Question

homework 1.2: limits involving infinity
score: 16/36 answered: 9/18
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question 10
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
ightarrowinfty}\frac{- 120 + 3x^{2}+66x - 3x^{3}}{-60 - 4x^{2}-32x}=
lim_{x
ightarrow-infty}\frac{- 120 + 3x^{2}+66x - 3x^{3}}{-60 - 4x^{2}-32x}=

Explanation:

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

For a rational function $\frac{f(x)}{g(x)}$ where $f(x)=- 120 + 3x^{2}+66x - 3x^{3}$ and $g(x)=-60 - 4x^{2}-32x$, the highest - power of $x$ in the denominator is $x^{2}$. Divide each term in the numerator and denominator by $x^{2}$:
\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{-120 + 3x^{2}+66x - 3x^{3}}{-60 - 4x^{2}-32x}&=\lim_{x ightarrow\infty}\frac{\frac{-120}{x^{2}}+3+\frac{66}{x}-3x}{\frac{-60}{x^{2}}-4-\frac{32}{x}} \end{align*}$$

\]

Step2: Evaluate the limit of each term

As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{-120}{x^{2}} = 0$, $\lim_{x
ightarrow\infty}\frac{66}{x}=0$, $\lim_{x
ightarrow\infty}\frac{-60}{x^{2}} = 0$, $\lim_{x
ightarrow\infty}\frac{32}{x}=0$.
So, $\lim_{x
ightarrow\infty}\frac{\frac{-120}{x^{2}}+3+\frac{66}{x}-3x}{\frac{-60}{x^{2}}-4-\frac{32}{x}}=\lim_{x
ightarrow\infty}\frac{0 + 3+0-3x}{0 - 4-0}= \lim_{x
ightarrow\infty}\frac{3-3x}{-4}=-\infty$

Step3: For $x

ightarrow-\infty$
Divide numerator and denominator by $x^{2}$: $\lim_{x
ightarrow-\infty}\frac{-120 + 3x^{2}+66x - 3x^{3}}{-60 - 4x^{2}-32x}=\lim_{x
ightarrow-\infty}\frac{\frac{-120}{x^{2}}+3+\frac{66}{x}-3x}{\frac{-60}{x^{2}}-4-\frac{32}{x}}$
As $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}\frac{-120}{x^{2}} = 0$, $\lim_{x
ightarrow-\infty}\frac{66}{x}=0$, $\lim_{x
ightarrow-\infty}\frac{-60}{x^{2}} = 0$, $\lim_{x
ightarrow-\infty}\frac{32}{x}=0$.
So, $\lim_{x
ightarrow-\infty}\frac{\frac{-120}{x^{2}}+3+\frac{66}{x}-3x}{\frac{-60}{x^{2}}-4-\frac{32}{x}}=\lim_{x
ightarrow-\infty}\frac{0 + 3+0-3x}{0 - 4-0}=\lim_{x
ightarrow-\infty}\frac{3-3x}{-4}=\infty$

Answer:

$\lim_{x
ightarrow\infty}\frac{-120 + 3x^{2}+66x - 3x^{3}}{-60 - 4x^{2}-32x}=-\infty$
$\lim_{x
ightarrow-\infty}\frac{-120 + 3x^{2}+66x - 3x^{3}}{-60 - 4x^{2}-32x}=\infty$