evaluate the limit
lim_{x→∞} (3x^3 - 9x^2 - 3x)/(9 - 7x - 7x^3)
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Accepted Answer

# 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 $x^{3}$.
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{3x^{3}-9x^{2}-3x}{9 - 7x-7x^{3}}&=\lim_{x
ightarrow\infty}\frac{\frac{3x^{3}}{x^{3}}-\frac{9x^{2}}{x^{3}}-\frac{3x}{x^{3}}}{\frac{9}{x^{3}}-\frac{7x}{x^{3}}-\frac{7x^{3}}{x^{3}}}\
&=\lim_{x
ightarrow\infty}\frac{3-\frac{9}{x}-\frac{3}{x^{2}}}{\frac{9}{x^{3}}-\frac{7}{x^{2}}-7}
\end{align*}
$$
## Step2: Apply limit rules
We know that $\lim_{x
ightarrow\infty}\frac{c}{x^{n}} = 0$ for any constant $c$ and positive integer $n$.
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{3-\frac{9}{x}-\frac{3}{x^{2}}}{\frac{9}{x^{3}}-\frac{7}{x^{2}}-7}&=\frac{\lim_{x
ightarrow\infty}(3)-\lim_{x
ightarrow\infty}\frac{9}{x}-\lim_{x
ightarrow\infty}\frac{3}{x^{2}}}{\lim_{x
ightarrow\infty}\frac{9}{x^{3}}-\lim_{x
ightarrow\infty}\frac{7}{x^{2}}-\lim_{x
ightarrow\infty}7}\
&=\frac{3 - 0-0}{0 - 0-7}
\end{align*}
$$
# Answer:
$-\frac{3}{7}$

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QUESTION IMAGE

Question

Explanation:

Step1: Divide by highest - power of x

Step2: Apply limit rules

Answer: