all variables and given/known data

I'm trying to show that a sequence does not have a limit, so that would mean proving the negation of the limit definition is true, right? Is this a correct negation of the definition of what it means for a sequence to have a limit?

2. Relevant equations

The definition of the limit of a sequence [tex](x_n)[/tex].
The sequence [tex](x_n)[/tex] converges to [tex]L[/tex] if given [tex]\epsilon > 0[/tex], [tex]\exists K(e) \in \mathbb{N} \ni[/tex] if [tex]n > K(e)[/tex], then [tex]|x_n-L| < \epsilon[/tex].

3. The attempt at a solution

The limit of a sequence [tex](x_n)[/tex] is not L if [tex]\exists \epsilon > 0 \ni \forall K \in \mathbb{N}[/tex], [tex]\existsn \in \mathbb{N} \ni n > K \ni |x_n - L| \geq \epsilon[/tex].
 
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Sep20-07, 02:39 PM   #2
 
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I think that is right, except it seems as if you have used too many N's, \in's or \ni's.
 
Sep20-07, 03:49 PM   #3
 
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Quote by antiemptyv View Post
1. The problem statement, all variables and given/known data

I'm trying to show that a sequence does not have a limit, so that would mean proving the negation of the limit definition is true, right? Is this a correct negation of the definition of what it means for a sequence to have a limit?

2. Relevant equations

The definition of the limit of a sequence [tex](x_n)[/tex].
The sequence [tex](x_n)[/tex] converges to [tex]L[/tex] if given [tex]\epsilon > 0[/tex], [tex]\exists K(e) \in \mathbb{N} \ni[/tex] if [tex]n > K(e)[/tex], then [tex]|x_n-L| < \epsilon[/tex].

3. The attempt at a solution

The limit of a sequence [tex](x_n)[/tex] is not L if [tex]\exists \epsilon > 0 \ni \forall K \in \mathbb{N}[/tex], [tex]\existsn \in \mathbb{N} \ni n > K \ni |x_n - L| \geq \epsilon[/tex].
Not if by "if [itex]n> K(e)[/tex] then [tex]|x_n_L|< \epsilon[/itex] you mean "for all n> N(e).
That only has to be true for some n> Ke)
 
Sep20-07, 08:17 PM   #4
 

negation of limit definition


Yes, it all seems right now I guess. Thanks! and oh yeah, i guess while editting, i left in a few extra symbols...
 
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