negation of limit definition

by antiemptyv
Tags: definition, limit, negation
 P: 34 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 $$(x_n)$$. The sequence $$(x_n)$$ converges to $$L$$ if given $$\epsilon > 0$$, $$\exists K(e) \in \mathbb{N} \ni$$ if $$n > K(e)$$, then $$|x_n-L| < \epsilon$$. 3. The attempt at a solution The limit of a sequence $$(x_n)$$ is not L if $$\exists \epsilon > 0 \ni \forall K \in \mathbb{N}$$, $$\existsn \in \mathbb{N} \ni n > K \ni |x_n - L| \geq \epsilon$$.
 HW Helper Sci Advisor P: 2,482 I think that is right, except it seems as if you have used too many N's, \in's or \ni's.
PF Patron
Thanks
Emeritus
P: 38,429
 Quote by antiemptyv 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 $$(x_n)$$. The sequence $$(x_n)$$ converges to $$L$$ if given $$\epsilon > 0$$, $$\exists K(e) \in \mathbb{N} \ni$$ if $$n > K(e)$$, then $$|x_n-L| < \epsilon$$. 3. The attempt at a solution The limit of a sequence $$(x_n)$$ is not L if $$\exists \epsilon > 0 \ni \forall K \in \mathbb{N}$$, $$\existsn \in \mathbb{N} \ni n > K \ni |x_n - L| \geq \epsilon$$.
Not if by "if $n> K(e)[/tex] then [tex]|x_n_L|< \epsilon$ you mean "for all n> N(e).
That only has to be true for some n> Ke)

P: 34

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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