1. **State the problem:** Find the limit $$\lim_{x\to -1} \frac{12x^3 + 12x^2}{x^4 - x^2}$$. 2. **Check direct substitution:** Substitute $x = -1$: $$\frac{12(-1)^3 + 12(-1)^2}{(-1)^4 - (-1)^2} = \frac{12(-1) + 12(1)}{1 - 1} = \frac{-12 + 12}{0} = \frac{0}{0}$$ which is an indeterminate form. So, we need to simplify. 3. **Factor numerator and denominator:** Numerator: $$12x^3 + 12x^2 = 12x^2(x + 1)$$ Denominator: $$x^4 - x^2 = x^2(x^2 - 1) = x^2(x - 1)(x + 1)$$ 4. **Simplify the fraction:** $$\frac{12x^2(x + 1)}{x^2(x - 1)(x + 1)}$$ Cancel common factors $x^2$ and $(x + 1)$: $$\frac{12\cancel{x^2}\cancel{(x + 1)}}{\cancel{x^2}(x - 1)\cancel{(x + 1)}}$$ 5. **Simplified expression:** $$\frac{12}{x - 1}$$ 6. **Evaluate the limit with simplified expression:** Substitute $x = -1$: $$\frac{12}{-1 - 1} = \frac{12}{-2} = -6$$ **Final answer:** The limit is $-6$.