$\lim \limits_{x \rightarrow 0 }\lfloor \cos(x) \rfloor= ?$

You're right.

The function $f(x)=\lfloor \cos x\rfloor$ is constant in a punctured neighborhood of $0$: for $x\in(-\pi/2,\pi/2)$, $x\ne0$, $$ \lfloor \cos x\rfloor=0 $$ because $0<\cos x<1$. Therefore $$ \lim_{x\to0}\lfloor \cos x\rfloor=0 $$

What the value of $\lfloor \cos x\rfloor$ is for $x=0$ is completely irrelevant as far as the limit is concerned, except for the fact you can conclude that $f$ is not continuous at $0$.