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L hopital

  1. Jun 24, 2007 #1
    Im trying to solve this problem using l'hopital but amm not sure how to do it

    X->infinite x^3 * e^(-x^2)

    soo this infinite * e^-infinite... but from there im not sure if you can use it to solve this...
  2. jcsd
  3. Jun 24, 2007 #2
    you have to put expression as the ratio of two functions first

    like this

    [tex] \frac{x^3}{e^{x^2}}[/tex]

    then you can apply l'hospitale
  4. Jun 24, 2007 #3
    yeah i thought about that but amm not sure how much is e^-infinite
  5. Jun 24, 2007 #4


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    This question makes no sense.

    As ice109 said, apply L'Hospital's rule. Do you know what L'Hospital's rule is? If so, apply it to this question.
  6. Jun 25, 2007 #5


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    That sounds like you are saying "how infinite is it"! What you mean is "what number is e^-infinite". (Strictly speaking that is also meaningless- you cannot evaluate a function of real numbers "at infinity", you can only take limits at infinity.) You know, I hope, that e^x increases without bound (i.e. "goes to infinity") as x goes to infinity. You should also know that "e^(-x2) MEANS 1/e^(x2). If A goes to infinity, what does 1/A go to?
    Last edited: Jun 25, 2007
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