1. **State the problem:** Find the limit $$\lim_{x \to -3} \frac{4x^2 + 12x}{x^2 + 4x + 3}$$. 2. **Check direct substitution:** Substitute $x = -3$ into numerator and denominator. Numerator: $$4(-3)^2 + 12(-3) = 4(9) - 36 = 36 - 36 = 0$$ Denominator: $$(-3)^2 + 4(-3) + 3 = 9 - 12 + 3 = 0$$ Since direct substitution gives $\frac{0}{0}$, an indeterminate form, we need to simplify. 3. **Factor numerator and denominator:** Numerator: $$4x^2 + 12x = 4x(x + 3)$$ Denominator: $$x^2 + 4x + 3 = (x + 3)(x + 1)$$ 4. **Simplify the expression:** $$\frac{4x(x + 3)}{(x + 3)(x + 1)} = \frac{4x}{x + 1}, \quad x \neq -3$$ 5. **Evaluate the limit of the simplified expression:** $$\lim_{x \to -3} \frac{4x}{x + 1} = \frac{4(-3)}{-3 + 1} = \frac{-12}{-2} = 6$$ 6. **Conclusion:** The limit is $6$. **Final answer:** C 6