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Limits and minimum

  1. Dec 31, 2007 #1
    hi all,
    like many here i have also picked up the spivak calculus text and run into some problems.
    i am reading the fifth chapter which introduces 'limits' and i can't get my head around the idea of using minimum (the concept was introduced in the problems of the first chapter and it reappears in chapter 5 without any formal discussion about it). i haven't figured out how to approach it yet. i guess thats what i mean to ask here. i don't think i am making any sense. probably an example will help. this is from the book, towards the end of the chapter, consider:

    f(x) = x^2 + x, and x approaches a,
    according to the definition in the book, we have to find a [tex]\delta[/tex] > 0 such that |x^2 + x - (a^2 + a)| < [tex]\epsilon[/tex]
    it then breaks down the function into x^2 and x, so that we need 2 [tex]\delta[/tex]s
    one for the x^2 and the other for x,
    if 0 < |x - a| < [tex]\delta[/tex]1, then |x^2 - a^2| < [tex]\epsilon[/tex]/2
    if 0 < |x - a| < [tex]\delta[/tex]2, then |x - a| < [tex]\epsilon[/tex]/2

    now, for some reason i don't know,

    [tex]\delta[/tex]1 = min (1, [tex]\epsilon[/tex]/2/2|a| + 1)
    [tex]\delta[/tex]2 = [tex]\epsilon[/tex]/2

    what is going on here?

    this is a very confusing first post and i am very sorry about it. (imagine the confusion in my head :)) perhaps i can try to clarify my question after a few responses. thanks in advance!
     
  2. jcsd
  3. Jan 1, 2008 #2

    HallsofIvy

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    Well, to have [itex]|x-a|<\epsilon/2[/tex] you must take [itex]\delta = \epsilon/2[/itex] so the that [itex]|x-a|< \epsilon/2[/itex] and [itex]|x-a|< \delta[/itex] are exactly the same! [itex]\delta_1= \epsilon/2[/itex]
    [itex]|x^2- a^2|< \epsilon/2[/itex] is a little harder. [itex]|x^2- a^2|= |(x-a)(x+a)|= |x-a||x+a|[/itex]. In order to have [itex]|x^2- a^2|= |x-a||x+a|< \epsilon/2[/itex] so we must have [itex]|x-a|< \epsilon/(2|x+a|)[/itex] (remember we want to be able to write [itex]|x-a|< \delta[/itex] so we solve for |x-a|). Of course, we need that number on the right to be a constant- not depend of a. We are talking about x close to a so suppose we take |x-a|< 1 (I chose 1 just because it is easy). That means that -1< x-a< 1. Now add 2a to both sides. Then 2a-1< x+ a< 2a+ 1. For any a, 2|a|+1 is the smaller of those so we know that |x+a|> 2|a|+1 and 1/|x+a|< 1/(2a+1). and so [itex]|x-a|< (1/(2|a|+1))(\epsilon/2)[/itex]. Taking [itex]\delta_2= 1[/itex] guarentees that [itex]|x-a|< (1/(2|a|+1))(\epsilon/2)[/itex].
    If we take [itex]\delta[/itex] to be the smaller of those two, that is, take [tex]\delta[/tex]1 = min (1, [tex]\epsilon[/tex]/2/2|a| + 1) both of those are true so if [itex]|x-a|< \delta[/itex],
    [itex]|x^2+x-(a^2+a)|= |(x^2-a^2)+ (x-a)|\le |x^2- a^2|+ |x-a|= |x-a||x+a|+ |x-a|[/itex]
    [itex]= (1/(2|a|+1))(\epsilon/2)|x+a|+ \epsilon/2\le \epsilon/2+ \epsilon/2= \epsilon[/itex].
     
  4. Jan 1, 2008 #3
    thanks for the response HallsofIvy, I think I follow it mostly. one more small query: why must we take the minimum to express delta? my uninformed guess is that it is to satisfy the absolute values of |x - a| form or, because of the power of x (x^2, x^3, etc).
    what do you think?
     
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