Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Epsilon Delta Limit.

  1. Feb 15, 2012 #1
    For part A, (described here: http://www.cramster.com/solution/solution/1157440) I don't understand why they say |x-2| < 1 and why [itex]\delta[/itex] = min{1,ε/5}

    In case you can't view the page:

    lim x2+2x-5 = 3, x [itex]\rightarrow[/itex] 2
    Let ε > 0 and L = 3.

    |x2 + 2x -5 -3| < ε
    |x2 + 2x - 8| < ε
    |x+4||x-2| < ε

    If |x-2| < 1, x [itex]\rightarrow[/itex] (1,3) [itex]\rightarrow[/itex] x+2 [itex]\in[/itex] (3,5)
    |x-2| < 1 [itex]\rightarrow[/itex] |x+2| < 5
    |x-2| < 1 [itex]\rightarrow[/itex] |x2 +2x - 8| = |x-4||x+2| < 5|x-4|

    For ε > 0, 5|x-4| < ε [itex]\rightarrow[/itex] |x-4| < [itex]\frac{ε}{5}[/itex]

    If [itex]\delta[/itex] = min{1,[itex]\frac{ε}{5}[/itex]}
    |x-2| < [itex]\delta[/itex] [itex]\rightarrow[/itex] |x2 + 2x - 8| < 5|x-4| < ε
     
    Last edited: Feb 15, 2012
  2. jcsd
  3. Feb 15, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi ƒ(x)! :smile:
    because if δ < 1, then |x+4| < 5

    and if δ < ε/5, then |x-2| < ε/5 (obviously!)

    so |x+4||x-2| < 5*ε/5 = ε :wink:

    (though i don't see why they say |x-2| < 1 either :redface:)
     
  4. Feb 15, 2012 #3
    Can you explain yourself a little further? I tend to be rather slow about this kind of thing.
     
  5. Feb 15, 2012 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    you want the product to be < ε

    one easy way to do this (there are others) is to make one of them < 1 and the other < ε

    but that won't work in this case, since |4+x| isn't going to be < 1

    however, it will be < 5,

    so we make one of them < 5 and the other < ε/5 :smile:
     
  6. Feb 15, 2012 #5
    So, for a different problem, part (c) on the link, I have:

    lim x^3 + 2x + 1 = 4, x --> 1

    Let ε > and L = 4

    |x-1| < δ, |x^3 +2x - 3| < ε

    But this one doesn't factor...[EDIT] wait, yes it does (I peeked at the solution, how do I factor this by myself?).

    |x-1||x^2 + x + 3| < ε
     
  7. Feb 16, 2012 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi ƒ(x)! :smile:

    (just got up :zzz: …)

    ok, so |x-1| < δ,

    and you can obviously choose x near 4 so that |x2 + x + 3| < 33 …

    so how would you finish the proof? :smile:
    you just have to guess …

    in an exam, they won't give you anything difficult, so start with ±1, and work your way upwards! :wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Epsilon Delta Limit.
Loading...