Prove the average quantum mechanical energy using l'Hôpital's rule

Question

I am trying to prove that taking the limit as $h\to0$ for the average quantum mechanical energy $$\dfrac{hν}{e^{hν/kBT}−1}$$ yields the average classical energy, $kBT$. How would you use l'Hôpital's rule for this?

Answer 1

Using the fact that the derivative of $e^{h\alpha}$ is $\alpha e^{h\alpha}$, we find that

\begin{align*} \lim_{h \to 0} \dfrac{h\nu}{e^{h\nu / kBT} - 1} &= \lim_{h \to 0} \dfrac{\nu}{\dfrac{\nu}{kBT} e^{h\nu/kBT}}\\ &= \dfrac{\nu}{\frac{\nu}{kBT}} \\ &= kBT \end{align*}

were we have used that $e^0 = 1$ to evaluate the limit.

Answer 2

Remember the fundamental limit $\lim_{x\to 0}\frac{e^x-1}{x}=1$ so that your limit is $kBT$.