There is a expression given in article arXiv:1505.01908 $$\bigg[n\big(\omega+\frac{\Omega}{2}\big)-n\big(\omega-\frac{\Omega}{2}\big)\bigg]g^R_{\omega-\frac{\Omega}{2}}g^A_{\omega+\frac{\Omega}{2}} +n\big(\omega-\frac{\Omega}{2}\big)g^A_{\omega-\frac{\Omega}{2}}g^A_{\omega+\frac{\Omega}{2}} -n\big(\omega+\frac{\Omega}{2}\big)g^R_{\omega-\frac{\Omega}{2}}g^R_{\omega+\frac{\Omega}{2}} \tag{45a} $$ here $n(\omega)=[e^{\beta\omega}-1]^{-1}$, $n'(\omega)=dn/d\omega$, and $$ g_\omega^{R(A)}=\frac{1}{\omega-\omega_q\pm i\alpha\omega} $$ they claim that in limit $\Omega\to 0$, (neglecting $\Omega^2$ terms), the above expression is equal to $$ \approx -\frac{\Omega}{2}n'(\omega)\bigg(g_\omega^A-g_\omega^R\bigg)^2 +n(\omega)\big[(g^A_{\omega})^2-(g^R_{\omega})^2\big] \tag{45b}$$
I wonder how exactly they get the final expression.
My Attempt: I can see that if we take limit $\Omega\to 0$ in the last two terms of $(45a)$, we get the last the last two terms of $(45b)$. And by using approximation for derivative $f'(x)=[f(x+h)-f(x-h)]/2h$, we get $$ \bigg[n\big(\omega+\frac{\Omega}{2}\big)-n\big(\omega-\frac{\Omega}{2}\big)\bigg]=\Omega \frac{dn(\omega)}{d\omega}$$ all in all, so far, I have
$$ \approx \Omega n'(\omega)\lim_{\Omega\to 0}\bigg(g^R_{\omega-\frac{\Omega}{2}}g^A_{\omega+\frac{\Omega}{2}}\bigg) +n(\omega)\big[(g^A_{\omega})^2-(g^R_{\omega})^2\big] \tag{A}$$ I am not sure how is the first term handled to get the expression given in $(45b)$