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Simple limit

  1. Apr 27, 2010 #1
    Hello, it is pretty obvious that the following limit is equal to zero:

    [tex]$Lim t \rightarrow \infty (\tmop{te}^{- t}) = 0$[/tex]

    For example, for t=100 it is [tex]100*e^{-100}[/tex]

    But how would you take this limit "rigorously"? I tried decomposing the function with a mclaurin series and [tex]te^-t[/tex] is equal to this series:

    [tex]$\sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1} t^n}{(n - 1) !}$[/tex]

    How would I actually evaluate this series for t->infinity???? Or is this the wrong approach?

    Also for a finite number of terms it appears that this series diverges...
  2. jcsd
  3. Apr 27, 2010 #2
    Errrr... L'hopital's rule. Sorry should have spent a while longer thinking about it before posting.
  4. Apr 28, 2010 #3


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    Write it as [itex]t/e^t[/itex] and use L'Hopital's rule as Nick R suggested.
  5. May 6, 2010 #4


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    Nick R = TS ;)

    A more direct proof: since [itex]e^x = 1 + x + x^2+ ... [/itex], it is obvious that [itex]e^x>x[/itex] for all [tex]x\in\mathbb{R}[/tex]. In other words, [itex]\frac{e^x}{x}>1[/itex]. Hence

    [tex]\frac{e^x}{x}=\frac{1}{2}\left(\frac{e^{x/2}}{x/2}\right)e^{x/2}>\frac{1}{2}e^{x/2}\to\infty[/tex] if [itex]x\to\infty[/itex].

    It follows that [itex]xe^{-x}=\frac{x}{e^x}\to 0[/itex] if [itex]x\to\infty[/itex].
  6. May 9, 2010 #5
    Ts = op?
  7. May 9, 2010 #6


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    I'm sorry, with TS I meant Topic (/Thread) Starter. Is OP (original poster?) more standard?
  8. May 9, 2010 #7
    We are mathematicians. We can call it whatever we want! But it is mandatory to use at least two of these:
    1) greek letter(s)
    2) subscript
    3) AlTeRnAtInG CaPs

    I recommend that we define [tex]\tau\sigma_{1}(399107)[/tex]:= {"Nick R"}
  9. May 9, 2010 #8


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    Hahaha that's a good one :rofl:
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