Limit used in stat mech, how to prove this?

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Homework Help Overview

The discussion revolves around a limit in statistical mechanics, specifically proving the expression involving the limit as \( dt \) approaches zero. The context is rooted in the properties of exponential functions and their behavior in limits.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the behavior of the limit as \( dt \) approaches zero and question the convergence rates of \( dt \) and \( 1/dt \). There is a reference to a known limit involving \( (1 + ax)^{1/x} \) and its relation to the exponential function, prompting discussions on the application of l'Hôpital's rule.

Discussion Status

The discussion is ongoing, with participants sharing insights on the limit and its proof. Some guidance has been offered regarding the application of l'Hôpital's rule, but there is no explicit consensus on the approach to take or the specific values involved.

Contextual Notes

There is a mention of forgetting previous knowledge related to limits and the conditions under which l'Hôpital's rule applies, indicating a potential gap in understanding that is being addressed in the discussion.

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Homework Statement



[tex] lim_{dt\rightarrow 0} [(1+ \alpha dt(e^{-ik}-1))^{1/dt}]^T = e^{\alpha (e^{-ik-1)T}[/tex]

It's a well known property in statistical physics, I'm not sure how to prove it

Homework Equations


The Attempt at a Solution



I know dt ->0

and 1/dt -> infinity

Which one converges faster? What test do I apply? I forgot everything.
 
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Yep, you forgot everything. lim x->0 of (1+ax)^(1/x) is e^a, right? You can prove that with l'Hopital. Is that what you forgot? What's 'a' in your problem?
 


a is a constant.

I thought l'hopital rule applies when you take the limit of some variable that appears in both the numerator and denominator.
lim x->0 of (1+ax)^(1/x) is e^a
why then is this the case?
 


[tex]\lim_{x \rightarrow 0}(1+ax)^{1/x} = \lim_{x \rightarrow 0}e^{\frac{1}{x}\log{(1+ax)}} = e^{\lim_{x\rightarrow 0 }\frac{\log{(1+ax)}}{x}}[/tex]

by the continuity of the exponential function. Now apply l'Hopital. This is a common manipulation.
 

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