I am dealing with the function
$$ f(D) = \left( 1 - \frac{(1-a^{1/D})(b^{1/D}-2(1-a^{1/D}))}{b^{2/D}} \right)^D $$
for some constants $0 < a, b < 1$ and I am interested in how this function behaves for $D \rightarrow \infty$. Apparently it holds that $\lim_{D \rightarrow \infty} f(D) = a$ as supported by Wolfram Alpha. However, Wolfram Alpha is unable to explain why this is the case. The only clue I have is that apparently, the Laurent series expansion of $f$ at $D = \infty$ is
$$ a + \frac{a\ln(a)(2\ln(a)-\ln(b))}{D} + \mathcal{O}\left( \frac{1}{D^2} \right) $$
At least this is what Wolfram Alpha claims. My question is whether this is correct and if so, how to derive this series expansion of $f$. Any help would be much appreciated.
Applying the logarithm, we get $$\begin{align*} \ln f(D) & = D \ln \left( 1 - \frac{(1-a^{1/D})(b^{1/D}-2(1-a^{1/D}))}{b^{2/D}} \right)\\ & = \frac{ \ln \left( 1 - \frac{(1-a^{1/D})(b^{1/D}-2(1-a^{1/D}))}{b^{2/D}} \right)}{\frac{1}{D}}. \end{align*}$$ Since both enumerator and denominator converge to $0$, we can use L'Hospitals rule:
$$\begin{align*} \lim_{D\to \infty} \ln f(D) & = \lim_{D\to \infty}\frac{1}{-\frac{1}{D^2}} \cdot-\frac{(-4+4a^{1/D}+b^{1/D})(a^{1/D} \ln a + \ln b - a^{1/D} \ln b)}{b^{2/D}D^2} \\ &= \frac{(-4+4\cdot 1 + 1)(1\cdot \ln a+\ln b - 1\cdot \ln b)}{1} = \ln a. \end{align*}$$ I am sure you can work out the derivatives yourself. If you are interested in the second term of the asymptotic expansion, the same strategy using L'Hospitals rule works if you subtract $a$ and multiply by $D$.