Question

homework 1.2: limits involving infinity
score: 16/36 answered: 10/18
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question 11
0/2 pts 8 49 details
determine the following limits. enter dne if a limit fails to exist, except in case of an infinite limit. if an infinite limit exists, enter ∞ or -∞, as appropriate.
lim(x→∞) (x³ - 21x - 20)/(-6x² - 69x + 3x³ + 180)=
lim(x→-∞) (x³ - 21x - 20)/(-6x² - 69x + 3x³ + 180)=
determine the equation of the horizontal asymptote that corresponds to the limit as x→∞.
equation of horizontal asymptote:
no horizontal asymptote corresponds to the limit as x→∞.
determine the equation of the horizontal asymptote that corresponds to the limit as x→-∞.
equation of horizontal asymptote:
no horizontal asymptote corresponds to the limit as x→-∞
add work

Explanation:

Step1: Divide numerator and denominator by highest - power of x

Divide both the numerator and denominator of the function $\frac{x^{3}-21x - 20}{-6x^{2}-69x + 3x^{3}+180}$ by $x^{3}$. We get $\lim_{x
ightarrow\pm\infty}\frac{1-\frac{21}{x^{2}}-\frac{20}{x^{3}}}{-\frac{6}{x}-\frac{69}{x^{2}} + 3+\frac{180}{x^{3}}}$.

Step2: Evaluate the limit as $x

ightarrow\infty$
As $x
ightarrow\infty$, $\frac{1}{x}
ightarrow0$, $\frac{1}{x^{2}}
ightarrow0$, and $\frac{1}{x^{3}}
ightarrow0$. So, $\lim_{x
ightarrow\infty}\frac{1-\frac{21}{x^{2}}-\frac{20}{x^{3}}}{-\frac{6}{x}-\frac{69}{x^{2}} + 3+\frac{180}{x^{3}}}=\frac{1 - 0-0}{0 - 0+3+0}=\frac{1}{3}$.

Step3: Evaluate the limit as $x

ightarrow-\infty$
As $x
ightarrow-\infty$, $\frac{1}{x}
ightarrow0$, $\frac{1}{x^{2}}
ightarrow0$, and $\frac{1}{x^{3}}
ightarrow0$. So, $\lim_{x
ightarrow-\infty}\frac{1-\frac{21}{x^{2}}-\frac{20}{x^{3}}}{-\frac{6}{x}-\frac{69}{x^{2}} + 3+\frac{180}{x^{3}}}=\frac{1 - 0-0}{0 - 0+3+0}=\frac{1}{3}$.

Step4: Determine the horizontal asymptote

Since $\lim_{x
ightarrow\infty}f(x)=\lim_{x
ightarrow-\infty}f(x)=\frac{1}{3}$, the equation of the horizontal asymptote for both $x
ightarrow\infty$ and $x
ightarrow-\infty$ is $y = \frac{1}{3}$.

Answer:

• $\lim_{x

ightarrow\infty}\frac{x^{3}-21x - 20}{-6x^{2}-69x + 3x^{3}+180}=\frac{1}{3}$

• $\lim_{x

ightarrow-\infty}\frac{x^{3}-21x - 20}{-6x^{2}-69x + 3x^{3}+180}=\frac{1}{3}$

• Equation of horizontal asymptote as $x

ightarrow\infty$: $y=\frac{1}{3}$

• Equation of horizontal asymptote as $x

ightarrow-\infty$: $y=\frac{1}{3}$