$$\lim_{x \to \infty}\left(\frac{x!}{2^{x} + 1}\right) = \infty$$ $$\lim_{x \to 0^-}\left(\frac{x!}{2^{x} + 1}\right) = \frac{1}{2}$$ More at x→0 from the left $$\lim_{x \to 0^+}\left(\frac{x!}{2^{x} + 1}\right) = \frac{1}{2}$$ More at x→0 from the right $$\lim_{x \to 1^-}\left(\frac{x!}{2^{x} + 1}\right) = \frac{1}{3}$$ More at x→1 from the left $$\lim_{x \to 1^+}\left(\frac{x!}{2^{x} + 1}\right) = \frac{1}{3}$$ More at x→1 from the right $$\lim_{x \to -\infty}\left(\frac{x!}{2^{x} + 1}\right) = \left(-\infty\right)!$$ More at x→-oo