1. We are asked to find the limit of the function $f(x)=\lfloor x \rfloor$ (the floor function) as $x \to 1$. 2. Recall that the floor function $\lfloor x \rfloor$ gives the greatest integer less than or equal to $x$. 3. Evaluate the left-hand limit as $x \to 1^{-}$: For values just less than 1, $\lfloor x \rfloor = 0$ because these values are less than 1 but greater than or equal to 0. 4. Evaluate the right-hand limit as $x \to 1^{+}$: For values just greater than 1, $\lfloor x \rfloor = 1$ because these values are greater than 1 but less than 2. 5. Check if left-hand limit equals the right-hand limit: Left limit = 0, Right limit = 1. 6. Since left limit $\neq$ right limit, the limit $\lim_{x \to 1} \lfloor x \rfloor$ does not exist. 7. Even though $f(1) = \lfloor 1 \rfloor = 1$, the limit from the left and right are not equal so the limit at $x=1$ does not exist. Final answer: $$\lim_{x \to 1} \lfloor x \rfloor \text{ does not exist.}$$