# Limit problem

So, I can't really find this limit:

$$\lim_{T \to \infty} \ 3Nk {(\epsilon/kT)}^2 \frac{e^{(\epsilon/kT)}}{{(e^{(\epsilon/kT)}-1)}^2}$$

This is actually the formula for the specific heat of an Einstein solid, which is pretty easy to derive but I haven't been able to calculate the limit to show it becomes Dulong-Petit at hight temperatures. Maybe I've been missing something simple... I don't see how you could use l'Hopital's rule.

note: I wasn't sure whether I should post this in one of the physics forums, but I figured it's more of a calculus problem at this point.

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I think you can use L'Hôpital's rule. First of all, I made the change of variables
$$y = \frac{\epsilon}{kT}$$
so that $y \to 0$ as $T \to \infty$. This yields
$$\lim_{y \to 0} 3Nk\frac{y^2 e^y}{(e^y - 1)^2}$$
which is of the form $\frac{0}{0}$. Using L'Hôpital's rule, this equals
$$\lim_{y \to 0} 3Nk \frac{2y e^y + y^2 e^y}{2(e^y -1)e^y} = 3Nk \lim_{y \to 0} \frac{2y + y^2}{2e^y -2}$$
which is once again of the form $\frac{0}{0}$. Applying LH again, we get
$$3Nk \lim_{y \to 0} \frac{2 + 2y}{2 e^y} = 3Nk \lim_{y \to 0} \frac{y +1}{e^y} = 3Nk \frac{0 +1}{1} = 3Nk$$

I hope this helps: hopefully I didn't make any mistakes!

I think you can use L'Hôpital's rule. First of all, I made the change of variables
$$y = \frac{\epsilon}{kT}$$
so that $y \to 0$ as $T \to \infty$. This yields
$$\lim_{y \to 0} 3Nk\frac{y^2 e^y}{(e^y - 1)^2}$$
which is of the form $\frac{0}{0}$. Using L'Hôpital's rule, this equals
$$\lim_{y \to 0} 3Nk \frac{2y e^y + y^2 e^y}{2(e^y -1)e^y} = 3Nk \lim_{y \to 0} \frac{2y + y^2}{2e^y -2}$$
which is once again of the form $\frac{0}{0}$. Applying LH again, we get
$$3Nk \lim_{y \to 0} \frac{2 + 2y}{2 e^y} = 3Nk \lim_{y \to 0} \frac{y +1}{e^y} = 3Nk \frac{0 +1}{1} = 3Nk$$

I hope this helps: hopefully I didn't make any mistakes!

That makes perfect sense. For some reason I forgot that there is also an inverse exponential in the numerator (I was thinking it was just an exponential). Thanks a bunch dude.