Confirm Precise Definition of a Limit solution

  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. 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. HallsofIvy

    HallsofIvy 40,673
    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?

     
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