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Math Help - Commutativity of limit using epsilon-delta

  1. #1
    Member Greengoblin's Avatar
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    Commutativity of limit using epsilon-delta

    I'm trying to write a delta-epsilon proof that:

    \lim_{x\to a}kf(x)=k\left(\lim_{x \to a}f(x)\right)=b\quad\text{(k is constant)}

    I know that \lim_{x\to a}kf(x) exists if \forall\epsilon>0\exists\delta>0 such that:

    0<|x-a|<\delta \Rightarrow |kf(x)-b|<\epsilon

    And k\left(\lim_{x\to a}f(x)\right) exists if \forall\epsilon>0\exists\delta>0 such that:

    0<|x-a|<\delta \Rightarrow k|f(x)-b|<\epsilon

    I don't understand how I can write a proof using an arbitrary function and limit. So far I have only been proving specific limits with the epsilon delta method, not writing any algebraic proofs like this. Can someone offer some help? Thanks

    Edit: I had a thought, but I'm not sure I wrote that second one out right, cause wouldn't I simply have to prove: |kf(x)-b|=k|f(x)-b| (which isn't true)?
    Last edited by Greengoblin; January 8th 2009 at 09:24 AM.
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  2. #2
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    We will assume k\not = 0.

    Since \lim_{x\to a}f(x) = b it means for any \epsilon > 0 we can find \delta > 0 so that 0 < |x-a| < \delta \implies |f(x) - b| < \epsilon.

    We want to show \lim_{x\to a} kf(x) = kb. Let \epsilon > 0 but then \tfrac{\epsilon}{|k|} > 0. This means by above there is \delta > 0 so that 0 < |x-a| < \delta \implies | f(x) - b | < \tfrac{\epsilon}{|k|}. Multiply by |k| to get |kf(x) - bk| < \epsilon whenever 0 < |x-a| < \delta.
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  3. #3
    Member Greengoblin's Avatar
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    Thanks alot, I understand it now.
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