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Math Help - Improper integral

  1. #1
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    Improper integral



    I need some help with the limit. I remember l'hospital's rule, but when I apply it; it really becomes a mess. I can't help thinking that I'm over thinking this... though it may just be as arduous as I think it is.

    edit: ignore the minus sign after [(e^(1/x))/x], rushed job in equation editor
    Last edited by ZeroAxiom; February 9th 2009 at 11:30 PM. Reason: accidental minus sign in the picture
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  2. #2
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    Quote Originally Posted by ZeroAxiom View Post


    I need some help with the limit. I remember l'hospital's rule, but when I apply it; it really becomes a mess. I can't help thinking that I'm over thinking this... though it may just be as arduous as I think it is.

    edit: ignore the minus sign after [(e^(1/x))/x], rushed job in equation editor
    This limit is +\infty. Another way to see this is to procede to the change of variable u=\frac{1}{x}: the integral becomes \int_0^1 \frac{e^{1/x}}{x}\frac{dx}{x^2}=\int_1^\infty u e^u du=+\infty, obviously (the function in the integral goes to +\infty when u\to\infty).
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