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Thread: Strange Limit Proof

  1. #1
    Member billa's Avatar
    Joined
    Oct 2008
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    100

    Strange Limit Proof

    I thought I understood how to show that a function has a limit using the limit definition, but my book is doing a problem in a way I can't follow. The question is "prove that the limit of x^2 as x approaches 2 is 4"
    The book starts with the epsilon equation
    $\displaystyle
    \left| {x^2 - 4} \right| < \varepsilon
    $

    and they show that
    $\displaystyle
    \sqrt {4 - \varepsilon } < x < \sqrt {4 + \varepsilon }
    $
    which I can understand (though somehow they assume epsilon is less than 4)

    Then they say that they want to find delta. They show that delta is
    $\displaystyle
    Min\{ 2 - \sqrt {4 - \varepsilon } ,\sqrt {4 + \varepsilon } - 2\}
    $
    Which I am now able to follow although it took awhile to see why this is true. Up to this point I can understand what is going on. This is all the "scratch work" if I am correct. Now they have to actually prove the limit and they say the following...

    "If delta has this or any smaller positive value, the inequality
    0<abs(x-2)<delta
    will place x between
    $\displaystyle
    \sqrt {4 - \varepsilon } \_and\_\sqrt {4 + \varepsilon }
    $
    This completes the proof."

    I don't understand how this completes the proof. I can prove the limit a different way but it is disheartening that I can't follow this. Can anyone explain how their work proves the limit. Thanks.
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  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by billa View Post
    I thought I understood how to show that a function has a limit using the limit definition, but my book is doing a problem in a way I can't follow. The question is "prove that the limit of x^2 as x approaches 2 is 4"
    The book starts with the epsilon equation
    $\displaystyle
    \left| {x^2 - 4} \right| < \varepsilon
    $

    and they show that
    $\displaystyle
    \sqrt {4 - \varepsilon } < x < \sqrt {4 + \varepsilon }
    $
    which I can understand (though somehow they assume epsilon is less than 4)

    Then they say that they want to find delta. They show that delta is
    $\displaystyle
    Min\{ 2 - \sqrt {4 - \varepsilon } ,\sqrt {4 + \varepsilon } - 2\}
    $
    Which I am now able to follow although it took awhile to see why this is true. Up to this point I can understand what is going on. This is all the "scratch work" if I am correct. Now they have to actually prove the limit and they say the following...

    "If delta has this or any smaller positive value, the inequality
    0<abs(x-2)<delta
    will place x between
    $\displaystyle
    \sqrt {4 - \varepsilon } \_and\_\sqrt {4 + \varepsilon }
    $
    This completes the proof."

    I don't understand how this completes the proof. I can prove the limit a different way but it is disheartening that I can't follow this. Can anyone explain how their work proves the limit. Thanks.
    Assuming I understand you correctly proving that works because

    $\displaystyle |x^2-4|<\varepsilon\implies -\varepsilon<x^2-4<\varepsilon$

    Etc. Is that what you are asking?
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  3. #3
    Member billa's Avatar
    Joined
    Oct 2008
    Posts
    100
    umm no lol, but ty still. What i don't understand is given the books definition of delta as the minimum of two expressions, why is this statement true

    0<abs(x-2)<delta
    will place x between


    and if that statement is true, why does it prove the limit?
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  4. #4
    Senior Member
    Joined
    Dec 2007
    From
    Melbourne
    Posts
    428
    A common definition for the limit is $\displaystyle \lim_{x\to a}f(x) = b \textrm{ iff } \forall \epsilon>0, \exists \delta>0 \textrm{ such that } |x-a|<\delta \implies |f(x)-b|<\epsilon
    $

    They show that delta is
    This is a definition of $\displaystyle \delta$ that exists and is >0 whenever $\displaystyle \epsilon>0$. So all we need to show now is that this definition of $\displaystyle \delta$ satisfies $\displaystyle |x-2|<\delta \implies |x^2-4|<\epsilon$ You can usually do this by reading how you got the definition for $\displaystyle \delta$ backwards, although this does not constitute a formal proof. To do it formally,

    Assume x>2 so we can replace |x-2| with x-2.
    $\displaystyle 0<x-2<min(2-\sqrt{4-\epsilon},\sqrt{4+\epsilon}-2)$
    $\displaystyle 2<x<min(4-\sqrt{4-\epsilon},\sqrt{4+\epsilon})$
    Find the minimum (Warning: the following is not standard notation)
    $\displaystyle 4-\sqrt{4-\epsilon}?\sqrt{4+\epsilon}$
    both of these are now >0 so we can safely square
    $\displaystyle 16-8\sqrt{4-\epsilon}+4-\epsilon?4+\epsilon$
    $\displaystyle 16-2\epsilon?8\sqrt{4-\epsilon}$
    Still both >0
    $\displaystyle 256-64\epsilon+4\epsilon^2?64(4-\epsilon)$
    $\displaystyle 256-64\epsilon+4\epsilon^2?256-64\epsilon$
    so ?=>
    and so $\displaystyle \sqrt{4+\epsilon}$ is the minimum, so
    $\displaystyle 2<x<\sqrt{4+\epsilon}$

    Repeating this process assuming x<2 produces $\displaystyle \sqrt{4-\epsilon}<x<2$
    so we can say that $\displaystyle \sqrt{4-\epsilon}<x<\sqrt{4+\epsilon}$

    It is very easy to show that this is equivalent to $\displaystyle |x^2-4|<\epsilon$

    though somehow they assume epsilon is less than 4
    When dealing with limits you only need to worry about the small values of epsilon, as how the function behaves when x = 9 doesn't affect the limit as $\displaystyle x \to 1$
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