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Math Help - help visualising a limit proof..

  1. #1
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    i tried to interpret it on a sketch

    http://img181.imageshack.us/img181/6498/13549423mx3.gif

    i cant see the animation of this proove
    Last edited by mr fantastic; January 5th 2009 at 05:38 AM. Reason: Merge
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  2. #2
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    If you take the definition of the limit:
    \forall\mathcal{O}(L)\;\exists\mathcal{P}(a)\;\for  all x\in D(f)\cap\mathcal{P}(a):f(x)\in\mathcal{O}(L)
    You may also understand this by delta-epsilon notation:
    |x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon
    where \mathcal{O}(L):=(L-\varepsilon;L+\varepsilon) and \mathcal{P}(a):=(a-\delta;a+\delta)\backslash\{a\}
    To prove that
    \lim_{x\to p}f(x)=p
    you need to use the definition:
    |x-p|<\delta\Rightarrow|f(x)-p|<\varepsilon
    where f(x)=x so I can rewrite:
    |x-p|<\delta\Rightarrow|x-p|<\varepsilon
    Now I can choose \delta=\varepsilon. So I prooved, that for all epsilons exists such delta that |x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon

    Another example: Proove that \lim_{x\to 1}(2x+2)=4.
    Solution: Again, I have to use definition: |x-1|<\delta\Rightarrow|(2x+2)-4|<\varepsilon. Now I have to find such delta, that satisfies epsilon. It is easy:
    |(2x+2)-4|=|2x-2|=|2|\cdot\underbrace{|x-1|}_{\delta}<\varepsilon
    There's another trick:
    |x-1|<\frac{\varepsilon}{2}=\delta
    So now, if you tell me some epsilon, I am able to tell you delta which satisfies your epsilon. If you tell \varepsilon=0.001 my delta is \delta=0.0005
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by lukaszh View Post
    If you take the definition of the limit:
    \forall\mathcal{O}(L)\;\exists\mathcal{P}(a)\;\for  all x\in D(f)\cap\mathcal{P}(a):f(x)\in\mathcal{O}(L)
    You may also understand this by delta-epsilon notation:
    |x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon
    where \mathcal{O}(L):=(L-\varepsilon;L+\varepsilon) and \mathcal{P}(a):=(a-\delta;a+\delta)\backslash\{a\}
    To prove that
    \lim_{x\to p}f(x)=p
    you need to use the definition:
    |x-p|<\delta\Rightarrow|f(x)-p|<\varepsilon
    where f(x)=x so I can rewrite:
    |x-p|<\delta\Rightarrow|x-p|<\varepsilon
    Now I can choose \delta=\varepsilon. So I prooved, that for all epsilons exists such delta that |x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon

    Another example: Proove that \lim_{x\to 1}(2x+2)=4.
    Solution: Again, I have to use definition: |x-1|<\delta\Rightarrow|(2x+2)-4|<\varepsilon. Now I have to find such delta, that satisfies epsilon. It is easy:
    |(2x+2)-4|=|2x-2|=|2|\cdot\underbrace{|x-1|}_{\delta}<\varepsilon
    There's another trick:
    |x-1|<\frac{\varepsilon}{2}=\delta
    So now, if you tell me some epsilon, I am able to tell you delta which satisfies your epsilon. If you tell \varepsilon=0.001 my delta is \delta=0.0005
    Of course are you are correct ! But for the benefit of the poster you might want to mention that \mathcal{P}(a) is a neighborhood of radius \delta around p...things like that
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