## Calculus (3rd Edition)

$\frac{3}{4}$
Divide the numerator and denominator by $x^{2}$, which is the highest power of x in the denominator. Then, we get $\lim\limits_{x \to \infty}\frac{3x^{2}+20x}{4x^{2}+9}=\lim\limits_{x \to \infty}\frac{3-\frac{20}{x}}{4+\frac{9}{x^{2}}}=\frac{3-0}{4+0}=\frac{3}{4}$