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?

I bet that your problem is, in fact, to compute
$$\lim _{x \to \color{red} \pi} \frac {\sin x} {x - \pi} ,$$
so either you have mistyped $\infty$ instead of $\pi$, or there is a typo in the text where you took this from.
In this case, L'Hospital's theorem could be used, but it's not necessary, because
$$\lim _{x \to \pi} \frac {\sin x} {x - \pi} = \lim _{x \to \pi} \frac {\sin x - 0} {x - \pi} = \lim _{x \to \pi} \frac {\sin x - \sin \pi} {x - \pi} = (\sin ') (\pi) = \cos \pi = -1 .$$
