Taylor series expansion and Laplace transform final value theorem

Question

I cant figure out how some transformations are made in one article on physics. Here is expression in s-domain and they want to find its asymptotic value. $$ \xi(s) = \nu_1(s+1)=\frac{1}{(s+1)} (1+\frac{2h(s+1)}{1-2g(s+1)}) (*) $$ Here is quote from the text: Asymptotic value may be obtained from equation $(*)$ by application of Laplace transform final value theorem. If the functions $(s+2)h(s+1)$ and $(s+2)(1-2g(s+2))$ (???) are expanded in Taylor series around s=0 retaining only first power terms. Using abbreviations: $$ g_2=dg/ds|_{s=1} $$ $$ h_1=2h(1) $$ $$ h_2=-d(2^{2-s}h(s))/ds|_{s=1} $$ we can get $$ \xi^{\infty}_1=\frac{g_1}{h_2} $$ $$ \Delta \xi^{\infty}=\frac{g_2}{h_2}-(ln2 - \frac{1}{2})\xi^{\infty}_1 -1$$ final value can be written as $$ \lim_{u \to \infty} \xi_1(u)= \xi^{\infty}_1-\Delta \xi^{\infty}e^{-u} $$ So, apparently the want to expand enumerator and denominator of (*) in Taylor series separetly (are they ? im terribly confused). And then apply final value theorem like: $$ \lim_{s \to 0} s\xi(s) $$ I'm trying to do it and cant get anything similar to their answer. Could anyone please help me ? also it is known that $h(0)=1/2; g(1)=1/2; g(1)=1$.