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Math Help - Sequential Characterization of limit

  1. #1
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    Sequential Characterization of limit

    So the limit as a sequence is defined as limn→∞ an = L , or more precisely as if for all ǫ > 0 there existsN ∈ N so that for all n ≥ N,
    |an − L| < ǫ.
    And then the sequential notion of continuity is defined as: for some function
    f: X → Y is sequentially continuous if whenever a sequence (xn) in X converges to a limit x, the sequence (f(xn)) converges to f(x).

    So can this idea be used to disprove a limit like as x->0, cos(1/x) -> no limit .
    For instance, if one supposes that two sequences x_1 = 1/(2npi) and some other sequence x_2 both tend to 0, as n->infinity, then can one state
    if f(x)=cos(1/x) then f(x_1)=1 and f(x_2) = 0 (suppose this is the case), which indicates that the limit does not exist.
    My understanding is that since both sequences tend to 0, we can use x_1 and x_2 as x in the function f by the above defn. And since f(x_1) and f(x_2) don't converge, the limit does not exist.
    Please correct me if my reasoning is wrong?

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  2. #2
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    Re: Sequential Characterization of limit

    Your notation is confusing. Let's say you are using a_n = \dfrac{1}{2n\pi}, b_n = \dfrac{2}{(2n-1)\pi}. Then \lim_{n\to \infty}a_n = \lim_{n\to\infty}b_n = \lim_{x\to 0}x = 0. Now, \lim_{n\to \infty} \cos\left(\dfrac{1}{a_n}\right) = 1 \neq 0 = \lim_{n\to \infty} \cos\left(\dfrac{1}{b_n}\right). Is that what you mean? They both converge, but not to the same limit. That is proof that the limit \lim_{x \to 0} \cos\left(\dfrac{1}{x}\right) does not exist.
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