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$$ \lim_{T \to \infty} \frac{1}{e^{\frac{\hbar\omega}{kT}}-1} = \frac{kT}{\hbar\omega} $$

I plugged the limit into mathematica and got "DirectedInfinity". Tried the trick of multiplying it by 1 and see if something more elucidating would come up but nothing.

The book I got this from says that this limit is for "high T", which I interpreted as T goes to infinity. Maybe that could be the problem, nonetheless, I don't see how this limit is done.

user17338
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  • You are correct. The limit as you have written it is $\infty$ provided $\hslash, k>0$. –  May 09 '13 at 17:58
  • Well I still would like to see how the statement in the book is true that at high T the expression reduces to kT/hw. – user17338 May 09 '13 at 18:01

2 Answers2

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The problem is that the right hand side goes to $\infty$ as well for $T\to\infty$. Thus, the claim of the book must really be formulated as follows:

$$\lim_{T\to\infty}\frac{\hbar\omega}{kT}\frac{1}{e^{\frac{\hbar\omega}{kT}}-1}=1.$$

To check this limit substitute $x = \frac{\hbar\omega}{kT}$ and write

$$\lim_{x\to 0}\frac{x}{e^x-1}.$$ By l'Hôpital this equals $$\lim_{x\to 0} \frac{1}{e^x} = 1.$$

Abel
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The limit doesn't converge as given. It looks more like an asymptotic approximation as $T\to\infty$: $$ \frac{1}{e^{\frac{\hbar\omega}{kT}}-1} \sim \frac{kT}{\hbar\omega} $$ The actual limit is $$ \lim_{T\to\infty}\frac1{T\left(e^{\frac{\hbar\omega}{kT}}-1\right)}=\frac{k}{\hbar\omega} $$ which follows from applying L'Hospital to $$ \lim_{x\to0}\frac{e^{ax}-1}{x}=a $$

robjohn
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