Finding the limit of N approaching infinity: $N(x^\frac{1}{N}-1)\approx\ln(x)+\frac{1}{2N}\ln(x)^2+...$

Question

I am having trouble understanding the linked exercise, final paragraph (not parts a or b)

Entropy Calc Problem

I understand this is a physics related exercise, however, my trouble comes in at the mathematical portion.

The exercise is from Thermodynamics and an Introduction to Thermostatistics by H.B Callen, 2nd ed.

For the first part (a) I ended up finding the entropy to be

$$\Delta S=\frac{C_v}{N}\ln\left(\frac{T_N}{T_0}\right)$$

Note: (b) was found to be 0, however, I don't feel that is correct and will inquire to the physics SE if need be.

The problem statement is to now find the entropy change in the limit as $N\to\infty$, for large N:

$$N(x^\frac{1}{N}-1)\approx\ln(x)+\frac{1}{2N}ln(x)^2+...$$

This is where I am having trouble. Is it as simple as plugging in $ln(x)$ into the entropy equation, or is there more to it? The book mentions this is a nontrivial limit, which makes me doubt my suggestion below:

$$\lim_{N\to\infty}\frac{C_v}{N}\ln\left(\frac{T_N}{T_0}\right)=C_v\left(\frac{\ln\left(\frac{T_N}{T_0}\right)}{\ln(x)}\right)$$

I am not sure where to continue with this, any help or guidance on this would be greatly appreciated.

Answer 1

Hint. You may just observe that, as $N \to \infty$, we have $$ x^{\large\frac1N}=e^{\large\frac{\ln x}N}=1+\frac{\ln x}N+\frac{\ln^2 x}{2N^2}+O\left(\frac1{N^3}\right) $$ giving directly

$$ N(x^\frac{1}{N}-1)\sim\ln x+\frac{\ln^2 x}{2N} $$

as announced. We have used the Taylor expansion $$ e^u=1+u+\frac{u^2}{2!}+O(u^3) $$ as $u \to 0$.