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Epsilon/Delta Limit Exercise

  1. Aug 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove each statement using the epsilon delta definition of limit.

    lim [tex](x^2-1)=3[/tex]
    x -> -2

    2. Relevant equations

    3. The attempt at a solution

    Given E > 0, we need D > 0 such that if [tex]|x-(-2)|<D[/tex] then [tex]|(x^2-4|<E[/tex].

    If [tex]|x+2|<1[/tex], then [tex]-1<x+2<1[/tex] [tex]-5<x-2<-3[/tex] [tex]|x-2|<5[/tex].

    Here's where I'm lost... my answer key says to take [tex]D=min{E/5,1}[/tex] but I don't understand why this is.

    Thanks in advance.
  2. jcsd
  3. Aug 22, 2011 #2


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    So if |x+2| < 1 what would be your overestimate for

    |x2 - 4| = |(x+2)(x-2)| ?

    This is what you are trying to make small. How close to -2 does x have to be?
  4. Aug 23, 2011 #3
    Sorry... I don't know how to figure that out.
  5. Aug 23, 2011 #4


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    You want to make [itex]|(x+2)(x- 2)|< \epsilon[/itex]. That's the same as [itex]|x+ 2|< \epsilon/|x- 2|[/itex]

    Now, you have calculated that -5< x- 2< -3 so that 3<|x- 2< 5. The one you really want is 3< |x- 2|. That way, 1/|x- 2|> 1/3 and so [itex]\epsilon/|x-2|> \epsilon/3[/itex].
    Last edited by a moderator: Aug 23, 2011
  6. Aug 23, 2011 #5


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    If |x+2| < 1 you have shown that |(x-2)| < 5. So

    |x2 - 4| = |(x+2)(x-2)| < 5|(x+2)|

    How small does |x+2| need to be to make this less than [itex]\epsilon[/itex]? Answer that and you will see where the book's answer comes from.
  7. Aug 23, 2011 #6
    I think I get it now... when I'm home from work I'll sit down, work through it and post the answer.

    Thanks again, your help is very appreciated guys.
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