# Just a limit

1. Nov 14, 2006

### quasar987

I'm having a blank here. How do I evaluate

$$\lim_{T\rightarrow 0^+} \ln(2e^{-\epsilon/kT}+1) \ \ \ \ \ \ \ (\epsilon>0, \ \ k>0)[/itex] ?!? Last edited: Nov 14, 2006 2. Nov 14, 2006 ### DeadWolfe Well, e^-(1/x) tends to 0 with x to 0, so our whole expression tends to log(0+1)=0. (Of course, we need k and epsilon positive, and we really should take a right-handed limit, since for negative T the expression blows up with T to 0.) Last edited: Nov 14, 2006 3. Nov 14, 2006 ### quasar987 What justifies your moving the limit inside the log? (I refined the expresion following your comments, thx) Last edited: Nov 14, 2006 4. Nov 14, 2006 ### slearch Continuity of ln away from 0. 5. Nov 14, 2006 ### quasar987 Oh I have it. It's that [tex]\lim_{x\rightarrow x_0} (f\circ g)(x)$$

will equal

$$f(\lim_{x\rightarrow x_0} g(x))$$

if the limit of g exists and it f is continuous as $\lim_{x\rightarrow x_0} g(x)$.

6. Nov 14, 2006