## Calculus (3rd Edition)

Dividing the numerator and denominator by x, we have $\lim\limits_{x \to \infty}\frac{x}{x+9}=\lim\limits_{x \to \infty}\frac{1}{1+\frac{9}{x}}=\frac{\lim\limits_{x \to \infty}1}{\lim\limits_{x \to \infty}1+\lim\limits_{x \to \infty}\frac{9}{x}}=\frac{1}{1+0}=1$