l'hopital's rule for expression involving square root Dear Colleagues, 
                I've been trying to apply l'Hopital's rule to find a limit, but have so far been defeated. Unfortunately, I'm not even clear on what general strategy to apply to get an answer.
I'm trying to solve : $\lim_{s \to 0} f(s)$ and $\lim_{s \to 0} \frac{d f(s)}{ds}$
for $f(s)= \frac{A_1(s)}{A_2(s)^{1/2}}$, 
$$A_2(s) = \frac{2C}{\epsilon(1-\epsilon^2)}[(1-s)^{(1+\epsilon)} -1 + (1+\epsilon)s-\frac{\epsilon(1+\epsilon)}{2}s^2]-\frac{2D}{\epsilon(1+\epsilon)(2+\epsilon)}[1-(1-s)^{2+\epsilon} - (2+\epsilon)s+\frac{\epsilon(1+\epsilon)(2+\epsilon)}{2}s^2]$$
and 
$$A_1(s) = \frac{C}{\epsilon(1-\epsilon)}[1-(1-s)^\epsilon -\epsilon s]-\frac{D}{\epsilon(1+\epsilon)}[(1-s)^{1+\epsilon} - 1 + (1+\epsilon)s]$$
One should note that $A_2'(s)=2A_1(s)$ and that $A_1''(0)=C-D$. My attempts to apply l'Hopital's rule haven't got me very far because of a pesky $A_2(s)^{1/2}$ term. For instance
$$
\begin{align}
\lim_{s \to 0} f(s) & = \lim_{s \to 0} \frac{A_1(s)}{A_2(s)^{1/2}} \\
                & = \lim_{s \to 0} \frac{A_1'(s)}{\frac{1}{2}A_2(s)^{-1/2}A_2'(s)}\\
                & = \lim_{s \to 0} \frac{A_1'(s)}{A_2(s)^{-1/2}A_1(s)}
\end{align}$$
While I can differentiate the $A_1(s)$ term until I reach a constant (for $s=0$), differentiating the $A_2(s)^{1/2}$ term always generates a fresh $A_2(s)^{-1/2}A_1(s)$ term whose convergence I am unable to determine. Can anyone see a way out?
Thanks in advance, RL
 A: Here is a naive attempt from a calculus beginner. Please feel free to point out my mistakes (and I am sure there will be some).
Observe that $A_2(0) = 0, A_1^{'}(0) = 0, A_1(0) = 0$ and hence $A_2^{'}(0) = 0$. Since $A_2^{\frac{1}{2}}(s) = \sqrt{A_2(s)}$, then we can work on $lim_{x \rightarrow 0} f(x)$ now. 
$\huge lim_{x \rightarrow 0}f(s) = \frac{A_1^{'}(s)A_2^{\frac{1}{2}}(s)}{A_1(s)}$. By L'Hopital Rule, because both $A_1^{'}(s)$ and $A_1(s)$ will converge to $0$ when $s \rightarrow 0$, hence, after differentiating the top and the bottom, we have $\huge lim_{s \rightarrow 0}f(s) = \frac{A_1^{''}(s)A_2^{\frac{1}{2}}(s) + \frac{1}{2}A_2^{-\frac{1}{2}}A_1^{'}(s)A_2^{'}(s)}{A_1^{'}(s)} = \frac{A_1^{''}(s)A_2(s)+\frac{1}{2}A_1^{'}(s)A_2^{'}(s)}{A_1^{'}(s)A_2^{\frac{1}{2}}(s)} = \frac{A_1^{''}(s)A_2(s) + A_1(s)A_1^{'}(s)}{A_1^{'}(s)A_2^{\frac{1}{2}}(s)}$. Again both the top and bottom converge to $0$ as $s$ does. Apply L'Hopital's Rule again:
$\huge \frac{A_1^{'''}(s)A_2(s) + A_1{''}(s)A_2{'}(s) + (A_1^{'}(s))^2 + A_1(s)A_1^{''}(s)}{A_1^{''}(s)A_2^{\frac{1}{2}} + \frac{1}{2}A_2^{-\frac{1}{2}}(s)A_1^{'}(s)} = \frac{A_1^{'''}(s)A_2(s) + (A_1^{'}(s))^2 + 3A_1(s)A_1^{''}(s)}{A_1^{''}(s)A_2^{\frac{1}{2}} + \frac{1}{2}A_2^{-\frac{1}{2}}(s)A_1^{'}(s)} = A_2^{\frac{1}{2}}(s) \frac{A_1^{'''}(s)A_2(s) + (A_1^{'}(s))^2 + 3A_1(s)A_1^{''}(s)}{A_1^{''}(s)A_2(s) + \frac{1}{2}A_1^{'}(s)}$
Once again both the top and the bottom converge to $0$. Apply L'Hopital's Rule and then we can observe that the next result will have $A_1^{''}(s)$ in the denominator, which cause the denominator not to converge to $0$ but in the nominator, we have $A_2^{-\frac{1}{2}}(s)$, which will converge to infinity. Hence I believe $lim_{s \rightarrow 0} f(s)$ does not exist. 
$\huge \frac{df(s)}{ds} = \frac{A_1^{'}(s)A_2^{\frac{1}{2}}(s) - \frac{1}{2}A_2^{-\frac{1}{2}}(s)A_1(s)A_2^{'}(s)}{A_2(s)} = \frac{A_1^{'}(s)A_2(s) - \frac{1}{2}A_1(s)A_2^{'}(s)}{A_2^{\frac{3}{2}}(s)} = \frac{A_1^{'}(s)A_2(s) - A_1^{2}(s)}{A_2^{\frac{3}{2}}(s)}$. 
Apply L'Hopital's Rule again, we have
$\huge \frac{A_1^{''}(s)A_2(s) + A_1^{'}(s)A_2^{'}(s) - 2A_1(s)A_1^{'}(s)}{\frac{3}{2}A_2^{\frac{1}{2}}(s)} = \frac{A_1^{''}(s)A_2(s)}{\frac{3}{2}A_2^{\frac{1}{2}}(s)}$. 
When $s \rightarrow 0$, this limit converge to $0$ and hence $\huge lim_{s \rightarrow 0} \frac{df(s)}{ds} = 0$
