There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty} \sin{\left(x! \right)} = \left\langle -1, 1\right\rangle$$ $$\lim_{x \to 0^-} \sin{\left(x! \right)} = \sin{\left(1 \right)}$$ More at x→0 from the left $$\lim_{x \to 0^+} \sin{\left(x! \right)} = \sin{\left(1 \right)}$$ More at x→0 from the right $$\lim_{x \to 1^-} \sin{\left(x! \right)} = \sin{\left(1 \right)}$$ More at x→1 from the left $$\lim_{x \to 1^+} \sin{\left(x! \right)} = \sin{\left(1 \right)}$$ More at x→1 from the right $$\lim_{x \to -\infty} \sin{\left(x! \right)} = \sin{\left(\left(-\infty\right)! \right)}$$ More at x→-oo