## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x \to -3}\dfrac{x^{2}+3x}{x^{2}-x-12}=\dfrac{3}{7}$
First apply direct substitution: $\lim\limits_{x \to -3}\dfrac{x^{2}+3x}{x^{2}-x-12}=\dfrac{(-3)^{2}+3(-3)}{(-3)^{2}-(-3)-12}=\dfrac{0}{0}$ $(Indeterminate$ $Form)$ Apply factorization to the numerator and to the denominator: $\lim\limits_{x \to -3}\dfrac{x^{2}+3x}{x^{2}-x-12}=\lim\limits_{x \to -3}\dfrac{x(x+3)}{(x-4)(x+3)}=\lim\limits_{x \to -3}\dfrac{x}{x-4}=...$ $...=\dfrac{-3}{-3-4}=\dfrac{-3}{-7}=\dfrac{3}{7}$