1. **State the problem:** We want to find the limit of the function $$f(x,y) = \begin{cases} \frac{x^3 - y^3}{x + y} & \text{if } x + y \neq 0 \\ 0 & \text{if } x + y = 0 \end{cases}$$ at the point $(0,0)$ and test if $f$ is continuous there. 2. **Recall the definition of continuity:** A function $f$ is continuous at $(0,0)$ if $$\lim_{(x,y) \to (0,0)} f(x,y) = f(0,0).$$ 3. **Evaluate $f(0,0)$:** Since $0 + 0 = 0$, we use the second case: $$f(0,0) = 0.$$ 4. **Simplify the expression for $x + y \neq 0$:** Factor the numerator using the difference of cubes formula: $$x^3 - y^3 = (x - y)(x^2 + xy + y^2).$$ So, $$f(x,y) = \frac{(x - y)(x^2 + xy + y^2)}{x + y}.$$ 5. **Check the limit along different paths:** - Along $y = -x$ (where denominator is zero), the function is defined as 0. - Along $y = 0$: $$f(x,0) = \frac{x^3 - 0}{x + 0} = \frac{x^3}{x} = x^2 \to 0 \text{ as } x \to 0.$$ - Along $y = mx$ for $m \neq -1$: $$f(x,mx) = \frac{x^3 - (mx)^3}{x + mx} = \frac{x^3 - m^3 x^3}{x(1 + m)} = \frac{x^3(1 - m^3)}{x(1 + m)} = \frac{x^2(1 - m^3)}{1 + m} \to 0 \text{ as } x \to 0.$$ 6. **Conclusion on the limit:** Since the limit is 0 along all paths and matches $f(0,0)$, the limit exists and equals 0. 7. **Test for continuity:** Since $$\lim_{(x,y) \to (0,0)} f(x,y) = f(0,0) = 0,$$ $f$ is continuous at $(0,0)$. **Final answer:** $$\lim_{(x,y) \to (0,0)} f(x,y) = 0,$$ and $f$ is continuous at $(0,0).$