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Another Delta-Epsilon Question

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that lim f(x) x --> 0 = lim f(x-a) x --> a.

    2. Relevant equations

    0 < |x-a| < d
    0 < |f(x)-a| < E

    3. The attempt at a solution

    Spivak's logic, again, is giving me trouble. However, I realize that his method probably has a better chance of being right than mine. His method follows:

    Suppose that lim f(x) x --> a = l, and let g(x) = f(x-a). Then for all E > 0, there is a d > 0 such that, for all x, if 0 < |x-a| < d, then |f(x)-l| < E. Now, if 0 < |y| < d, then 0 < |(y+a) - a| < d, so |f(y+a)-l| < E. But this last inequality can be written |g(y) - l| < E. So lim g(y) y --> 0 = l.

    The bold portion is where I get lost. How can he rewrite |f(y+a)-l| < E as |g(y)-l| < E? Doesn't this suggest that f(y+a) = g(y)? However, wouldn't that conflict with the original definition of g(x) = f(x-a)? Wouldn't this definition mean g(y) = f(y-a), not f(y+a)? Where's my error?
     
  2. jcsd
  3. Jul 19, 2010 #2
    if its true for all x, its true for all y such that y=x-a for some fixed a, then we relabel the hypothesis on x with y+a. (hopefully if i'm interpreting your problem correctly)
     
  4. Jul 19, 2010 #3
    I think you are, but I don't understand how the bold part is possible.
     
  5. Jul 19, 2010 #4
    Hmmm it does look a bit fishy, even though it really shouldn't. Anyways I think it's a lot easier to start with lim f(x) as x -> 0, and then simply replace x with the obvious substitution. I mean that's really the intuitive idea. If x is going to 0, might as well replace it with x-a, and let x go to a instead. This kind of thing comes up a few times, e.g. characterizations of continuity, equivalent definitions of derivative, and so on.
     
  6. Jul 20, 2010 #5
    yes, i don't think this is Spivak's exact reasoning. you're assuming incorrectly either 'x' is near 'a' (it should be 'zero') or lim f(x-a) x-->a followed by g(x)=f(x-a), which is a little backwards and doesn't follow your reasoning.
     
  7. Jul 20, 2010 #6
    The thing is, Spivak specifically asks for a rigorous proof, which I'm assuming must mean deltas and epsilons.

    I'm sorry but I really can't tell what you're saying, not being intentionally dense, just...unintentionally dense.
     
  8. Jul 20, 2010 #7
    Oops, I forgot about that part about writing out the definition of the limit. So starting over, start with lim f(x) as x -> 0 = L, write out the definition, and make the obvious substitution. That is a rigorous proof, there is no need to define an extra function g.
     
  9. Jul 20, 2010 #8
    sorry i'm a little obtuse. just follow what snipez said. start with: let f(x)-->L as x-->0 (not x-->a).
     
  10. Jul 21, 2010 #9
    I'm still not getting this.

    lim f(x) as x-->0 = l
    |f(x) - l| < E
    |x| < d

    I'm trying to get |(x - a) - a| < d in order to get |f(x - a) - l| < E, right? But it seems like I can't get |(x - a) - a|, given that it =/= |x|. What's wrong, here?
     
  11. Jul 21, 2010 #10
    Well gettting 0 < |(x-a)-a| < d doesn't make much sense. In terms of the definition of the limit, writing that means you intend (x-a) to approach a, but you want x to approach a. If you don't see this, look at the definition of the limit again.

    It's really quite simple, and I'm not sure if you're taking a shortcut in not writing out the full definition of the limit. At the very least you should write 0 < |x| < d IMPLIES |f(x) - L| < E. You already understand that we need to get |f(x-a) - L| < E as our conclusion. We're given 0 < |x| < d IMPLIES |f(x) - L| < E. Can you see what we need to replace x by to get from the hypothesis to the conclusion?
     
  12. Jul 21, 2010 #11
    There's one of my misunderstandings. I'd been thinking about the definition as f(...) where ... as a whole approaches a, not just a single variable.

    Well, it seems like we'd have to replace x with (x - a), but I don't see how that's possible.
     
  13. Jul 21, 2010 #12
    I'm not exactly sure what your misunderstanding is. I mean lim f(x-a) x --> a = L means given E > 0 there exists d > 0 such that 0 < |x-a| < d IMPLIES |f(x-a) - L | < E, which of course is just an application of the definition of the limit.

    Explain to me why you think we can't replace x with x - a. One thing that's important to keep in mind is that we really do have an implication: 0 < |x| < d IMPLIES |f(x) - L| < E. This logical statement has to be taken in its entirety, and the implication cannot be ignored. I'll be happy to clarify in any case.
     
  14. Jul 21, 2010 #13
    i think you're confusing the dummy variables in the limit.

    (*) if f(y)-->L as y-->0 then f(x-a)-->L as x-->a

    but what is x?

    you are given several things. you are given the e,d-definition of a limit which is the hypothesis in (*) and you are given |x-a|<d in the conclusion
    therefore all you need is to show is that |f(x-a)-L|<e necessarily follows.
     
    Last edited: Jul 21, 2010
  15. Jul 21, 2010 #14
    Good point xaos, I thought of bringing that up but I thought the implication issue was the touchy part. At least when I first encountered this type of proof, I could see that replacing x with x - a seemed to be the right approach, but I was also unsure of why I could do that. Of course you could replace x with y - a if you're not used to working with dummy variables (psychologically it's helpful at first). But at first I was still unsure because I focused too much on why I could go from 0 < |x| < d to 0 < |x-a| < d. The error of course is in not treating the implication statement in its entirety.
     
  16. Jul 21, 2010 #15
    Let me see if I have this right: is your point that:

    x as x --> 0 = (x - a) as a --> 0

    and therefore substituting (x - a) for x is a valid move to make? (I think I can see how the rest follows).

    I guess the main issue here might be that I don't yet have a good feel for what constitutes "rigor" vs. what constitutes "hand-waving", since the two often seem very similar to me.
     
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