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Math Help - Exemplify why this definition of a limit does not hold

  1. #1
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    Exemplify why this definition of a limit does not hold

    I've been thinking in circles over this problem so I thought I'd give this forum a try - I hope someone can help.

    I've been given a definition:
    For all epsilon>0 , there exists a natural number N such that if |Xn-L| < epsilon, then n is greater than or equal to N.

    I need to either find an infinite sequence Xn which converges but does not satisfy the conditon above OR
    find a sequence Xn and L which do satisfy the condition above but Xn does not converge to L.

    I'd be grateful if anybody could talk me through what features such a sequence should have.

    Thanks
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  2. #2
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    I'll give you the sequence and it's limit from the new definition, you show that it 'converges' to that limit:

    a_n = n \ \forall n \in \mathbb{N} and L = 0
    Hint: Take N = 1 for every \epsilon

    After doing that, see if you can do it for any sequence.
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  3. #3
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    Thanks for your reply, although I still can't get my head around it.

    Does this sequence guarantee n \geq N when |x_n|< \epsilon but {x_n} does not converge?

    or

    Would this show that |x_n|< \epsilon does not guarantee n \geq 1 ?

    If the latter, does that apply for all possible epsilon greater than zero?

    Thanks.
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  4. #4
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    Well, is there any time where n < N?
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  5. #5
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    Ah, okay we have N=1 so n will always be greater than, or equal to it.

    I guess my problem is, how does this work for all possible values of epsilon? e.g what if epsilon = 1/2, i.e. for n=1,
    |1|< \frac{1}{2} => n \geq 1

    and how does this converge to 0 when n goes to infinity?

    I appreciate your patience but I'm still missing something here.
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  6. #6
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    You have (roughly speaking) a proposition of the form P \Rightarrow Q where here, P means |x_n-L| < \epsilon and Q means n \ge N
    By definition, if P is false then the proposition is true, so you don't care about values of n for which |x_n-L| \ge \epsilon.
    Now assume P is true. Is Q always true?
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  7. #7
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    Ahh I've answered my own question Xn= n does satisfy the condition for L= 0 but does not converge to L which is what I was initially looking for.
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  8. #8
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    Correct.
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