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:<math>\begin{align}\lim_{n \rightarrow 0} \dfrac{Z^n - 1}{n} &= \lim_{n \rightarrow 0} \dfrac{e^{n \ln Z} - 1}{n}\\
&= \lim_{n \rightarrow 0} \dfrac{n \ln Z + {1 \over 2!} (n \ln Z)^2 + \dots}{n}\\ &= \ln Z ~~.\end{align}</math>
Alternatively, one can use the fact that the derivative of <math>Z^n</math> with respect to <math>n</math> is <math>Z^n \ln Z</math>
:<math>\begin{align}\lim_{n \rightarrow 0} \dfrac{Z^n - 1}{n} &= \left.\dfrac{\partial Z^n}{\partial n}\right |_{n=0} \\&= (Z^n \ln Z)|_{n=0} \\
&= \ln Z ~~.\end{align}</math>
==References==
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