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Using the squeeze theorem to solve a limit

  1. Sep 2, 2011 #1
    1. The problem statement, all variables and given/known data
    I have a limit, which when substitution is used the indeterminate form 0/0 occurs. I've been asked to solve the limit by using the squeeze theorem. I really have no idea where to take this.


    2. Relevant equations
    Lim (x[itex]^2[/itex]-9)(x-3[itex]/[/itex]|x-3|)
    x[itex]\rightarrow[/itex]3


    3. The attempt at a solution
    I've tried so many attempts at this problem, posting all my work would be massive.
     
  2. jcsd
  3. Sep 2, 2011 #2

    SammyS

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    Are you sure the problem is what you state, which is equivalent to [itex]\displaystyle \lim_{x\to3}\left((x^2-9)\left(x-\frac{3}{|x-3|}\right)\right)\,,[/itex] or is it [itex]\displaystyle \lim_{x\to3}\left((x^2-9)\frac{x-3}{|x-3|}\right)\,?[/itex]
     
  4. Sep 2, 2011 #3
    It's your second one. I'm a bit confused by the format used on the forum.
     
  5. Sep 2, 2011 #4

    SammyS

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    It appears to me that simplifying the expression using algebra works better than using the squeeze theorem, but ...

    Certainly, [itex]\displaystyle (x^2-9)\frac{x-3}{|x-3|}\ge0[/itex] for x > 0.

    Can you show that [itex]\displaystyle (x^2-9)\frac{x-3}{|x-3|}< 8|x-3|[/itex] for 2 < x < 4 ?
     
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