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

  • Context: Graduate 
  • Thread starter Thread starter quasar987
  • Start date Start date
  • Tags Tags
    Limit
Click For Summary

Discussion Overview

The discussion revolves around evaluating the limit \lim_{T\rightarrow 0^+} \ln(2e^{-\epsilon/kT}+1), focusing on the conditions where \epsilon>0 and k>0. The scope includes mathematical reasoning and limit evaluation techniques.

Discussion Character

  • Mathematical reasoning, Technical explanation

Main Points Raised

  • One participant expresses uncertainty about how to evaluate the limit.
  • Another participant suggests that as T approaches 0 from the positive side, e^{-\epsilon/kT} tends to 0, leading to the conclusion that the limit approaches \ln(0+1)=0, while noting the necessity of positive values for k and \epsilon.
  • A participant questions the justification for moving the limit inside the logarithm function.
  • Another participant cites the continuity of the logarithm function away from 0 as a reason for this manipulation.
  • One participant references a limit property involving the composition of functions, indicating that the limit can be taken inside the function under certain conditions.
  • Another participant notes that this definition of continuity is commonly found in textbooks.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the evaluation method, as there are differing views on the justification for moving the limit inside the logarithm. The discussion remains unresolved regarding the overall evaluation of the limit.

Contextual Notes

Participants emphasize the importance of the conditions on k and \epsilon being positive and the need to consider the right-handed limit as T approaches 0.

quasar987
Science Advisor
Homework Helper
Gold Member
Messages
4,796
Reaction score
32
I'm having a blank here. How do I evaluate

[tex]\lim_{T\rightarrow 0^+} \ln(2e^{-\epsilon/kT}+1) \ \ \ \ \ \ \ (\epsilon>0, \ \ k>0)[/itex]<br /> <br /> ?!?[/tex]
 
Last edited:
Physics news on Phys.org
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:
What justifies your moving the limit inside the log?

(I refined the expresion following your comments, thx)
 
Last edited:
Continuity of ln away from 0.
 
Oh I have it. It's that

[tex]\lim_{x\rightarrow x_0} (f\circ g)(x)[/tex]

will equal

[tex]f(\lim_{x\rightarrow x_0} g(x))[/tex]

if the limit of g exists and it f is continuous as [itex]\lim_{x\rightarrow x_0} g(x)[/itex].
 
Most textbooks define continuity this way.
 

Similar threads

  • · Replies 29 ·
Replies
29
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K