1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A simpler epsilon-delta proof

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

    Prove that

    \lim_{x \to 2} x^2 + 5x -2 = 12

    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

    0<|x-2|<\delta \Rightarrow |f(x) - 12| < \varepsilon

    = x^2+5x-2-12\\
    = (x+7)(x-2)

    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


    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})
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: A simpler epsilon-delta proof
  1. Epsilon-Delta Proof (Replies: 3)

  2. Delta epsilon proof (Replies: 5)