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L'hospital's rule

  1. Nov 18, 2007 #1
    i have to use l'hospital's rule to compute the limit of:

    lim (1-e^(3x))/sinx
    x to 0

    i got -3 bt i'm not sure

    and what type is it? is it ''infinity/infinity'' type?
  2. jcsd
  3. Nov 18, 2007 #2
    Looks like you applied L'Hopitals rule right to me - :)

    When you say you're not sure what "type" it is, I'm assuming you mean what kind of indeterminate form? If you look at the function in your limit, actually evaluated at the limit value (x = 0), what kind of form is it? Surely not infinity over infinity... :)
  4. Nov 18, 2007 #3
    It is common to taylor expand those kind of functions that become small so you can get something like this:
    (1-exp(3x))/sinx =(1-1-3x)/x +0(x^2)=-3.
    I dont think you can use de L'Hopital here because you have a pole.
  5. Nov 18, 2007 #4

    Ben Niehoff

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    This is incorrect and confusing.

    Taylor expansion and l'Hopital's rule are actually equivalent. And f/g does not have a pole at zero, because the limit is finite. 1/g has a pole, and that is precisely why you use l'Hopital's in the first place.
  6. Nov 19, 2007 #5

    Gib Z

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    Great post =]
  7. Nov 19, 2007 #6
    oh ok so wold the type be ''0/0'' ?
  8. Nov 19, 2007 #7

    Gib Z

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    Yes =]
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