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Tough limtis, answers involve e

  1. Feb 20, 2013 #1
    limit as n→∞ of [itex]\frac{(2t)^n}{n!}[/itex] and [itex]\frac{(-t)^n}{n!}[/itex]

    Answers are e2t-1 and e-t-1 but I don't know how to work them out, thanks.

    edit: btw these are series
     
  2. jcsd
  3. Feb 20, 2013 #2

    Dick

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    If those are series you should fix up your question showing the limits of the series. Do you know ##e^a=\Sigma^\infty_0 \frac{a^n}{n!}##?
     
  4. Feb 20, 2013 #3

    haruspex

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    Starting at n=1, it would seem? That would be unusual.
     
  5. Feb 21, 2013 #4

    HallsofIvy

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    These are easy if you know that
    [tex]\sum_{n=0}^\infty \frac{x^n}{n!}= e^x[/tex]
     
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