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Just another limit problem

  1. Apr 8, 2008 #1
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  2. jcsd
  3. Apr 8, 2008 #2
    You should have 0/0 not 6/0. In any case, try a trick similar to what you did with the radical in the denominator.
     
  4. Apr 8, 2008 #3
    Well if you plug the four in initially you get:

    2-2/(3-3)=0/0

    Your trick did not work nor does using the other radical as Mathdope suggests.

    But, never fear, 0/0 limits call for one man.....


    L'Hopital!

    Since it is indeterminate form, it is eligible for L'Hopital's rule. That should make it work.
     
  5. Apr 8, 2008 #4

    Dick

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    Multiplying by 2+sqrt(x) does so work. You can then cancel the factor that's going to zero in the numerator and the denominator.
     
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