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Using the precise definition of a limit (epsilon & delta)

  1. Jun 23, 2015 #1
    I'm trying to practise, precise definition of a limit (epsilon & delta)
    Just to check I'm along the right lines here's a previous question to the one I'm stuck on

    If epsilon > 0 then there is delta >0 ..... All that introduction stuff, then
    Lim x-> 2 (3x-1) =5
    Hence
    |x-2| < delta then |3x - 6| < epsilon
    >> |x-2|< epsilon/2
    >> delta = epsilon/2
    >> epsilon = 3delta

    On my other problem I have the same setup but f(x) = 4x^2 +2 & L=18
    So
    |4x^2 -16| < epsilon |x-2|<delta
    I've tried /4 but
    I'm just not sure how I can remove the squared from the left of epsilon
    Any help would be great
     
  2. jcsd
  3. Jun 23, 2015 #2

    SammyS

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    I should have mentioned the above typo.

    Linear functions work out nicely. Quadratic functions require somewhat more effort.

    I assume that for the limit involving 42 +2, you also have x → 2 .
     
  4. Jun 23, 2015 #3
    Awh sorry missed that. Yeah it's
    Lim. (4x^2 +2) = 18
    X--> 2
     
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