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Lim as b approaches infinite -e^(-b) = 0 ? 0__o

  1. Apr 1, 2012 #1
    I'm doing improper integrals and my book is saying that

    lim as b approaches infinite of
    [tex] -e^{-b} + 1[/tex]
    [tex] = 1 [/tex]

    This doesn't make any sense to me. I would think that it would be

    [tex] -e^{-b} + 1[/tex]
    [tex] = - ∞ + 1[/tex]
    [tex] = - ∞ [/tex]

    Help is appreciated.
     
  2. jcsd
  3. Apr 1, 2012 #2
    [tex] -e^{-b} = \frac{-1}{e^b}[/tex]Thus you have a fraction whose denominator is getting really large.
     
  4. Apr 1, 2012 #3

    rock.freak667

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    As b→∞, eb→∞ as well. That part you know.

    The part that you are missing is that e-b=1/eb.

    So as eb tends towards infinity, you have 1/eb tending towards zero.

    (If you need to check, calculate 1/e,1/e2,1/e5.1/e100 and you will see it gets smaller and smaller until it approaches zero)
     
  5. Apr 1, 2012 #4
    [itex]b[/itex] is getting really big, right? Doesn't [itex]e^{-b} = \frac{1}{e^b}[/itex].

    Do you see it, now?
     
  6. Apr 1, 2012 #5

    SammyS

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    [itex]\displaystyle e^{-b}= \frac{1}{e^{b}}\ \ \to\ \ \frac{1}{\infty}\,. [/itex] (Please excuse the offensive notation.)
     
  7. Apr 1, 2012 #6
    Offensive notation excused. :redface:

    Thanks all for the quick reply.
     
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