I know how to solve this using the squeeze theorem, but I am supposed to solve only using L'Hôpital's rule
$$\lim\limits_{x \to \infty} \frac{\sin(x)}{x-\pi}$$
I tried: $$\lim\limits_{x \to \infty} \frac{\sin(x)}{x-\pi} = \lim\limits_{x \to \infty} \frac{d/dx[\sin(x)]}{d/dx[x-\pi]} = \lim\limits_{x \to \infty} \frac{\cos(x)}{1}$$
From here I am stuck because the rule no longer applies and using $\infty$ for $x$ doesn't not help to simplify.
Logically the limit is $0$ because $\sin(x)$ can only be $-1$ to $1$, but this is using squeeze theorem.
Is still there any way to solve this without using the squeeze theorem?