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Math Help - Fairly simple limit proof?

  1. #1
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    Fairly simple limit proof?

    How would I prove that

    \lim\limits_{x\to\infty}xe^{-x}=0 ?

    I've only done eps-delta proofs for finite limit points, not to infinity.
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  2. #2
    Super Member Deadstar's Avatar
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    Are you familiar with L'Hopitals rule?
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    Yes. I think I may know how to solve it then. -- Wait, but LHopitals rule is for finding the limit, not proving it?
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    Quote Originally Posted by scorpion007 View Post
    Yes. I think I may know how to solve it then. -- Wait, but LHopitals rule is for finding the limit, not proving it?
    Surely if you find that the limit is equal to zero by using a valid theorem like l'Hopital's Rule, then you have proved that the limit is equal to zero! But context might be important ..... Where has the question come from?
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  5. #5
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    It arose indirectly from me solving an integral of the form \int_0^\infty xe^{-x} where I had to evaluate the limit of a similar expression, and was curious if I could prove it using a method like e-d proofs.
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  6. #6
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    Quote Originally Posted by scorpion007 View Post
    It arose indirectly from me solving an integral of the form \int_0^\infty xe^{-x} where I had to evaluate the limit of a similar expression, and was curious if I could prove it using a method like e-d proofs.
    In that case, why use a cannon when a flyswatter will do?
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  7. #7
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    Only because I was curious if I could prove it -- not because it was strictly necessary to do so.
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  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by scorpion007 View Post
    How would I prove that

    \lim\limits_{x\to\infty}xe^{-x}=0 ?

    I've only done eps-delta proofs for finite limit points, not to infinity.

    A some kind of try:

    By definition...

    For every \epsilon>0 exist x>1 so that:

    |xe^{-x}|<\frac{1}{x}<\epsilon
    Last edited by Also sprach Zarathustra; June 27th 2010 at 05:53 PM.
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