Solving $\lim\limits_{x \to \infty} \frac{\sin(x)}{x-\pi}$ using L'Hôpital's rule 
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?

L'Hopital's Rule only works when the limit is $0/0$ or $\infty/\infty$. In this case, the limit is undefined over $\infty$, so you cannot use L'Hopital's Rule on this problem (at least the way it's written.)
