Two different answers for limit at zero The following problem arises when calculating the result of Theorem 2 (part (4)) in Takács (1962) Introduction to the Theory of Queues (page 211).

Calculate
  $$\lim_{s\to 0^{+}} \left[
    \frac{ 2\bigl( \Pi_0'(s) \bigr)^2 }{ \bigl( \Pi_0(s) \bigr)^3 }
    -
    \frac{ \Pi_0''(s)}{ \bigl( \Pi_0(s) \bigr)^2 }
\right]
$$
  given $$\lim_{s \to 0^{+}} s^{n+1} \Pi_0^{(n)}(s) = (-1)^n n!\,\mathrm e^{-\lambda\alpha}$$ for all non-negative integers $n$.

Remark: The function $\Pi_0(s)$ is the Laplace transform of 
$$P_0(t) = \exp\left( -\lambda\int_{0}^{t}[1-H(x)]\,\mathrm dx \right)$$ for a cumulative distribution function $H(x)$ on the non-negative reals, and $\alpha$ is the mean of $H(x)$.
My question: I can obtain two different answers for the limit, the second being the negative of the first. What did I do wrong?
Solution 1 (obtains the same result as Takács, 1962)
\begin{align*}
    \frac{ 2\bigl( \Pi_0'(s) \bigr)^2 }{ \bigl( \Pi_0(s) \bigr)^3 }
    -
    \frac{ \Pi_0''(s) }{ \bigl( \Pi_0(s) \bigr)^2 }
=
    \frac{ 2 \Pi_0(s) \bigl( s^2 \Pi_0'(s) \bigr)^2 }{ \bigl( s \Pi_0(s) \bigr)^4 }
    -
    \frac{ s^3 \Pi_0''(s) }{ s \bigl( s \Pi_0(s) \bigr)^2 }
\end{align*}
so
\begin{align*}
\lim_{s\to 0^{+}} \left[
    \frac{ 2\bigl( \Pi_0'(s) \bigr)^2 }{ \bigl( \Pi_0(s) \bigr)^3 }
    -
    \frac{ \Pi_0''(s) }{ \bigl( \Pi_0(s) \bigr)^2 }
\right]
&=
\lim_{s\to 0^{+}} \left[
    \frac{ 2 \Pi_0(s) \bigl( -e^{-\lambda\alpha} \bigr)^2 }{ \bigl( e^{-\lambda\alpha} \bigr)^4 }
    -
    \frac{ 2e^{-\lambda\alpha} }{ s \bigl( e^{-\lambda\alpha} \bigr)^2 }
\right]
\\ &=
\lim_{s\to 0^{+}} 2e^{2\lambda\alpha} \left[
    \Pi_0(s)
    -
    \frac{ e^{-\lambda\alpha} }{ s }
\right]
\end{align*}
Solution 2
\begin{align*}
    \frac{ 2\bigl( \Pi_0'(s) \bigr)^2 }{ \bigl( \Pi_0(s) \bigr)^3 }
    -
    \frac{ \Pi_0''(s) }{ \bigl( \Pi_0(s) \bigr)^2 }
=
    \frac{ 2\bigl( s^2 \Pi_0'(s) \bigr)^2 }{ s \bigl( s \Pi_0(s) \bigr)^3 }
    -
    \frac{ \Pi_0(s) s^3 \Pi_0''(s) }{ \bigl( s \Pi_0(s) \bigr)^3 }
\end{align*}
so
\begin{align*}
\lim_{s\to 0^{+}} \left[
    \frac{ 2\bigl( \Pi_0'(s) \bigr)^2 }{ \bigl( \Pi_0(s) \bigr)^3 }
    -
    \frac{ \Pi_0''(s) }{ \bigl( \Pi_0(s) \bigr)^2 }
\right]
&=
\lim_{s\to 0^{+}} \left[
    \frac{ 2\bigl( -e^{-\lambda\alpha} \bigr)^2 }{ s \bigl( e^{-\lambda\alpha} \bigr)^3 }
    -
    \frac{ 2 \Pi_0(s) e^{-\lambda\alpha} }{ \bigl( e^{-\lambda\alpha} \bigr)^3 }
\right]
\\ &=
\lim_{s\to 0^{+}} 2e^{2\lambda\alpha} \left[
    \frac{ e^{-\lambda\alpha} }{ s }
    -
    \Pi_0(s)
\right]
\end{align*}
Assuming that I haven't done something silly with the algebra, my guess is that has to do with even versus odd powers of $s$ vis-a-vis $-s$. In the first answer, after multiplying by powers of $s$, the denominators are even powers ($4$ and $2$). But in the second answer the denominators are odd powers ($3$ and $3$). So in some sense, in the first answer I could replace $s$ with $-s$ and everything is the same, but in the second answer I have a "$-$" left over.
Many thanks in advance.
 A: The issue in both Takács' and your derivation is to assume that the following proposition holds:

If functions $f_1, f_2, g_1, g_2$ satisfy $f_1(x) \sim f_2(x)$ and $g_1(x) \sim g_2(x)$ as $x → 0^+$, and $\lim\limits_{x → 0^+} (f_1(x) - g_1(x))$ exists, then $\lim\limits_{x → 0^+} (f_2(x) - g_2(x))$ exists and$$
\lim_{x → 0^+} (f_1(x) - g_1(x)) = \lim_{x → 0^+} (f_2(x) - g_2(x)).$$

This proposition, however, is not necessarily true, e.g. if$$
f_1(x) = \frac{1}{x} + 1,\ f_2(x) = \frac{1}{x} + 2,\  g_1(x) = g_2(x) = \frac{1}{x}. \quad \forall x > 0
$$
Thus Takács might have derived a correct result using an incorrect method.
A: Building on the answer by @Saad into something that I can apply ...
Lemma. Suppose that for functions $f_1(x)$, $f_2(x)$, $g_1(x)$, and $g_2(x)$ the following hold:


*

*$\displaystyle \lim_{x \to a} \bigl\lvert f_1(x) - f_2(x) \bigr\rvert = 0$

*$\displaystyle \lim_{x \to a} \bigl\lvert g_1(x) - g_2(x) \bigr\rvert = 0$

*$\displaystyle \lim_{x \to a} \bigl\lvert f_1(x) - g_1(x) \bigr\rvert$ exists
Then $\displaystyle \lim_{x \to a} \bigl\lvert f_2(x) - g_2(x) \bigr\rvert$ exists and
$$
  \lim_{x \to a} \bigl\lvert f_2(x) - g_2(x) \bigr\rvert
  =
  \lim_{x \to a} \bigl\lvert f_1(x) - g_1(x) \bigr\rvert
$$
Proof. For any given $x$ we have
$$ \begin{align*}
  \bigl\lvert f_2(x) - g_2(x) \bigr\rvert
&=
  \bigl\lvert f_2(x) - f_1(x) + g_1(x) - g_2(x) + f_1(x) - g_1(x) \bigr\rvert
\\ &\leq
  \bigl\lvert f_1(x) - f_2(x) \bigr\rvert + 
  \bigl\lvert g_1(x) - g_2(x) \bigr\rvert +
  \bigl\lvert f_1(x) - g_1(x) \bigr\rvert
\end{align*} $$
by the triangle inequality so
$$ 
  \bigl\lvert f_2(x) - g_2(x) \bigr\rvert
  -
  \bigl\lvert f_1(x) - g_1(x) \bigr\rvert
\leq
  \bigl\lvert f_1(x) - f_2(x) \bigr\rvert + 
  \bigl\lvert g_1(x) - g_2(x) \bigr\rvert
  \label{upper}
$$
Likewise
$$ \begin{align*}
  \bigl\lvert f_1(x) - g_1(x) \bigr\rvert
&=
  \bigl\lvert f_1(x) - f_2(x) - g_1(x) + g_2(x) + f_2(x) - g_2(x) \bigr\rvert
\\ &\leq
  \bigl\lvert f_1(x) - f_2(x) \bigr\rvert + 
  \bigl\lvert g_1(x) - g_2(x) \bigr\rvert +
  \bigl\lvert f_2(x) - g_2(x) \bigr\rvert
\end{align*} $$
so
$$
  \bigl\lvert f_1(x) - g_1(x) \bigr\rvert
  -
  \bigl\lvert f_2(x) - g_2(x) \bigr\rvert
  \leq
  \bigl\lvert f_1(x) - f_2(x) \bigr\rvert + 
  \bigl\lvert g_1(x) - g_2(x) \bigr\rvert
  \label{lower}
$$
Combining the inequalities yields
$$
  -\left( 
    \bigl\lvert f_1(x) - f_2(x) \bigr\rvert + 
    \bigl\lvert g_1(x) - g_2(x) \bigr\rvert
  \right)
\leq
  \bigl\lvert f_1(x) - g_1(x) \bigr\rvert
  -
  \bigl\lvert f_2(x) - g_2(x) \bigr\rvert
\leq
  \bigl\lvert f_1(x) - f_2(x) \bigr\rvert + 
  \bigl\lvert g_1(x) - g_2(x) \bigr\rvert
$$
Hence by the squeeze theorem
$$
  \lim_{x \to a}
  \Bigl(
  \bigl\lvert f_1(x) - g_1(x) \bigr\rvert
  -
  \bigl\lvert f_2(x) - g_2(x) \bigr\rvert
  \Bigr)
  =
  0
$$
and result follows from existence of 
$\displaystyle \lim_{x \to a} \bigl\lvert f_1(x) - g_1(x) \bigr\rvert$. $\Box$
The insight is that with this result, I can draw conclusions about
$$
  \lim_{s \to 0^{+}}
  \left\lvert
    \frac{ 2\bigl( \Pi_0'(s) \bigr)^2 }{ \bigl( \Pi_0(s) \bigr)^3 }
    -
    \frac{ \Pi_0''(s)}{ \bigl( \Pi_0(s) \bigr)^2 }
   \right\rvert
$$
emphasizing the absolute value, but I will need something more to establish the $\pm$ aspect.
