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Homework Help: Epsilon-Delta definition of a Limit: Question

  1. Sep 6, 2006 #1
    This may be a dumb question, but:

    Using the epsilon-delta definition of a limit, when x is approaching the maximum of a parabola, like with: [tex]\lim_{x\rightarrow0}-x^2[/tex], obviously any value of epsilon puts [tex]L+\epsilon[/tex] outside the range of f(x). So, I take it in this case only [tex]L-\epsilon[/tex] is considered (which made sense to me graphically since you now have two points where the function intersects this line)?

    I just need verification (or correction) on this because every example and problem in the two textbooks I have has given/displayed an L and an epsilon such that [tex]L-\epsilon[/tex] and [tex]L+\epsilon[/tex] both lie within the range of f(x).

    -GeoMike-
     
    Last edited: Sep 6, 2006
  2. jcsd
  3. Sep 6, 2006 #2
    no. you find the limit at a max/min just like you would at any other point. i think the only instance where the direction you approach a certain valuie matters is when you approcah from the right or left, not from above/below.
     
  4. Sep 6, 2006 #3

    0rthodontist

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    x always does take values between L + e and L - e for L = 0 and e an arbitrary positive number. The values all happen to be between L and L - e, with none greater than L, but this is extra information that you don't need to evaluate the limit.
     
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