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Thread: 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; Jan 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:
    $\displaystyle \forall\mathcal{O}(L)\;\exists\mathcal{P}(a)\;\for all x\in D(f)\cap\mathcal{P}(a):f(x)\in\mathcal{O}(L)$
    You may also understand this by delta-epsilon notation:
    $\displaystyle |x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon$
    where $\displaystyle \mathcal{O}(L):=(L-\varepsilon;L+\varepsilon)$ and $\displaystyle \mathcal{P}(a):=(a-\delta;a+\delta)\backslash\{a\}$
    To prove that
    $\displaystyle \lim_{x\to p}f(x)=p$
    you need to use the definition:
    $\displaystyle |x-p|<\delta\Rightarrow|f(x)-p|<\varepsilon$
    where $\displaystyle f(x)=x$ so I can rewrite:
    $\displaystyle |x-p|<\delta\Rightarrow|x-p|<\varepsilon$
    Now I can choose $\displaystyle \delta=\varepsilon$. So I prooved, that for all epsilons exists such delta that $\displaystyle |x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon$

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

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