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Limit evaluation

  1. Jul 22, 2008 #1
    1. calculate the limit of the following function as m[tex]\rightarrow[/tex]0

    [tex]\frac{\beta J}{sinh^{2}(\beta J m)}[/tex]-[tex]\frac{\beta 2 J (s+1/2)^{2}}{sinh^{2}(\beta 2 J m (s+1/2))}[/tex]

    2. [tex]\frac{\beta J}{sinh^{2}(\beta J m)}[/tex]-[tex]\frac{\beta 2 J (s+1/2)^{2}}{sinh^{2}(\beta 2 J m (s+1/2))}[/tex]

    3. I tried lupitals law after expressing the function in a 0/0 I also tried to expand sinhx to 1+x but I get infinity and the answer should be finit
  2. jcsd
  3. Jul 22, 2008 #2


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    sinh(x)=x+x^3/3!+... You just need the first term. Each term in your difference goes to infinity like a different constant times 1/m^2. The limit isn't finite. Unless m is not just a multiplicative factor. What is it?
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