1. **State the problem:** Find the limit $$\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$$ algebraically. 2. **Recall the formula and rules:** To find limits involving rational expressions where direct substitution leads to an indeterminate form like $$\frac{0}{0}$$, factor numerator and denominator and simplify. 3. **Factor numerator and denominator:** $$x^3 - 8 = (x - 2)(x^2 + 2x + 4)$$ (difference of cubes) $$x^2 - 4 = (x - 2)(x + 2)$$ (difference of squares) 4. **Rewrite the limit expression:** $$\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)}$$ 5. **Cancel common factor:** $$\lim_{x \to 2} \frac{\cancel{(x - 2)}(x^2 + 2x + 4)}{\cancel{(x - 2)}(x + 2)} = \lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 2}$$ 6. **Evaluate the limit by direct substitution:** $$\frac{2^2 + 2 \cdot 2 + 4}{2 + 2} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3$$ **Final answer:** $$3$$