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Limit help

  1. Dec 24, 2007 #1
    [tex]\lim_{n\rightarrow\infty}\int_0^1 e^{t^n}dt[/tex]. Im not really sure where to start to evaluate this limit, but I probably have enough tricks up my sleeve to solve it if someone knowledgeable is able to point me in the right direction. My usual integral tricks seem to fail here.

    Cheers.
     
  2. jcsd
  3. Dec 24, 2007 #2
    I think you can get [tex]e^{t^n}[/tex] as close as you want to [tex]e^0=1[/tex] on [0,d), for any 0 < d < 1. Then take the limit as d -> 1.
     
  4. Dec 24, 2007 #3

    arildno

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    You might do it this way:
    [tex]e^{t^{n}}=\sum_{i=0}^{\infty}\frac{t^{ni}}{i!}[/tex]
    Thus, switching about the summation limit and the integral (acceptable for any fixed n), we get:
    [tex]\int_{0}^{1}e^{t^{n}}dt=\sum_{i=0}^{\infty}\frac{t^{ni+1}}{(ni+1)i!}\mid_{0}^{1}=\sum_{i=0}^{1}\frac{1}{(ni+1)i!}[/tex]
    This should converge to 1 as n jumps into the air,
    since for i>=1, we have:
    [tex]\frac{1}{(ni+1)i!}<\frac{1}{n}*\frac{1}{i!}[/tex]
     
    Last edited: Dec 24, 2007
  5. Dec 24, 2007 #4
    Cool, thanks for the help! Also, I think you mean it converges to 0? The replies have made me wonder about changing the limits on the integral to be from a to b, since it seems that the 0 to 1 case converges quite nicely, I should have thought to do a taylor series :blushing:, oops, and im sure there is some nice upper limit for which this converges.

    Thanks again!


    Oops, nope it is 1!
     
    Last edited: Dec 24, 2007
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