Limit evaluation

  • #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
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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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