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Lhopital's rule limit help

  1. Dec 7, 2007 #1
    1. The problem statement, all variables and given/known data

    Lim[x to infinity] (X^1000+10)/(e^x+1)

    It seems like the answer is zero but I'm not exactly sure why it is.
    if you use lhopital's rule you get both numerator and demoniator huge numbers
    (or is it right that if you take the derivative for both of them you get a constant
    on the top and infinity at the bottom?)

    Please help me!!

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 7, 2007 #2
    from what i'm seeing

    [tex]\lim_{x \rightarrow \infty}\frac{x^{1000}+1}{e^{x}+1}[/tex]


    so we have infinity/infinity and thus we can apply L'Hopital's method. well from what you said, it's true, we will get huge numbers on the numerator but if we do this 1000 times, we'll eventually get

    [tex]\lim_{x \rightarrow \infty}\frac{1000!}{e^{x}}[/tex]

    our limit is 0 b/c the derivative of e is simply itself.
    Last edited: Dec 7, 2007
  4. Dec 7, 2007 #3
    l'hopital's rule is if the numerator and denominator both go to infinity or 0 then if you take the limit of there ration it's equal to the limit of the ratio of there derivatives. In this case both numerator and denominator go to infinity so you can use this rule.
  5. Dec 7, 2007 #4
  6. Dec 8, 2007 #5

    Gib Z

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    Homework Helper

    Exponentials grow much faster than any finite polynomial. It's not so easy to actually calculate when the denominator becomes larger than the numerator (You need to use the Lambert W function), but as long as we know exponentials grow faster than polynomials, as x --> infinity the quotient goes to 0.
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