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?

Unfortunately L'Hopital would not help since this does not satisfy the assumptions. 
Note that 
$$
\left|\frac{\sin x}{x-\pi}\right|\leq\frac{1}{|x-\pi|}\to 0
$$
as $x\to\infty$. 
What is the point not to use the squeeze theorem?
