How do I do this limit?

Question

I have the following equation

$$ S = 3Nk\left[ ln \left( \frac{e^{\Theta/T}}{e^{\Theta/T}-1} \right) + \frac{\Theta/T}{e^{\Theta/T}-1} \right]$$

And I need to evaulate it in the limit as $T \rightarrow 0$.

I have used the approximation $e^{\Theta/T} \approx 1 + \frac{\Theta}{T}$ (Taylor Expansion to first order).

Using this and rearranging I get that $S \rightarrow 3Nk$ but this is incorrect.

The correct answer is:

$$S \rightarrow 3Nk\frac{\Theta}{T}e^{-\Theta/T} \rightarrow 0$$

Can you explain to me why? I must be missing something here.

EDIT: As one of you pointed out, my professor clearly meant: $$S \sim 3Nk\frac{\Theta}{T}e^{-\Theta/T} \rightarrow 0$$

You are using the expansion of the exponential at 0 while you should be interested in the limit at +oo since Theta/T goes to +oo when T goes to 0. — Did, Mar 23 '14 at 17:25
Furthermore, in the "correct answer" you cite, the intermediate step is absurd since it depends on T and a limit when T goes to 0 cannot depend on T. — Did, Mar 23 '14 at 17:27
@Did INDEED! This comes from my lecturer's notes. It made me wonder too; it is absurd. — turnip, Mar 23 '14 at 17:31
@Did Perhaps, although I can't seem to see how the intermidiate step is obtained still. — turnip, Mar 23 '14 at 17:36

Did · Accepted Answer · 2014-03-23T17:40:57.063

1

Hints:

Let $x=\Theta/T$ with $\Theta\gt0$. When $T\to0^+$, $x\to+\infty$.
When $x\to+\infty$, $\mathrm e^x/(\mathrm e^x-1)\to1$ hence $\log(\mathrm e^x/(\mathrm e^x-1))\to0$.
When $x\to+\infty$, $x/(\mathrm e^x-1)\to0$.
Ergo?

edited Mar 23 '14 at 17:40

answered Mar 23 '14 at 17:29

Did

1 Answers1