## University Calculus: Early Transcendentals (3rd Edition)

$\lim\limits_{x \to -3} \frac{(x+3)}{x^{2}+4x+3} == -\frac{1}{2} = -0.5$
Simplify: $\frac{(x+3)}{x^{2}+4x+3} = \frac{(x+3)}{(x+3)(x+1)} = \frac{1}{(x+1)}$ $\lim\limits_{x \to -3} \frac{(x+3)}{x^{2}+4x+3} =\lim\limits_{x \to -3} \frac{1}{x+1}$ $= -\frac{1}{2} = -0.5$