Limit used in stat mech, how to prove this?

[Nusc]

Homework Statement

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

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.

 

[Dick]

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?

 

[Nusc]

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?

 

[snipez90]

\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}}

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