## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x \to 3}\frac{x^{2}-3x}{x^{2}-9}=1/2$
For the numerator $x^{2}-3x= x(x-3)$ For the denominator $x^{2}-9= (x-3)(x+3)$ Cancel (x-3) We have $\lim\limits_{x \to 3}\frac{x}{x+3}=3/(3+3)=1/2$