1. **Problem:** Find the limit $$\lim_{x \to 0} x \lfloor \cos x \rfloor$$ where $$\lfloor . \rfloor$$ is the greatest integer function. 2. **Recall:** The greatest integer function $$\lfloor y \rfloor$$ gives the greatest integer less than or equal to $$y$$. 3. **Evaluate inside the floor:** As $$x \to 0$$, $$\cos x \to 1$$. 4. Since $$\cos 0 = 1$$, and for values of $$x$$ close to 0, $$\cos x < 1$$ but very close to 1, so $$\lfloor \cos x \rfloor = 0$$ for $$x \neq 0$$ near zero because $$\cos x$$ is slightly less than 1. 5. Therefore, near zero, $$x \lfloor \cos x \rfloor = x \times 0 = 0$$. 6. At $$x=0$$, $$x \lfloor \cos x \rfloor = 0 \times 1 = 0$$. 7. Hence, $$\lim_{x \to 0} x \lfloor \cos x \rfloor = 0$$. **Final answer:** (b) 0