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Homework Help: Limit and l'hopital's rule

  1. Feb 21, 2010 #1
    lim as x approaches 4 from the right of [3(x-4)]^(x-4)

    when you plug in 4 it becomes a 0^0. So I attempted to manipulate it so it would become a 0/0 form so I could l'hopital it. (sorry if I keep spelling it wrong)

    so lim(x-4)ln(3x-12)=lny
    then I l'hopitaled it:
    lim (x-4)(1/3x-12)*3+ln(3x-12)=lny
    lim 1+ln(3x-12)=lny
    lim e + (3x-12)=y
    so y=e

    I do not know if my logic makes sense or not. If anyone sees any flaws in my logic, I would really like to know.
  2. jcsd
  3. Feb 21, 2010 #2
    why don't you try this:


    lhopital this until you reach something that is defined.
  4. Feb 21, 2010 #3
    hmmm ok so I get that the limit instead aproaches 0, however I look on my calculator and it should instead approach -infinity.
    I had that: ln(3x-12)/1/x-4
    so l'hopital it to get:
    1/3x-12 * 3/(1/(x-4)^2)=0/0
    so I l'hopitaled it again:
    so that the limit approaches 0. So where did I go wrong?
  5. Feb 21, 2010 #4


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    Use some more parentheses when you write this stuff, ok? It's hard to figure out whether 1/3x-12 is supposed to (1/(3x))-12 or (1/3)x-12 or what you really want which is 1/(3x-12). Your first l'Hopital form isn't 0/0. It's infinity/infinity. But if you do some algebra on it and simplify, you'll see you don't have l'Hopital it again. And if you are getting 0 that means ln of your original limit is 0. What's the original limit?
  6. Feb 21, 2010 #5
    ok so if its: 1/(3x-12)*3/[1/(x-4)^2]
    I simplified it to get: 1/(x-4)/(1/(x-4)^2)
    so x-4 basically
    which gets me...infinity?
  7. Feb 21, 2010 #6


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    x approaches 4. Is that really infinity?
  8. Feb 21, 2010 #7
    0...gah I don't get this. Since on my calc it looks like it should approach 1
  9. Feb 21, 2010 #8
    when x --> 4+

    however y approaches 1

  10. Feb 22, 2010 #9


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    This is NOT L'Hopital. On the left you have just the limit of the derivative of ln y.

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