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Demonstration analysis and good books?

  1. May 2, 2011 #1
    When I try to demonstrate that lim (x^2)=4
    x--->2

    I found a different delta (delta=min{2-sqrt(epsilon-4),sqrt(epsilon+4)-2}, towards the one that is written in Demidovich┬┤s book (delta=epsilon/5). Could someone help me?
    Could someone tell me, too, a good algebra, calculus and real analysis book? Is it ok if i try analysis next semester absent a more advanced calculus (without calculus 2)? I need some books that explain polar coordinates too (good ones for beginner).
    Thanks and sorry for my mathematical notation.
     
    Last edited: May 2, 2011
  2. jcsd
  3. May 3, 2011 #2

    HallsofIvy

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    You want [itex]|x^2- 4|= |(x- 2)(x+ 2)|= |x+ 2||x- 2|< \epsilon[/itex].
    So we can write [itex]|x- 2|< \epsilon/|x+ 2|[/itex]
    But we need a value, [itex]\delta[/itex] that does NOT depend on x so that is [itex]|x- 2|< \delta[/itex] then [itex]|x- 2|< \epsilon/|x+ 2|[/itex]. That means we want [itex]\delta< \epsilon/|x+2|[/itex].

    If |x- 2|< 1 (The "1" is chosen just because it is simple. Any positive number would do.) then -1< x- 2< 1 so 1< x< 3 and then 3< x+2< 5 so that 3< |x+ 2|< 5. Then
    [tex]\frac{1}{5}< \frac{1}{|x+2|}< 1/3[/tex]
    and so
    [tex]\frac{\epsilon}{5}< \frac{\epsilon}{|x+2|}[/tex]

    That is, if
    [tex]|x- 2|< \frac{\epsilon}{5}[/tex]
    and |x- 2|< 1 as assumed above to get this, we will have
    [tex]|x-2|< \frac{\epsilon}{|x+2|}[/tex]
    so
    [tex]|x-2||x+2|= |x^2- 4|< \epsilon[/tex]
    as need.

    In other words, take [itex]\delta[/itex] to be the smaller of [itex]\epsilon/5[/itex] or 1.

    Analysis is basically the theory behind the Calculus. I would NOT take an analysis course without having completed the Calculus sequence.
     
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