Asymptotic behavior of the coefficients of expansion of $\sqrt{1-4a(1-a) \cdot x+(2a-1)\sqrt{1-4a(1-a) \cdot x^2}}$ as a power series Consider the following generating function of some sequence $(f_n)_n$
$$
G(f_n;x) = \sqrt{1-4a(1-a) \cdot x+(2a-1)\sqrt{1-4a(1-a) \cdot x^2}}
$$
for $0\leq a\leq1/2$. How can I deduce the asymptotic formula of the coefficients corresponding to the above generating function? 
The first five coefficients of the expansion around zero are:
\begin{align}
&x^0:\ \sqrt{2a}\\
&x^1:\ -\sqrt{2}(1-a)\sqrt{a}\\
&x^2:\ -\sqrt{2}(1-a)^2\sqrt{a}-\sqrt{2}(1-a)\sqrt{a}(2a-1)\\
&x^3:\ -3\sqrt{2}(1-a)^3\sqrt{a}-3\sqrt{2}(1-a)^2\sqrt{a}(2a-1)\\
&x^4:\ -15\sqrt{2}(1-a)^4\sqrt{a}-18\sqrt{2}(1-a)^3\sqrt{a}(2a-1)-12\sqrt{2}(1-a)^2a^{3/2}(2a-1)\\
&\ \ \ \  \ \ \ \;\ -3\sqrt{2}(1-a)^2\sqrt{a}(2a-1)^2\\
&x^5:\ -105\sqrt{2}(1 - a)^5\sqrt{a} - 
 150 \sqrt{2} (1 - a)^4 \sqrt{a} (2a-1) - 
 60 \sqrt{2} (1 - a)^3 a^{3/2} (2a-1) - 
 45 \sqrt{2} (1 - a)^3 \sqrt{a} (2a-1)^2
\end{align}

Let $G_a(x):=\sqrt{1-4ab \cdot x+(a-b)\sqrt{1-4ab \cdot x^2}},\;$ where $b:=1-a$. Then we find also that
$G_a(x) = \sqrt{a/2}(f_a(x)+f_a(-x))-\sqrt{b/2}(f_a(x)-f_a(-x)),\;$
where $f_a(x):=\sqrt{1+x\sqrt{4ab}}.$
The coefficient of $x^n$ in $G_a(x)$ is asymptotic to
$\;-\sqrt{c_a(4ab)^n/(2n^3\pi)},\;$ where $c_a=a$ if $n$ is even, else $b$ if $n$ is odd.
