Evaluate Limit: [tex]\lim_{T\rightarrow 0^+} \ln(2e^{-\epsilon/kT}+1)[/itex]

[quasar987]

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

\lim_{T\rightarrow 0^+} \ln(2e^{-\epsilon/kT}+1) \ \ \ \ \ \ \ (\epsilon&gt;0, \ \ k&gt;0)[/itex]<br /> <br /> ?!?

 

[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.)

 

[quasar987]

What justifies your moving the limit inside the log?

(I refined the expresion following your comments, thx)

 

[slearch]

Continuity of ln away from 0.

 

[quasar987]

Oh I have it. It's that

\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).

 

[DeadWolfe]

Most textbooks define continuity this way.