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Thread: 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:

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

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

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

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

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

    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: $\displaystyle |kf(x)-b|=k|f(x)-b|$ (which isn't true)?
    Last edited by Greengoblin; Jan 8th 2009 at 08:24 AM.
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  2. #2
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    We will assume $\displaystyle k\not = 0$.

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

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