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Problems with limits at infinity within improper integrals

  1. Jun 23, 2013 #1
    1. The problem statement, all variables and given/known data


    ##\int_{2}^{\infty} ue^{-u} du##


    3. The attempt at a solution

    What I did was find the family of functions described by the indefinite integral ##\int ue^{-u} du## then found the limit as b increases without bound. $$=\lim_{b\rightarrow \infty} -ue^{-u}-e^{-u}\mid_{2}^{b}$$ $$=\lim_{b\rightarrow \infty} \Big(-be^{-b}-e^{-b}\Big)~ -~ \Big(-2e^{-2}-e^{-2}\Big)$$ $$=\lim_{b\rightarrow \infty} e^{-b} \Big(-b-1\Big) + 3e^{-2}$$

    My problem is trying to remember when ##e^{-b}## is approaching 0 and ##\Big(-b-1\Big)## is getting infinitely more negative...do they "cancel" each other out? Does that part of the limit approach 0 instead of ∞? And then does that mean that the definite integral converges at ##3e^{-2}## ?
     
  2. jcsd
  3. Jun 23, 2013 #2

    tiny-tim

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    hi ninfinity! :smile:
    you're asking whether be-b -> 0

    that's b/eb

    what does eb/b -> ? :wink:
     
  4. Jun 23, 2013 #3
    Isn't that ∞/∞? That doesn't make sense.
     
  5. Jun 23, 2013 #4

    tiny-tim

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    what does b3/b2 –> as b -> ∞ ? :wink:
     
  6. Jun 23, 2013 #5

    Mark44

    Staff: Mentor

    ∞/∞ isn't a number; it's an indeterminate form, as is 0/0 and several others. If you take a limit and get one of these indeterminate forms, it means you aren't done yet.
     
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