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Epsilon Delta Proofs, finding bounds

  1. Aug 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that lim x->3 of (x[tex]^{2}[/tex]+x-5=7

    2. Relevant equations
    0<x-c<[tex]\delta[/tex] and |f(x)-L|<[tex]\epsilon[/tex]

    3. The attempt at a solution
    The preliminary analysis.
    The first equation in the relevant equations becomes

    And the second equation becomes

    Now this is were it gets confusing. The book says "The factor |x-3| can be made as small as we wish, and we know that |x+4| will be about 7." Then it says that this comes from setting |x-3|[tex]\leq[/tex]1
    Then from that I would get x[tex]\leq[/tex]4
    So what do I do then? The book continues on with
    [tex]\leq[/tex]|x-3|+|7| (Triangle Inequality)
    So this means that 8 is the bound. But my problem is what factor do I set [tex]\leq[/tex]1? Also, whats going on in the steps immediately above? I can do the delta epsilon proofs that don't require this step, but I don't understand this. I have seen it explained graphically it didn't really help.
  2. jcsd
  3. Aug 27, 2010 #2
    i think you're making it too hard.

    i'm going to use d for delta and e for epsilon

    there are two things to consider -- so you make two choices for d;

    |(x^2 +x - 5) - 7| = |(x+4)| |(x-3)|

    look at the first factor.
    If i pick d<1 then for |x-3|< d = 1 i get |x+4|<8, right?
    then take the second
    If i pick d<e/8 i get |x-3|< d = e/8.

    so let d be smaller than both so both conditions are true
    |(x^2 +x - 5) - 7| = |(x+4)| |(x-3)| < 8 * (e/8) = e.
  4. Aug 27, 2010 #3
    If you solve |x-3|<1, you would get 4, where does the 8 come from? Are you not showing a step or something? I really need to know what's going on there and it doesn't show it in the book either.
  5. Aug 27, 2010 #4


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    Not sure where you're getting this, but you should get -1 < x-3 < 1, from which it follows that 2 < x < 4 if that's what you mean. Anyway, you're missing the point though; by limiting the size of |x-3| we can set an upper bound on the size of |x+4|. To do this, we require that |x-3| be less then some fixed number, say, 1. Then, if |x-3| < 1, we have that -1 < x-3 < 1 or 6 < x+4 < 8, which means that |x+4| < 8. And that's how you get the 8.
  6. Aug 27, 2010 #5
    How do you jump from -1 < |x-3| < 1 to 6 < x+4 < 8? I don't see how one implies the other.
  7. Aug 28, 2010 #6


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    Start with -1 < x-3 < 1 and add 7 to each term. It's pretty simple.
  8. Aug 28, 2010 #7
    So I just take the inequality and add the value at the limit? Or what? And how do I know what factor to do this with?
  9. Aug 28, 2010 #8


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    Why don't you show us what you think rather than have us guess at what you mean? I'm not sure exactly where you're getting caught up and I can't really help you until I know.

    Review the definition of a limit and it should be clear. You need to understand this definition before you'll be able to use in any sort of capacity.
  10. Aug 28, 2010 #9
    I don't understand why you're adding 7. I knew the definition of a limit. But my problem is why are you adding 7? And how do I know to set the inequality up with the |x-3| instead of the |x+4|. I know the whole f(x)-L<[tex]\delta[/tex] thing, and then the 0 < x-c < [tex]\epsilon[/tex]. And then if you give me a delta, I can find any epsilon that will be less than the delta. That's the definition of a limit. That's also the point of these proofs. Now that you know that I know what I'm talking about. Why add the limit, which is 7, and |x-3|, and not the |x+4|. As |x+4||x-3| is the function, so it would be |x+4||x-3|-7<[tex]\delta[/tex]. But then I as the book says and what you guys are saying, I'm supposed to do, -1<|x-3|<1. Add 7(the limit? why), and then why |x-3|, and not |x+4|.
  11. Aug 28, 2010 #10


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    Okay, so it sounds like you're confused on a number of levels. The first thing that I'll address concerns the definition of a limit to some extent or another. Clearly, for any given ε > 0, we need to make |x-3||x+4| < ε by limiting the size of |x-3| and |x+4|; however, from the definition of a limit, we can only directly limit the size of one of these terms. So, let's examine the definition of a limit as x tends to 3 ...

    Definition: limx→3f(x) means that for every ε > 0, we can find a δ > 0, such that for all x, if 0 < |x-3| < δ, then |f(x) - L| < ε.

    Since |x-3| is the only term that we can really limit the size of directly, we're going to see if we can put an upper bound on how big |x+4| is by making |x-3| sufficiently small. This explains why we're working with |x-3| and not |x+4|; the definition of a limit as x tends to 3 tells us how big |x-3| is, but it doesn't tell us much of anything (directly) about the size of |x+4|. By the way, based on your post, it sounds like you're confused about some aspects concerning the definition of a limit and exactly what it means. Even if you don't think that you need to, and even if you don't really need to (the two are different things), it is still a good exercise to go back and work through that definition. It will only help you understand things better.

    Second, I'll explain why we add 7 (which is really quite simple). First, we limit the size of |x-3| by making it so that 0 < |x-3| < 1. One consequence of the definition of an absolute value is that -1 < x-3 < 1 from the first inequality (note: There are no absolute value signs around x-3). Now, based on the size of |x-3|, we want to see if we can say something about how big |x+4| is. To do this, note that I need to add 7 to x-3 in order to change it into x+4. This is where the 7 comes from. Now, since we need to add 7 to the x-3 term, we also need to add 7 to the remaining terms in order for the inequality to hold; thus, we have -1 + 7 < x-3 + 7 < 1 + 7 or 6 < x+4 < 8. This means that |x+4| < 8.

    Hopefully this clears up some of your questions.
  12. Aug 29, 2010 #11
    Ok, I understand it now, except one thing, and you probably explained it a few times, but WHY |x-3|? Why is this the only factor that we can find the limit to directly? I think the thing that confused me is because x-c is x-3, and that's less than [tex]\delta[/tex], that's also a factor. But if I'm wrong here, please correct me if I'm wrong or still misunderstanding something. Sorry for being so stubborn, but the only thing I don't understand now is why |x-3|, and not |x+4|, or is it |x-3| because that's the x-c part of the definition of a limit. That's what I'm asking, things just got mixed up. Sorry.
  13. Aug 29, 2010 #12
    You're finding the limit as x approaches a value. For your problem, x --> 3. Hence, you set c=3 in the generalized inequality 0<x-c<[tex] \delta. [/tex] To put in more formal mathematical words, we want to prove that (as jgens said above):

    "[tex] \forall \epsilon > 0, \ \exists \delta > 0 \ \ni (\forall x)(0 < |x - c| < \delta \ \rightarrow \ |f(x) - L| < \epsilon).[/tex]"
  14. Aug 29, 2010 #13


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    The reason that we work directly with the term |x-3| and not the term |x+4| is due entirely to the definition of a limit. The definition says that we can have 0 < |x-3| < δ, but it says nothing directly about the size of |x+4|. This is why we work first with the term |x-3| to find an upper bound for |x+4|.
  15. Aug 29, 2010 #14
    Ah, thanks guys, I just got things mixed but because x-3 is also a term in the function. Thanks.
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