It is known that $$\Gamma'(1)=-\gamma$$where $\Gamma(x)=(x-1)!$ at integer $x$ and $\gamma$ is the Euler-Mascheroni constant. We can redefine this as $$\Pi'(0)=-\gamma$$Where $\Pi(x)=x!$. By L'hopital's rule $$\lim_{k\rightarrow0}\frac{\Pi(k)}k=\lim_{k\rightarrow0}\Pi'(k)=\Pi'(0)=-\gamma$$But when I graph the function $\frac{\Pi(k)}k$, there is a singularity at $k=0$. What is wrong here? I do know that $\frac{\Pi(k)}k=\Gamma(k)$ but I don't understand how this makes an issue.
I am very sorry if there is something obvious here that I am missing.
L'Hopital's rule does not apply to the quotient $\Pi(k)/k$, because it is not an indeterminate form: from your definition, $$\Pi(x) = \Gamma(x+1) \implies \Pi(0) = \Gamma(1) = 0! = 1,$$ whereas your denominator is $0$.