Can't L'Hôpital this one

$\lim_{x\to0} \frac{x! + \alpha x -1}{x^{2}}$ We can rewrite the factorial function as its definition, the $\Gamma(x)$ function $\lim_{x\to0} \frac{\Gamma(x+1) + \alpha x -1}{x^{2}}$ Because the limit is in the form $\frac{0}{0}$ , (note that $\Gamma(1) = 1$ ), we can us L'Hôpital's rule $\lim_{x\to0} \frac{\Gamma^{\prime}(x+1) + \alpha}{2x}$ For the limit to give a finite result, it must not diverge. The limit will not diverge if the fraction is in the form $\frac{0}{0}$ , (the form $\frac{\infty}{\infty}$ is unattainable in this case). Thus, $\Gamma^{\prime}(x+1) = -\alpha$ Evaluating at $x=0$ $\Gamma^{\prime}(1) = -\alpha$ Evoking the definition of the $\Gamma(x)$ function leads to $\Gamma(x) = \int_{0} ^{\infty} e^{-t}\;t^{x-1} \;\mathrm{d}t$ Taking the derivative on both sides with respect to $x$ , (I am ignoring the Leibniz Rule of differentiation under the integral sign. For more info, look at notable resources below) $\Gamma^{\prime}(x) = \int_{0} ^{\infty} e^{-t}\;t^{x-1}\;\ln(t) \;\mathrm{d}t$ Evaluating at $x=1$ $\Gamma^{\prime}(1) = \int_{0} ^{\infty} e^{-t}\;\ln(t) \;\mathrm{d}t$ This is one of the integral definitions of the Euler-Mascheroni constant, thus $\Gamma^{\prime}(1) = -\gamma = -\alpha$ $\alpha = \gamma = \boxed{0.57721}$ In order to evaluate the limit, which is now $\lim_{x\to0} \frac{\Gamma^{\prime}(x+1) + \gamma}{2x}$ we will evoke L'Hôpital's rule once more $\lim_{x\to0} \frac{\Gamma^{\prime\prime}(x+1)}{2}$ $\frac{\Gamma^{\prime\prime}(1)}{2}$ Using the $\Gamma^{\prime}(x)$ from earlier, which was $\Gamma^{\prime}(x) = \int_{0} ^{\infty} e^{-t}\;t^{x-1}\;\ln(t) \;\mathrm{d}t$ Taking the derivative once more on both sides with respect to $x$ $\Gamma^{\prime\prime}(x) = \int_{0} ^{\infty} e^{-t}\;t^{x-1}\;\ln^{2}(t) \;\mathrm{d}t$ Setting $x=1$ $\Gamma^{\prime\prime}(1)= \int_{0} ^{\infty} e^{-t}\;\ln^{2}(t) \;\mathrm{d}t$ $\Gamma^{\prime\prime}(1) = \gamma^{2} + \frac{\pi^{2}}{6}$

Thus our limit is then equal to $\frac{\gamma^{2} + \frac{\pi^{2}}{6}}{2} \approx 0.989$

Notable Resources:

In depth analysis of the derivative of the Gamma Function

Gamma Function

Euler-Mascheroni constant