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An Error in Spivak's Calculus?

  1. Aug 10, 2011 #1
    An Error in Spivak's Calculus? [problem resolved]

    EDIT: Thanks for your help, guys. I see where I went wrong.

    At one point in the discussion of limits, he says:

    If |x-a| < 1, then |x| < |a| + 1.

    As far as I can tell, this is true if x and a are both positive, but not otherwise.

    Am I missing something here?

    (And no, there isn't really any other information regarding the x-value or the value of a.)
     
    Last edited: Aug 11, 2011
  2. jcsd
  3. Aug 10, 2011 #2

    disregardthat

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    It's true. |x|-|a| <= |x-a| < 1, so |x| < |a| + 1
     
  4. Aug 10, 2011 #3

    gb7nash

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    This is a result of the triangle inequality. We know for any real numbers a,b that:

    |a-b| >= |a| - |b| (do you know how to prove this using triangle inequality?)

    Now looking back at the problem, Since |x-a| < 1, this implies that |x| - |a| <= |x-a| < 1, so...
     
  5. Aug 10, 2011 #4
    Wow, thanks for clearing that up. I'm not sure what I was thinking before, but it makes sense now.
     
  6. Aug 11, 2011 #5
    |x - a| < 1
    -1 + a < x < 1 + a
    |x| < |1 + a| <= 1 + |a|
     
  7. Aug 11, 2011 #6
    Thanks, your explanation is terrific.
     
  8. Aug 11, 2011 #7
    This is quite a common trick in epsilon delta proofs.

    |x| = |x + a - a| <= |x - a| + |a| < |a| + 1

    this version requires only the standard triangle inequality without the need to break the absolute value up.
     
  9. Aug 11, 2011 #8

    I just realised there is an error in my proof :) sorry about that.

    |x| = |x + a - a| <= |x - a| + |a| < 1 + |a|

    This one posted above is abolutely correct.
     
  10. Aug 11, 2011 #9
    I only meant to do it so that a better understanding was possible but I erred.
     
  11. Aug 18, 2011 #10
    Isn't there an error (or some sloppiness, at least) in the subsequent paragraph though: shouldn't

    |x-a| < ε/(2|a|+1)​

    be

    |x-a| ≤ ε/(2|a|+1)​
     
  12. Aug 18, 2011 #11

    disregardthat

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    In what context? What subsequent paragraph are you talking about, what does it say?
     
  13. Aug 19, 2011 #12
    On the same page (93) as the rest of what's being discussed here (text line 18).
     
  14. Aug 19, 2011 #13
    I think you're right in that less than or equal to is sufficient. But when discussing epsilon - delta limits, it is simply easier to use strict inequalities without changing the end result too much.
     
  15. Aug 19, 2011 #14

    disregardthat

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    No, it's <. We require |x-a| < ε/(2|a|+1). You should copy the text, not everyone has the textbook.
     
  16. Aug 19, 2011 #15
    Yes, that's exactly what's going on. Thanks.
     
  17. Aug 19, 2011 #16
    The question is about the textbook (as your answer demonstrates).
     
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