## Calculus (3rd Edition)

Divide the numerator and denominator by $x^{4}$, which is the highest power of x occurring in the denominator. Then, we get $\lim\limits_{x \to \infty}\frac{3x^{2}+20x}{2x^{4}+3x^{3}-29}=\lim\limits_{x \to \infty}\frac{\frac{3}{x^{2}}+\frac{20}{x^{3}}}{2+\frac{3}{x}-\frac{29}{x^{4}}}=\frac{0+0}{2+0-0}=0$