An Additional Apostrophe

$\large = \lim_{x\to0} \dfrac{1}{\Gamma(x)}$ Because $\lim_{x\to0} \Gamma(x) = \infty$ , we get
$\lim_{x\to0} \dfrac{1}{\Gamma(x)} = \frac{1}{\infty} = \boxed{0}$

If we define $x! := \Gamma\left(x+1\right)$ , then by definition of continuity, $\lim\limits_{x \to 0} \dfrac{x}{x!} = \dfrac{0}{1} = 0$ ,
because $f\left(x\right) = x$ and $g\left(x\right) = x!$ are continuous at every $x \in \mathbb{C}$ meaning that $\dfrac{f\left(x\right)}{g\left(x\right)}$ is continuous at every $x \in \mathbb{C}$ such that $g\left(x\right) \neq 0$ .

Hence, the answer is $\boxed{0}$ .

Since x! = 0 and 0/1 is equal to zero, we can say that this limit equal to 0 by implementing "0" in expression.

P. S. I'm not good in formatting text, especially math symbols, so I hope you understand my explanation