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Just a limit

  1. Nov 14, 2006 #1

    quasar987

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    I'm having a blank here. How do I evaluate

    [tex]\lim_{T\rightarrow 0^+} \ln(2e^{-\epsilon/kT}+1) \ \ \ \ \ \ \ (\epsilon>0, \ \ k>0)[/itex]

    ?!?
     
    Last edited: Nov 14, 2006
  2. jcsd
  3. Nov 14, 2006 #2
    Well, e^-(1/x) tends to 0 with x to 0, so our whole expression tends to log(0+1)=0. (Of course, we need k and epsilon positive, and we really should take a right-handed limit, since for negative T the expression blows up with T to 0.)
     
    Last edited: Nov 14, 2006
  4. Nov 14, 2006 #3

    quasar987

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    What justifies your moving the limit inside the log?

    (I refined the expresion following your comments, thx)
     
    Last edited: Nov 14, 2006
  5. Nov 14, 2006 #4
    Continuity of ln away from 0.
     
  6. Nov 14, 2006 #5

    quasar987

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    Oh I have it. It's that

    [tex]\lim_{x\rightarrow x_0} (f\circ g)(x)[/tex]

    will equal

    [tex]f(\lim_{x\rightarrow x_0} g(x))[/tex]

    if the limit of g exists and it f is continuous as [itex]\lim_{x\rightarrow x_0} g(x)[/itex].
     
  7. Nov 14, 2006 #6
    Most textbooks define continuity this way.
     
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