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Thread: Limit_problem ass4_7

  1. #1
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    Smile Limit_problem ass4_7

    How I solve this ?

    $\displaystyle lim_{x \to \infty} (x^{18}e^{-0.0001x})$
    Last edited by gilyos; Dec 28th 2009 at 09:07 AM.
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  2. #2
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    Smile

    hi
    $\displaystyle \lim_{x\to \infty }\frac{x}{8\exp(0.0001x)}=0$
    hope it's right.
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  3. #3
    Super Member bigwave's Avatar
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    Quote Originally Posted by gilyos View Post
    How I solve this ?

    $\displaystyle lim_{x \to \infty} (x^18e^{-0.0001x})$
    do you mean

    $\displaystyle
    x^{18}e^{-0.0001x}
    $
    instead of

    $\displaystyle x^18e^{-0.0001x}$
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  4. #4
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    I didn't wrote this right

    $\displaystyle lim_{x \to \infty} (x^{18}e^{-0.0001x}) $
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  5. #5
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    This is an indeterminate form - $\displaystyle \frac{\infty}{\infty}$. You can use L'Hopital 18 times and you will be left with a scalar in the numerator, giving you the result that this limit is 0.
    Last edited by Defunkt; Dec 28th 2009 at 10:09 AM.
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  6. #6
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    But you should recognize that any thing of the form $\displaystyle P(x)e^{-x}$, where P(x) is any polynomial, will go to 0 as x goes to infinity. $\displaystyle e^{-x}$ "dominates" any polynomial.
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