Limit Difference Cubes 87270E

1. **State the problem:** Find the limit $$\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$$. 2. **Recognize the form:** Direct substitution gives $$\frac{2^3 - 8}{2 - 2} = \frac{8 - 8}{0} = \frac{0}{0}$$, which is an indeterminate form. 3. **Use factorization:** Recall the difference of cubes formula: $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$. 4. **Apply the formula:** Here, $$x^3 - 8 = (x - 2)(x^2 + 2x + 4)$$. 5. **Rewrite the limit:** $$\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{x - 2}$$ 6. **Cancel common factors:** $$\lim_{x \to 2} \frac{\cancel{(x - 2)}(x^2 + 2x + 4)}{\cancel{(x - 2)}} = \lim_{x \to 2} (x^2 + 2x + 4)$$ 7. **Evaluate the limit:** Substitute $$x = 2$$: $$2^2 + 2 \times 2 + 4 = 4 + 4 + 4 = 12$$. **Final answer:** $$12$$.