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Unusual Limit involving e

  1. Jan 18, 2016 #1
    This was just very basic, I have accepted it in just a heartbeat, but when I tried to chopped it and examined one by one, somethings fishy is happening, this just involved [tex]\int_{0}^{\infty}x e^{-x}dx=1.[/tex]

    Well, when we do Integration by parts we will have [tex]let[/tex] [tex] u = x [/tex] [tex] du = dx [/tex] [tex]dv = e^{-x}dx[/tex] [tex]v = -e^{-x}[/tex] so that would give us [tex]\int_{0}^{\infty}x e^{-x}dx= - x e^{-x} + \int_{0}^{\infty} e^{-x}dx [/tex] evaluating the second term will just yield 1, so, [tex]\int_{0}^{\infty}x e^{-x}dx= - x e^{-x} + 1,[/tex] the second term should yield zero by employing limits as x approaches infinity, [tex] \lim_{x\rightarrow \infty }xe^{-x}=0 [/tex] but whoa, it didn't. I will yield instead [tex]1/e[/tex] Where did I went wrong?
     
  2. jcsd
  3. Jan 19, 2016 #2

    Mark44

    Staff: Mentor

    It's not clear what you're asking.
    ##\lim_{x \to \infty} xe^{-x} = 0## is correct. Are you getting 1/e for this limit? If so, that's where you went wrong.
     
  4. Jan 19, 2016 #3
    I get it now, I am imputing the wrong variable from my computer, indeed lim xe^{-x} as x approaches infinity indeed gives 0.
     
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