Answer
$\lim\limits_{x \to 1} \frac{3x+9}{x^2+4x+3} = \frac{3}{2}$
Work Step by Step
Using the rules of limits, we find:
$$\lim\limits_{x \to 1} \frac{3x+9}{x^2+4x+3} \\= \lim\limits_{x \to 1} \frac{3(x+3)}{(x+3)(x+1)} \\= \lim\limits_{x \to 1} \frac{3}{x+1} = \frac{3}{1+1} = \frac{3}{2} $$
