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Homework Help: Limit proof help please!

  1. Feb 3, 2006 #1
    Hi,

    I'm currently working on a proof problem and I'm just totally clueless on what I've been asked to prove.

    The question is:

    Define the real valued function f by f(x) = x^3 and prove that f(x) --> -8 as x --> -2. [ use the epsilon and delta definition of a limit ]

    So far this is my attempt but I'm stuck where I'm trying to find epsilon. I'm not even sure my attempt is even correct :frown: Can someone help? Thanks in advance!

    http://www.tinypic.com/view/?pic=n49nqv"
     
    Last edited by a moderator: Apr 22, 2017
  2. jcsd
  3. Feb 4, 2006 #2

    VietDao29

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    Homework Helper

    Now you are looking for a [tex]\delta[/tex] in terms of [tex]\epsilon[/tex], such that:
    If [tex]0 < |x - (-2)| = |x + 2| < \delta[/tex] then [tex]|x ^ 3 - (-8)| < \epsilon[/tex], for some given [tex]\epsilon > 0[/tex], right?
    Now work backward, assume that:
    [tex]x ^ 3 - (-8) < \epsilon[/tex], we will try to rearrange it to make it look like: |x + 2| < something, and then we can let [tex]\delta = \mbox{something}[/tex], and finish our proof. Do you get it?
    [tex]|x ^ 3 + 8| < \epsilon[/tex], you need |x + 2|, so let's factor it.
    [tex]\Leftrightarrow |x ^ 3 + 2 ^ 3| < \epsilon[/tex]
    [tex]\Leftrightarrow |(x + 2) (x ^ 2 - 2x + 4)| < \epsilon[/tex]
    [tex]\Leftrightarrow |x + 2| \times |x ^ 2 - 2x + 4| < \epsilon[/tex], everything is positive, divide both sides by |x2 - 2x + 4|
    [tex]\Leftrightarrow |x + 2| < \frac{\epsilon}{|x ^ 2 - 2x + 4|} = \frac{\epsilon}{|(x - 1) ^ 2 + 3|} < \mbox{what ?}[/tex].
    Can you go from here? :smile:
     
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