Computing $\lim_{\varepsilon\to 0^{+}}\psi(\varepsilon)/\Gamma(\varepsilon)$ with asymptotic expansions I have the following limit of which I want to compute:
\begin{equation}
\lim_{\varepsilon\to 0^{+}} \frac{\psi(\varepsilon)}{\Gamma(\varepsilon)}.
\end{equation}
For $\varepsilon\approx 0$ and $\varepsilon\neq 0$ I have the following limiting forms
\begin{equation}
\tag{1}
\psi(\varepsilon)=-\frac{1}{\varepsilon}-\gamma+O(\varepsilon)
\end{equation}
and
\begin{equation}
\tag{2}
\frac{1}{\Gamma(\varepsilon)}=\varepsilon+O(\varepsilon^{2}).
\end{equation}
If I multiply $(1)$ and $(2)$ together we get
\begin{align}
\tag{3}
\frac{\psi(\varepsilon)}{\Gamma(\varepsilon)}
&=
-1-\frac{O(\varepsilon^{2})}{\varepsilon}
-\gamma\varepsilon-\gamma O(\varepsilon^{2})
+\varepsilon O(\varepsilon)+O(\varepsilon)O(\varepsilon^{2})\\
&=
-1-O(\varepsilon)
-\gamma\varepsilon-\gamma O(\varepsilon^{2})
+O(\varepsilon^{2})+O(\varepsilon^{3}).
\end{align}
In the limit, all of the terms with $\varepsilon$ approach zero such that we arrive at
\begin{equation}
\lim_{\varepsilon\to0^{+}} \frac{\psi(\varepsilon)}{\Gamma(\varepsilon)} =
-1.
\end{equation}
I have checked this answer against WolframAlpha which yields the same result. Despite getting the same result, I have doubts as to if this is a sound approach to computing the limit.

My question is this: Is the use of asymptotic expansions in this manner proper (i.e. is this a valid method to computing my limit)? Or does it just happen to work out in this example?


One could take the following method:
\begin{align}
\frac{\psi(x)}{\Gamma(x)} &= \frac{\Gamma(x) \, \psi(x)}{\Gamma^{2}(x)} = \frac{\Gamma'(x)}{\Gamma^{2}(x)} \\
&= - \frac{d}{dx} \left(\frac{1}{\Gamma(x)}\right).
\end{align}
Now make use of the Taylor series expansion of the inverse of the Gamma function, namely,
$$\frac{1}{\Gamma(x)} \approx x + \gamma \, x^{2} + \left(\frac{\gamma^{2}}{2} - \frac{\pi^{2}}{12}\right) \, x^{3} + \mathcal{O}(x^{4})$$
then
$$\frac{\psi(x)}{\Gamma(x)} \approx - \left(1 + 2 \gamma \, x + \left(\frac{3 \, \gamma^{2}}{2} - \frac{\pi^{2}}{4}\right) \, x^{2} + \mathcal{O}(x^{3}) \right).$$
Taking the limit as $x \to 0$ leads to
$$\lim_{x \to 0} \frac{\psi(x)}{\Gamma(x)} = -1.$$
