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Limit used in stat mech, how to prove this?

  1. Sep 18, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]
    lim_{dt\rightarrow 0} [(1+ \alpha dt(e^{-ik}-1))^{1/dt}]^T = e^{\alpha (e^{-ik-1)T}
    [/tex]

    It's a well known property in statistical physics, I'm not sure how to prove it
    2. Relevant equations



    3. The attempt at a solution

    I know dt ->0

    and 1/dt -> infinity

    Which one converges faster? What test do I apply? I forgot everything.
     
  2. jcsd
  3. Sep 18, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Limit

    Yep, you forgot everything. lim x->0 of (1+ax)^(1/x) is e^a, right? You can prove that with l'Hopital. Is that what you forgot? What's 'a' in your problem?
     
  4. Sep 19, 2009 #3
    Re: Limit

    a is a constant.

    I thought l'hopital rule applies when you take the limit of some variable that appears in both the numerator and denominator.
    lim x->0 of (1+ax)^(1/x) is e^a
    why then is this the case?
     
  5. Sep 19, 2009 #4
    Re: Limit

    [tex]\lim_{x \rightarrow 0}(1+ax)^{1/x} = \lim_{x \rightarrow 0}e^{\frac{1}{x}\log{(1+ax)}} = e^{\lim_{x\rightarrow 0 }\frac{\log{(1+ax)}}{x}}[/tex]

    by the continuity of the exponential function. Now apply l'Hopital. This is a common manipulation.
     
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