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Another limit with l'hospital

  1. May 19, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] \lim_{u \to 1} \frac {(u-1)^{3}}{u^{-1}-u^{2}+3u-3} [/tex]

    3. The attempt at a solution

    So the first derivative is:

    [tex] \lim_{u \to 1} \frac {3(u-1)^{2}}{-u^{-2}-2u+3} [/tex]

    this limit also gives a 0/0 situation so take the derivative again:

    [tex] \lim_{u \to 1} \frac{6(u-1)}{2u^{-3}-2} [/tex]

    and this limit as well gives a 0/0 again, so would I do it again? or whats the next step? The book doesn't give me any instruction to this situation just to the 2nd derivative.

    If I do take the derivative again it is -1, is that correct?
  2. jcsd
  3. May 19, 2010 #2


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    There's no need to use L'Hopital's rule here. Just start by multiplying both numerator and denominator by [itex]u[/itex], and then factor the denominator (you already know [itex]u=1[/itex] is a factor, so it should be easy)
  4. May 19, 2010 #3
    right well its from the section in my book to learn l'hopital's rule, so that is why I am using it here.
  5. May 19, 2010 #4


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    In that case, yes, just differentiate one more time to get -1.
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