# Limit evaluation

1. Jul 22, 2008

### ramyfishler

1. calculate the limit of the following function as m$$\rightarrow$$0

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

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

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. Jul 22, 2008

### Dick

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?