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Easy L'Hopitals question

  1. Feb 17, 2004 #1
    Howdy, I'm stumped by a seemingly easy L'Hopitals limit question.

    [tex]\lim_{t\rightarrow 0}\ \frac{\ln t}{t^2-1}[/tex]

    I said the limit doesn't exist, but the asker claims it does. I tried various transformations so that I could use L'Hopitals theorm but with no success. I keep getting [tex]\frac{- \infty}{-1}[/tex] so I can't take the derivative of the top and the bottom to get the answer.

    Any help would be greatly appreciated.
  2. jcsd
  3. Feb 17, 2004 #2


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    I think that the problem statement has a minor(?) error, it should be limit for t->1, not 0. Then L'Hospital's can be used (to get 1/2).
  4. Feb 17, 2004 #3
    That's what I thought too, but the question was worded as I stated it...

    Anyway, thanks for the input.
  5. Feb 18, 2004 #4


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    As stated, L'Hopital's rule does not apply: the numerator goes to infinity and the denominator goes to 1 so the limit does not exist.

    The limits as x->1 or as x-> infinity both exist and can be done by L'Hopital. As mathman said, the limit at 1 is 1/2. The limit at
    infinity is 0.
  6. Feb 20, 2004 #5
    (-infinity)/(-1) is not an indeterminate form....it equals positive infinity

    so the limit can be computed by simply plugging zero and then getting positive infintity
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