1. **State the problem:** Determine if the limit $$\lim_{(x,y) \to (0,0)} \frac{x^2 - 3x}{4x - xy}$$ exists. 2. **Recall the rule:** For a two-variable limit to exist, the limit must be the same regardless of the path taken to approach $(0,0)$. 3. **Simplify the expression:** $$\frac{x^2 - 3x}{4x - xy} = \frac{x(x - 3)}{x(4 - y)}$$ 4. **Cancel common factor $x$ (for $x \neq 0$):** $$\frac{\cancel{x}(x - 3)}{\cancel{x}(4 - y)} = \frac{x - 3}{4 - y}$$ 5. **Evaluate the limit as $(x,y) \to (0,0)$:** $$\lim_{(x,y) \to (0,0)} \frac{x - 3}{4 - y} = \frac{0 - 3}{4 - 0} = \frac{-3}{4}$$ 6. **Check if the limit depends on the path:** Since the simplified expression no longer depends on $y$ in the numerator and the denominator approaches 4, the limit is consistent. **Final answer:** The limit exists and equals $-\frac{3}{4}$. **Answer choice:** B. -3/4