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

  • Thread starter quasar987
  • Start date
  • Tags
    Limit
In summary, the conversation discusses the evaluation of the limit of a logarithmic expression. The limit is taken as T approaches 0 from the right, and is written as the composition of two functions. The speaker justifies moving the limit inside the logarithm by citing the continuity of the logarithm function away from 0. This is a commonly accepted definition of continuity in most textbooks.
  • #1
quasar987
Science Advisor
Homework Helper
Gold Member
4,807
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]

?!?
 
Last edited:
Physics news on Phys.org
  • #2
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:
  • #3
What justifies your moving the limit inside the log?

(I refined the expresion following your comments, thx)
 
Last edited:
  • #4
Continuity of ln away from 0.
 
  • #5
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].
 
  • #6
Most textbooks define continuity this way.
 

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

1. What is the limit of the given expression as T approaches 0 from the positive side?

The limit is ln(2) or approximately 0.693147. This can be obtained by substituting 0 for T in the expression and simplifying.

2. How does the value of T affect the limit of the expression?

As T approaches 0, the value of the expression approaches ln(2). As T increases, the value of the expression decreases and approaches 0.

3. Can the limit be evaluated for negative values of T?

No, the limit is only defined for positive values of T as the expression contains a natural logarithm, which is only defined for positive values.

4. What is the significance of the constant e in the expression?

The constant e, also known as Euler's number, is a mathematical constant that is approximately equal to 2.71828. In this expression, it is used in the denominator of the exponential term, which helps to approach the limit value of ln(2) as T approaches 0.

5. How does the value of the constant epsilon affect the limit of the expression?

As the value of epsilon increases, the value of the expression decreases and approaches ln(2) as T approaches 0. This is because the exponential term in the expression becomes closer to 0, making the entire expression closer to ln(2).

Similar threads

Replies
29
Views
2K
Replies
9
Views
1K
  • Calculus
Replies
4
Views
2K
Replies
1
Views
2K
Replies
5
Views
1K
Replies
4
Views
1K
  • Calculus
Replies
2
Views
1K
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
11
Views
828
  • Calculus
Replies
3
Views
1K
Back
Top