exercises 4.6 limits at infinity and asymptotes
score: 21.5/31 answered: 10/14
question 11
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evaluate the limit
$ \lim _ { x \
ightarrow + \infty } \frac { 3 x ^ { 3 } - 11 x ^ { 2 } - 6 x } { 10 - 2 x - 5 x ^ { 3 } } $
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Accepted Answer

# Explanation:
## Step1: Divide numerator and denominator by $x^{3}$
When $x\to+\infty$, divide each term in the numerator $3x^{3}-11x^{2}-6x$ and denominator $10 - 2x-5x^{3}$ by $x^{3}$.
$$
\begin{align*}
\lim_{x\to+\infty}\frac{3x^{3}-11x^{2}-6x}{10 - 2x-5x^{3}}&=\lim_{x\to+\infty}\frac{\frac{3x^{3}}{x^{3}}-\frac{11x^{2}}{x^{3}}-\frac{6x}{x^{3}}}{\frac{10}{x^{3}}-\frac{2x}{x^{3}}-\frac{5x^{3}}{x^{3}}}\
&=\lim_{x\to+\infty}\frac{3-\frac{11}{x}-\frac{6}{x^{2}}}{\frac{10}{x^{3}}-\frac{2}{x^{2}}-5}
\end{align*}
$$
## Step2: Use the limit rule $\lim_{x\to+\infty}\frac{c}{x^{n}} = 0$ ($c$ is a constant and $n>0$)
We know that $\lim_{x\to+\infty}\frac{c}{x^{n}}=0$ for $c\in R$ and $n > 0$.
$$
\begin{align*}
\lim_{x\to+\infty}\frac{3-\frac{11}{x}-\frac{6}{x^{2}}}{\frac{10}{x^{3}}-\frac{2}{x^{2}}-5}&=\frac{\lim_{x\to+\infty}(3-\frac{11}{x}-\frac{6}{x^{2}})}{\lim_{x\to+\infty}(\frac{10}{x^{3}}-\frac{2}{x^{2}}-5)}\
&=\frac{\lim_{x\to+\infty}3-11\lim_{x\to+\infty}\frac{1}{x}-6\lim_{x\to+\infty}\frac{1}{x^{2}}}{10\lim_{x\to+\infty}\frac{1}{x^{3}}-2\lim_{x\to+\infty}\frac{1}{x^{2}}-\lim_{x\to+\infty}5}\
&=\frac{3 - 0-0}{0 - 0-5}
\end{align*}
$$

# Answer:
$-\frac{3}{5}$

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Question

Explanation:

Step1: Divide numerator and denominator by \(x^{3}\)

Step2: Use the limit rule \(\lim_{x\to+\infty}\frac{c}{x^{n}} = 0\) (\(c\) is a constant and \(n>0\))

Answer: