calculus - If $lim_{xtoinfty}frac{sin x}{x} = 0$ is zero, why does it not work with L'hospital's way?

I just got a simple question regarding the use of L'Hopitals method for finding limits. Usually L'Hopitals method can be used to find limits like

$$\lim_{x\to0}\frac{\sin x}{x} = \lim_{x\to 0}\dfrac{\dfrac{d}{dx} \sin x}{\dfrac{d}{dx} x} = \lim_{x\to 0}\cos x$$

Here if we plug $0$, we can find the limit of the original function $\dfrac{\sin x}{x}$ at $0$ using the $\cos x$ function. Put $0$ in, and you will get $1$, which is correct. However, if we replace $x$ with $\infty$, we don't get the right limit.

$$\cos x$$
$$\cos(\infty)$$

Which is not right for the limit of the original function, as

$$\lim_{x\to\infty}\frac{\sin x}{x} = 0$$ Using the new function which we get via L'Hopital's method does not help get that. Is this like a special case? In what cases could then L'Hopital's way not work?