Two different answers for limit at zero

Question

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.

Answer 1

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.