Computing $\lim_{\varepsilon\to 0^{+}}\psi(\varepsilon)/\Gamma(\varepsilon)$ with asymptotic expansions

Question

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?

Answer 1

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 $$\lim_{x \to 0} \frac{\psi(x)}{\Gamma(x)} = -1.$$