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Epsilon-delta inequality help

  1. Jan 28, 2007 #1
    1. The problem statement, all variables and given/known data

    Show that, if [tex]0<\delta<1[/tex] and [tex]|x-3|<\delta[/tex], then [tex]|x^{2}-9|<7\delta[/tex]

    2. Relevant equations

    3. The attempt at a solution

    I know that I need to transform one of the inequalities into the form of the other to prove it, but I don't see how. I can plug in values and of course it works since the limit of x^2 as x approaches 3 is 9, but I don't see how to show it algebraically. Please help
    Last edited: Jan 28, 2007
  2. jcsd
  3. Jan 28, 2007 #2


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    |x²-9| = |x+3||x-3|<|x+3|d

    From there, how can you use the triangle inequality and the hypothesis of |x-3|<d and 0<d<1 to get the result?
  4. Jan 28, 2007 #3


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    Note that if |x-3|< [itex]\delta[/itex] and and [itex]\delta[/itex]< 1, then |x-3|< 1 so -1< x-3< 1. What does that tell you about x+ 3 and |x+3|?
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