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Homework Help: Epsilon Delta proof help

  1. Oct 3, 2006 #1
    I'm supposed to prove that lim x -> -2, x^2+3x+7 = 5

    Here's what I have:

    |x^2 +3x+7 – 5| < ε

    |x+2| < δ


    |x^2+3x+2| -> |(x+2)(x+1)| < ε whenever |x+2| < δ

    |x+1||x+2| < δ |x+1|

    |x+1||x+2| < δ|x+1| < ε

    ε / |x+1| > δ , as x -> -2, |x+1| -> 3,

    therefore: ε / -1 > δ

    But I'm not sure if I did it correctly, since its our first time using quadratic. Help, please?
     
  2. jcsd
  3. Oct 3, 2006 #2

    StatusX

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    So what's your final answer? Remember, given an ε>0, you're supposed to find a δ such that ... Thus δ should be written as a function of ε. So what is your δ?
     
  4. Oct 4, 2006 #3
    I thought that was enough...where did I go wrong?
     
  5. Oct 4, 2006 #4

    StatusX

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    I'll ask again. What is delta?
     
  6. Oct 4, 2006 #5
    I get delta = min {1, e/0}
     
  7. Oct 4, 2006 #6

    StatusX

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    e/0? That doesn't make sense. You're pretty close to the answer in your first post.

    This part is mostly correct. You've shown that if |x+2|<δ and δ<ε/|x+1|, then |(x2+3x+7)-5|<ε. What you need to do is find δ in terms of ε only (not x) such that if |x+2|<δ then |(x2+3x+7)-5|<ε. How can you replace |x+1| in the expression you found with a constant that maintains the inequality?
     
  8. Oct 4, 2006 #7
    We need to find such a number that restricts x to lie in some interval centered at 2, right?
     
  9. Oct 4, 2006 #8

    StatusX

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    Well, yea, that's what delta is. The problem is finding delta.
     
  10. Oct 4, 2006 #9
    since it's a value close to -2, can we assume that it is within a distance 1 from -2, that is, -1<x+2<1 , -3 < x < -1

    Here's where I was confused: Do I add 1 since the other value was x + 1? so, I'd be -2 < x + 1 < 0 ?or x + 1 < 0 ?
     
  11. Oct 4, 2006 #10
    Then it'd be delta = e / 0
     
  12. Oct 4, 2006 #11

    matt grime

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    e/0 does not make any sense.
     
  13. Oct 4, 2006 #12

    StatusX

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    You can just enforce that δ<1/2, so that you don't need to worry about |x+1|=0.
     
  14. Oct 4, 2006 #13
    It should be e / 1, since the limit is approaching -2, then -1 < x+2 < 1, or x+2 < 1, so delta = 1 and e / 1 ?
     
  15. Oct 4, 2006 #14

    StatusX

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    No, because -2-δ<x<-2+δ, so x may be less than -2. If you set δ<=1/2 as I suggested (ie, whatever expression you might end up with, just put δ=min{1/2,...}), then you know that 1/2<=|x+1|<=3/2 (the bound might end up even better than this, but this is all you need).
     
    Last edited: Oct 4, 2006
  16. May 21, 2009 #15
    How and why did you write the line:

    |x+1||x+2|<δ|x+1|

    It seems to come out of nowhere. I don’t understand.
     
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