details\nno additional details were added for this assignment.\nhw6 limits at infinity and asymptotes…

Explanation:

Step1: Divide by highest - power term

Divide both the numerator and denominator by $x^{2}$, the highest - power of $x$ in the denominator. So we have $\lim_{x\rightarrow\infty}\frac{\frac{9x}{x^{2}}+\frac{6}{x^{2}}}{\frac{3x^{2}}{x^{2}}-\frac{4x}{x^{2}}+\frac{8}{x^{2}}}=\lim_{x\rightarrow\infty}\frac{\frac{9}{x}+\frac{6}{x^{2}}}{3-\frac{4}{x}+\frac{8}{x^{2}}}$.

Step2: Use limit properties

We know that $\lim_{x\rightarrow\infty}\frac{c}{x^{n}} = 0$ for any constant $c$ and positive integer $n$. So, $\lim_{x\rightarrow\infty}\frac{9}{x}=0$, $\lim_{x\rightarrow\infty}\frac{6}{x^{2}} = 0$, $\lim_{x\rightarrow\infty}\frac{4}{x}=0$, and $\lim_{x\rightarrow\infty}\frac{8}{x^{2}} = 0$. Then $\lim_{x\rightarrow\infty}\frac{\frac{9}{x}+\frac{6}{x^{2}}}{3-\frac{4}{x}+\frac{8}{x^{2}}}=\frac{0 + 0}{3-0 + 0}$.

Answer:

$0$