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A simpler epsilon-delta proof

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that

    [tex]
    \begin{equation*}
    \lim_{x \to 2} x^2 + 5x -2 = 12
    \end{equation*}
    [/tex]

    2. Relevant equations



    3. The attempt at a solution

    We want to prove that given [tex]\varepsilon > 0[/tex], there exists a [tex]\delta[/tex] such that

    [tex]
    0<|x-2|<\delta \Rightarrow |f(x) - 12| < \varepsilon
    [/tex]

    [tex]
    \begin{equation*}
    f(x)-12\\
    = x^2+5x-2-12\\
    = (x+7)(x-2)
    \end{equation*}
    [/tex]

    So I have an (x-2) term there in the epsilon part. I don't know how to apply that information so that I can choose a delta. Suggestions please!
     
  2. jcsd
  3. Oct 19, 2009 #2

    lurflurf

    User Avatar
    Homework Helper

    write x+7 as (x-2)+9
     
  4. Oct 19, 2009 #3
    sorry, i don't follow

    so when i write x+7 as (x-2)+9 i get

    f(x) - 12
    = (x-2)^2 + 9(x-2)

    and it looks like it might be useful, but i don't know how to use it.
     
  5. Oct 19, 2009 #4
    can you use the fact that the the limit distributes of addition and products?
     
  6. Oct 19, 2009 #5
    i'm sorry i'm still confused. i have no idea where to go.
     
  7. Oct 19, 2009 #6
    never mind, i think i figured it out, but not with factoring it like that.

    i got [tex]
    \delta = min(1,\frac{\varepsilon}{10})
    [/tex]
     
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