- #1

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**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**