1. Not finding help here? Sign up for a free 30min 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!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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

Have something to add?



Similar Discussions: Precise definition of a limit
Loading...