evaluate the limit\nlim_{x→∞} (3x^3 - 9x^2 - 3x)/(9 - 7x - 7x^3)\nquestion help: ▶ video

Explanation:

Step1: Divide by highest - power of x

Divide both the numerator and denominator by $x^{3}$ since the highest - power of $x$ in the denominator is $x^{3}$. [ \begin{align*} \lim_{x\rightarrow\infty}\frac{3x^{3}-9x^{2}-3x}{9 - 7x-7x^{3}}&=\lim_{x\rightarrow\infty}\frac{\frac{3x^{3}}{x^{3}}-\frac{9x^{2}}{x^{3}}-\frac{3x}{x^{3}}}{\frac{9}{x^{3}}-\frac{7x}{x^{3}}-\frac{7x^{3}}{x^{3}}}\ &=\lim_{x\rightarrow\infty}\frac{3-\frac{9}{x}-\frac{3}{x^{2}}}{\frac{9}{x^{3}}-\frac{7}{x^{2}}-7} \end{align*} ]

Step2: Apply limit rules

We know that $\lim_{x\rightarrow\infty}\frac{c}{x^{n}} = 0$ for any constant $c$ and positive integer $n$. [ \begin{align*} \lim_{x\rightarrow\infty}\frac{3-\frac{9}{x}-\frac{3}{x^{2}}}{\frac{9}{x^{3}}-\frac{7}{x^{2}}-7}&=\frac{\lim_{x\rightarrow\infty}(3)-\lim_{x\rightarrow\infty}\frac{9}{x}-\lim_{x\rightarrow\infty}\frac{3}{x^{2}}}{\lim_{x\rightarrow\infty}\frac{9}{x^{3}}-\lim_{x\rightarrow\infty}\frac{7}{x^{2}}-\lim_{x\rightarrow\infty}7}\ &=\frac{3 - 0-0}{0 - 0-7} \end{align*} ]

Answer:

$-\frac{3}{7}$