1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor

    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?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Limits and minimum
  1. Limit of (Replies: 13)

  2. Limit of this (Replies: 21)

  3. No limit (Replies: 3)

  4. Limit ? (Replies: 2)

  5. Absolute Minimum (Replies: 3)