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Math Help - Sequence Convergence Proofs vs. Continuity Function Proofs

  1. #1
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    Sequence Convergence Proofs vs. Continuity Function Proofs

    Definition of converge:
    For each epsilon > 0 there exists a number N such that n>N implies
    |sn -s|<epsilon.

    In this case the N you chose was usually a max {constant, f(epsilon)}.

    Definition of continuity
    For each epsilon > 0 there exists a delta>0 such that x element dom(f) and
    |x-x0| < delta implies |f(x)-f(x0)|< epsilon.

    In this case the delta you choose is usually a min{constant, f(epsilon)}

    I was just wondering how to think about why for one you need a min and the other a max.

    Thanks
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  2. #2
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    Frankly, I do not think what you wrote above is correct.
    Perhaps you can fill out what you wrote with more detail.
    There is no way to know what you mean by \left\{ \mbox{constant},f(\varepsilon ) \right\}.
    Please tell us.
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  3. #3
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    Well to prove that 4n^3 + 3n / (n^3 - 6) = 4 you choose an
    N = max{2,squareroot(54/epsilon)}

    To prove that 2x^2 + 1 is continuous, you choose a
    delta = min{1, epsilon/(2(2|x0|+1))}

    Hopefully that's more clear.
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  4. #4
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    While those exact examples may not work in all cases. However, I do get the idea.

    In the case of the sequence, the subscript is approaching infinity.
    Therefore, we take the maximum N to insure all conditions hold.

    The case of continuity is a bit more abstract. In choosing the \delta we are creating a bound on x, we want to be able to say that \left| {x - x_0 } \right| < \delta \quad  \Rightarrow \quad \left| {f(x) - f\left( {x_0 } \right)} \right| < \varepsilon .
    Knowing that \delta < 1 means x_0  - 1 < x < x_0  + 1.
    This means we can construct bounds on f(x).
    So we want \delta to be at most 1 (not more).

    I hope that helps somewhat.
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