# I Limit of cosh and sinh

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1. Mar 30, 2016

### thegirl

Hi I was wondering how you get this when taking the limit of T going to 0

From this expression of S:

Please help I don't see how ln infinity goes to uB/KbT (used u to represent the greek letter. And how does the other expression of sinh and cosh approach 1?

2. Mar 30, 2016

### phyzguy

Both sinh(x) and cosh(x) approach e^x as x->infinity. Do you see why? So the 1 term becomes negligible compared to the cosh term. Does this help?

3. Mar 30, 2016

### thegirl

is that because when x aproaches infinity cosh(x) = infinity and as e^x also approaches infinity when x approaches infinity it can be said that lim(x-> infinity) cosh(x) = (lim x-> infinity)e^x???? Also, thanks for replying!!

4. Mar 30, 2016

### Ssnow

As $T\rightarrow 0$ we have that $\ln{\left(1+2\cosh{\left(\frac{\mu B}{k_{B}T}\right)}\right)}\sim \ln{\left(1+ e^{\left(\frac{\mu B}{k_{B}T}\right)}\right)}\sim \frac{\mu B}{k_{B}T}$ ...

5. Mar 30, 2016

### phyzguy

$$cosh(x) = \frac{e^x + e^{-x}}{2}$$

As x->infinity the e^(-x) term approaches zero and becomes negligible compared to the e^x term.