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Homework Help: Precise definition of a limit

  1. Oct 1, 2006 #1
    Hello,

    I'm having trouble with a step;I was wondering if someone could shed some light on me. Thanks.

    Prove that lim x->-2 (x^2-1)=3

    0<|f(x)-L|<epsilon whenever 0<|x-a|<delta

    0<|(x^2-1)-3|<epsilon whenever 0<|x-(-2)<delta
    = |x^2-4|<epsilon
    = |(x-2)(x+2)|<epsilon
    = |x-2||x+2|<epsilon
    = If |x-2|<C (C=constant) then |x-2||x+2|<C|x+2|
    = C|x+2|<epsilon = |x+2|<epsilon/C

    Assume |x+2|<1 so -1<x+2<1 = -3<x<-1 = -5<x-2<-3

    After this step -5<x-2<-3, I'm inclined to set |x-2|<-3, but book states that |x-2|<5. I'm lost at this step, should it be|-5|? Anyways, any explanations would be great.

    Thanks,
    Chris
     
  2. jcsd
  3. Oct 2, 2006 #2

    Galileo

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    You've shown that: |x+2|<1 <=> -5<x-2<-3
    This implies that |x-2|<5, simply because x-2<-3 means that surely x-2<5, so that -5<x-2<5, or |x-2|<5.

    You wrote |x-2|<-3, which is never true becuase the left side is positive and the right side isn't. Did you mean |x-2|<3?
     
  4. Oct 2, 2006 #3

    quasar987

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    Small detail that is important:

    You only want to prove that "|f(x)-L|<epsilon whenever 0<|x-a|<delta".

    In other words, f(x) can be = L.
     
  5. Oct 2, 2006 #4
    Perhaps this way would work:

    Prove that lim x->-2 (x^2-1)=3
    let E > 0 be given
    and:
    |f(x) - L|
    = |x^2 - 4|
    = |(x - 2)(x + 2)|
    = |x - 2||x + 2|
    if 0 < |x + 2| < 1 then,
    -3 < x < -1
    |x - 2| < 5
    < 5|x + 2|
    therefore:
    if 0 < |x + 2| < 1 and 5|x + 2| < E
    then by transitivity of <(less than), |f(x) - L| < E
    or reworded:
    if D = minimum(1, E/5), then |f(x) - L| < E

    (where E = epsilon; D = delta)
     
  6. Oct 2, 2006 #5
    Sorry this part:
    if 0 < |x + 2| < 1 then,
    -3 < x < -1
    |x - 2| < 5
    is a "side part"
     
  7. Oct 3, 2006 #6
    Alright this kinda clarifies my question, so -5 is like the min? Which would grant -5<x-2<5--since epsilon can never be E<0 - the absolute value is necessary? Sorry if I come off a bit slow, but it's just that I'm trying to justify every step so I completely understand the concept.
     
    Last edited: Oct 3, 2006
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