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!

Confirm Precise Definition of a Limit solution

  1. Jul 3, 2008 #1
    As the subject says, I have "proved" a function, but because this type of problem seems to have no set answer, I'd like some opinions on whether I walked through it properly or not.

    1. The problem statement, all variables and given/known data
    lim(x=> -2) (x^2-1) = 3

    2. Relevant equations
    Precise Definition of a Limit

    3. The attempt at a solution
    Part 1. Assume a value for [tex]\delta[/tex]
    Since 0 < |x + 2|<[tex]\delta[/tex],

    |f(x) - 3| < [tex]\epsilon[/tex]
    |f(x) - 3| = |x^2 - 1 - 3|
    = |x^2 - 4| = |x + 2||x - 2|< [tex]\epsilon[/tex]

    Let C = |x - 2|, which leads to C|x + 2| < [tex]\epsilon[/tex]

    |x + 2| = [tex]\delta[/tex] < [tex]\frac{\epsilon}{C}[/tex]

    Applying the Precise Definition of a Limit,

    For a [tex]\delta[/tex] value [tex]\frac{\epsilon}{C}[/tex] greater than zero, there exists [tex]\epsilon[/tex] greater than zero such that if 0 < |x + 2|<[tex]\delta[/tex], then |f(x) - 3| < [tex]\epsilon[/tex].

    |f(x) - 3| < [tex]\epsilon[/tex]
    |f(x) - 3| = |x^2 - 1 - 3|
    = |x^2 - 4| = |x + 2||x - 2|< [tex]\epsilon[/tex]

    Re-apply C and [tex]\delta[/tex] to get

    C x [tex]\frac{\epsilon}{C}[/tex] < [tex]\epsilon[/tex]

    Hence, by the Precise Definition of a limit, said limit does exist.

    Much thanks in advance. Comments and criticisms are always welcome.
     
  2. jcsd
  3. Jul 3, 2008 #2
    It looks like you're doing this backwards. Your proof should start like "Given [itex]\epsilon > 0[/itex] there exists [itex]\delta > 0[/itex] such that if [itex]0<|x+2|<\delta[/itex], then [itex]|f(x)-3|<\epsilon[/itex]..." Now show that this is true using the $\delta$ that you found.
     
  4. Jul 3, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You can't do that. |x-2| is a variable, not a constant
    what you need is a BOUND on |x-2|. If C is close to -2, say between -1 and -3, how large or how small can |x-2| be?

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

Have something to add?
Similar Discussions: Confirm Precise Definition of a Limit solution
Loading...