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I Limit of cosh and sinh

  1. Mar 30, 2016 #1
    Hi I was wondering how you get this when taking the limit of T going to 0
    Screen Shot 2016-03-30 at 16.44.47.png
    From this expression of S:
    Screen Shot 2016-03-30 at 16.49.30.png

    Please help I don't see how ln infinity goes to uB/KbT (used u to represent the greek letter. And how does the other expression of sinh and cosh approach 1?
  2. jcsd
  3. Mar 30, 2016 #2


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    Both sinh(x) and cosh(x) approach e^x as x->infinity. Do you see why? So the 1 term becomes negligible compared to the cosh term. Does this help?
  4. Mar 30, 2016 #3
    is that because when x aproaches infinity cosh(x) = infinity and as e^x also approaches infinity when x approaches infinity it can be said that lim(x-> infinity) cosh(x) = (lim x-> infinity)e^x???? Also, thanks for replying!!
  5. Mar 30, 2016 #4


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    As ##T\rightarrow 0## we have that ##\ln{\left(1+2\cosh{\left(\frac{\mu B}{k_{B}T}\right)}\right)}\sim \ln{\left(1+ e^{\left(\frac{\mu B}{k_{B}T}\right)}\right)}\sim \frac{\mu B}{k_{B}T}## ...
  6. Mar 30, 2016 #5


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    [tex] cosh(x) = \frac{e^x + e^{-x}}{2}[/tex]

    As x->infinity the e^(-x) term approaches zero and becomes negligible compared to the e^x term.
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