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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?

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I assume you mean $e^{h\nu / kBT} - 1$ for the denominator? In that case, the derivative will simply give a factor of $\nu / kBT$ in the denominator, and cancellation gives the desired result. – user61527 Dec 18 '13 at 7:40
Yes, but how do you get the desired result. – user34039 Dec 18 '13 at 7:41
Can you take the derivatives of the numerator and denominator individually? Then what does L'Hospital's rule tell you? – user61527 Dec 18 '13 at 7:44
OKay, I remember now Thanks! – user34039 Dec 18 '13 at 7:44

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.

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

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But how does that help? – user34039 Dec 18 '13 at 8:17
Let $x=\frac{h\nu}{kBT}$ (for $h\to 0$ then $x\to 0$) so that your function is $kBT\frac{x}{e^x-1}=\frac{kBT}{\frac{e^x-1}{x}}$ and $kBT$ is a constant.... – alexjo Dec 18 '13 at 8:23

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