Question

hw 4 - limits to infinity section 1.3: problem (1 point) results for this submission entered answer preview dne dne the answer above is not correct. find the following limit. notes: enter \dne\ if the limit does not exist. (lim_{x \to -infty} \frac{2x^3 + 6x + 3}{-2x^2 + 6x + 6}) = dne preview my answers submit answers your score was recorded. your score was successfully sent to the lms. you have attempted this problem 1 time. you received a score of 0% for this attempt. your overall recorded score is 0%. you have 5 attempts remaining.

Explanation:

Step1: Identify the degrees of numerator and denominator

The numerator is \(2x^{3}+6x + 3\), the degree of the numerator (the highest power of \(x\)) is \(n = 3\). The denominator is \(-2x^{2}+6x + 6\), the degree of the denominator is \(m=2\).

Step2: Apply the rule for limits at infinity for rational functions

For a rational function \(\lim_{x
ightarrow\pm\infty}\frac{a_{n}x^{n}+a_{n - 1}x^{n-1}+\cdots+a_{0}}{b_{m}x^{m}+b_{m - 1}x^{m - 1}+\cdots+b_{0}}\), if \(n>m\), then:

• If \(x

ightarrow+\infty\) and \(a_{n}\) and \(b_{m}\) have the same sign, the limit is \(+\infty\); if they have opposite signs, the limit is \(-\infty\).

• If \(x

ightarrow-\infty\) and \(n\) is odd:

• If \(a_{n}>0\), as \(x

ightarrow-\infty\), \(x^{n}
ightarrow-\infty\) (since \(n\) is odd), and if \(b_{m}\) is a non - zero constant (or a polynomial of lower degree), we can analyze the leading terms.
For our function \(\lim_{x
ightarrow-\infty}\frac{2x^{3}+6x + 3}{-2x^{2}+6x + 6}\), we can divide both the numerator and the denominator by the highest power of \(x\) in the denominator, which is \(x^{2}\) (note that \(x
ightarrow-\infty\), so \(x^{2}>0\) and \(\frac{x^{3}}{x^{2}}=x\), \(\frac{x}{x^{2}}=\frac{1}{x}\), \(\frac{1}{x^{2}}\) etc.)

\[

$$\begin{align*} \lim_{x ightarrow-\infty}\frac{2x^{3}+6x + 3}{-2x^{2}+6x + 6}&=\lim_{x ightarrow-\infty}\frac{\frac{2x^{3}}{x^{2}}+\frac{6x}{x^{2}}+\frac{3}{x^{2}}}{\frac{-2x^{2}}{x^{2}}+\frac{6x}{x^{2}}+\frac{6}{x^{2}}}\\ &=\lim_{x ightarrow-\infty}\frac{2x+\frac{6}{x}+\frac{3}{x^{2}}}{-2+\frac{6}{x}+\frac{6}{x^{2}}} \end{align*}$$

\]

As \(x
ightarrow-\infty\), \(\frac{6}{x}
ightarrow0\), \(\frac{3}{x^{2}}
ightarrow0\), and \(\frac{6}{x}
ightarrow0\), \(\frac{6}{x^{2}}
ightarrow0\). So we have:

\[
\lim_{x
ightarrow-\infty}\frac{2x+0 + 0}{-2+0 + 0}=\lim_{x
ightarrow-\infty}\frac{2x}{-2}=\lim_{x
ightarrow-\infty}(-x)
\]

Since \(x
ightarrow-\infty\), then \(-x
ightarrow+\infty\). So the limit is \(+\infty\) (or we can say the limit does not exist in the sense of a finite number, but the behavior is that it goes to \(+\infty\)). Wait, actually, when \(n>m\) and \(n\) is odd:

The leading term of the numerator is \(2x^{3}\), as \(x
ightarrow-\infty\), \(x^{3}
ightarrow-\infty\), so \(2x^{3}
ightarrow-\infty\). The leading term of the denominator is \(-2x^{2}\), as \(x
ightarrow-\infty\), \(x^{2}
ightarrow+\infty\), so \(-2x^{2}
ightarrow-\infty\). But we can also use the rule: when \(n>m\), \(\lim_{x
ightarrow\pm\infty}\frac{\text{numerator}}{\text{denominator}}\) has the same sign as \(\frac{a_{n}}{b_{m}}\) times the sign of \(x^{n - m}\) as \(x
ightarrow\pm\infty\). Here \(n - m=1\) (odd), \(a_{n}=2\), \(b_{m}=-2\). \(\frac{a_{n}}{b_{m}}=\frac{2}{-2}=-1\). As \(x
ightarrow-\infty\), \(x^{1}
ightarrow-\infty\). So \(-1\times(-\infty)=+\infty\).

Answer:

\(+\infty\) (or we can also express it as the limit does not exist in the finite sense, but the correct behavior is that it approaches \(+\infty\) as \(x
ightarrow-\infty\))