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Limit Calc Test Problem

  1. Sep 5, 2010 #1
    1. The problem statement, all variables and given/known data

    This is a problem off a multiple choice practice test:
    lim t->3 ( 1/(t^2-3t) - 2/(t^2-9) =

    The solutions are:
    (a) 0 (b) -1/9 (c) -1/18 (d)1 (e) 1/3

    The correct answer is (c).

    Can someone explain to me how to solve it? Any help would be very appreciated. You don't need to explain how to do limits. I just need to know how to manipulate the equation so that it's not indeterminate.

    2. Relevant equations


    3. The attempt at a solution

    I tried multiplying by the conjugate with no success.
  2. jcsd
  3. Sep 5, 2010 #2


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    Staff Emeritus
    Science Advisor

    The first thing I would do is actually subtract the two fractions:
    [tex]\frac{1}{t^2- 3t}- \frac{2}{t^2- 9}= \frac{1}{t(t- 3)}- \frac{2}{(t- 3)(t+ 3)}[/tex]

    Clearly the "common denominator" is t(t- 3)(t+ 3):
    [tex]\frac{t+ 3}{t(t- 3)(t+ 3)}- \frac{2t}{t(t- 3)(t+ 3)}[/tex][tex]= \frac{t+ 3- 2t}{t(t- 3)(t+ 3)}= \frac{-t+ 3}{t(t- 3)(t+ 3)}= -\frac{t- 3}{t(t- 3)(t+ 3)}[/itex]
  4. Sep 5, 2010 #3
    Thank you!!
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