1. **State the problem:** Find the limit $$\lim_{x \to 2} \frac{x^2 - 4}{\sqrt[3]{x - 3} + 1}$$. 2. **Recall the formula and rules:** The limit of a function as $x$ approaches a value is the value the function approaches. If direct substitution leads to an indeterminate form like $\frac{0}{0}$, we simplify the expression. 3. **Evaluate the numerator and denominator at $x=2$: ** - Numerator: $2^2 - 4 = 4 - 4 = 0$ - Denominator: $\sqrt[3]{2 - 3} + 1 = \sqrt[3]{-1} + 1 = -1 + 1 = 0$ Since direct substitution gives $\frac{0}{0}$, we simplify. 4. **Factor the numerator:** $$x^2 - 4 = (x - 2)(x + 2)$$ 5. **Rewrite the limit:** $$\lim_{x \to 2} \frac{(x - 2)(x + 2)}{\sqrt[3]{x - 3} + 1}$$ 6. **Use substitution:** Let $t = \sqrt[3]{x - 3}$, so $t^3 = x - 3$ and $x = t^3 + 3$. When $x \to 2$, then $t = \sqrt[3]{2 - 3} = \sqrt[3]{-1} = -1$. Rewrite numerator and denominator in terms of $t$: - Numerator: $(t^3 + 3 - 2)(t^3 + 3 + 2) = (t^3 + 1)(t^3 + 5)$ - Denominator: $t + 1$ 7. **Rewrite the limit in $t$: ** $$\lim_{t \to -1} \frac{(t^3 + 1)(t^3 + 5)}{t + 1}$$ 8. **Factor $t^3 + 1$ using sum of cubes:** $$t^3 + 1 = (t + 1)(t^2 - t + 1)$$ 9. **Simplify the expression:** $$\frac{(t + 1)(t^2 - t + 1)(t^3 + 5)}{t + 1} = (t^2 - t + 1)(t^3 + 5)$$ 10. **Evaluate the limit by substituting $t = -1$: ** - $t^2 - t + 1 = (-1)^2 - (-1) + 1 = 1 + 1 + 1 = 3$ - $t^3 + 5 = (-1)^3 + 5 = -1 + 5 = 4$ Multiply: $$3 \times 4 = 12$$ **Final answer:** $$\lim_{x \to 2} \frac{x^2 - 4}{\sqrt[3]{x - 3} + 1} = 12$$