1. We are asked to evaluate the limit $$\lim_{x \to -1} \frac{x^2 + 4x + 3}{x^2 - 3x - 4}$$. 2. First, factor both numerator and denominator: $$x^2 + 4x + 3 = (x + 1)(x + 3)$$ $$x^2 - 3x - 4 = (x - 4)(x + 1)$$ 3. Substitute these into the limit expression: $$\lim_{x \to -1} \frac{(x + 1)(x + 3)}{(x - 4)(x + 1)}$$ 4. Cancel the common factor $x + 1$ (noting $x \to -1$ but $x \neq -1$ in limit): $$\lim_{x \to -1} \frac{\cancel{(x + 1)}(x + 3)}{(x - 4)\cancel{(x + 1)}} = \lim_{x \to -1} \frac{x + 3}{x - 4}$$ 5. Now substitute $x = -1$: $$\frac{-1 + 3}{-1 - 4} = \frac{2}{-5} = -\frac{2}{5}$$ 6. Therefore, the limit is $$-\frac{2}{5}$$. --- Since the user requested only even-numbered problems and to solve only the first, the answer corresponds to problem 2.