1. **State the problem:** We need to find the left-hand limit as $x$ approaches $-\frac{1}{3}$ from the left of the function $f(x) = \left\lfloor \frac{1}{x} \right\rfloor$. 2. **Recall the floor function:** The floor function $\left\lfloor y \right\rfloor$ gives the greatest integer less than or equal to $y$. 3. **Analyze the behavior of $\frac{1}{x}$ as $x \to -\frac{1}{3}^-$:** - Approaching $-\frac{1}{3}$ from the left means $x < -\frac{1}{3}$ but close to it. - For values just less than $-\frac{1}{3}$, $x$ is slightly more negative, so $x < -\frac{1}{3}$. - Since $x$ is negative and close to $-\frac{1}{3}$ from the left, $\frac{1}{x}$ will be a large negative number but slightly greater than $-3$ (because $\frac{1}{-\frac{1}{3}} = -3$). 4. **Evaluate the limit:** - For $x$ just less than $-\frac{1}{3}$, $\frac{1}{x}$ is slightly greater than $-3$ (e.g., $-2.999$). - The floor of a number slightly greater than $-3$ but less than $-2$ is $-3$. 5. **Conclusion:** $$\lim_{x \to -\frac{1}{3}^-} \left\lfloor \frac{1}{x} \right\rfloor = -3$$