- #1
ramyfishler
- 3
- 0
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
[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