Question

hw 4 - limits to infinity section 1.3: problem 7 (1 point) 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} = \square\\)

Explanation:

Step1: Analyze the degrees of numerator and denominator

The numerator \(2x^3 + 6x + 3\) has a degree of 3 (the highest power of \(x\) is 3), and the denominator \(-2x^2 + 6x + 6\) has a degree of 2 (the highest power of \(x\) is 2). When finding the limit as \(x\to -\infty\) (or \(x\to\infty\)) of a rational function, if the degree of the numerator is greater than the degree of the denominator, we can also consider the leading terms (the terms with the highest degrees) of the numerator and the denominator.

First, we can divide both the numerator and the denominator by the highest power of \(x\) in the denominator, which is \(x^2\) (since the denominator's degree is 2). But we need to be careful with the sign when \(x\) is negative (because we are taking the limit as \(x\to -\infty\)). Recall that \(\sqrt{x^2}=\vert x\vert=-x\) when \(x < 0\), but when we divide by \(x^n\) where \(n\) is an integer, for \(x\to -\infty\), \(x^n\) is positive when \(n\) is even and negative when \(n\) is odd.

Let's divide numerator and denominator by \(x^2\):

\[

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

\]

Step2: Evaluate the limit of each term

Now, we analyze the limit of each term as \(x\to -\infty\):

• For the term \(2x\) in the numerator: As \(x\to -\infty\), \(2x\to -\infty\) (because when \(x\) becomes a large negative number, multiplying by 2 still keeps it a large negative number).
• For the terms \(\frac{6}{x}\) and \(\frac{3}{x^2}\) in the numerator: As \(x\to -\infty\), \(\frac{6}{x}\to 0\) (because the denominator becomes a large negative number, so the fraction approaches 0) and \(\frac{3}{x^2}\to 0\) (since \(x^2\) becomes a large positive number, the fraction approaches 0).
• For the terms in the denominator: \(-2\) is a constant, \(\frac{6}{x}\to 0\) as \(x\to -\infty\), and \(\frac{6}{x^2}\to 0\) as \(x\to -\infty\). So the denominator approaches \(-2 + 0+0=-2\).

So now we have:

\[
\lim_{x\to -\infty}\frac{2x+\frac{6}{x}+\frac{3}{x^2}}{-2+\frac{6}{x}+\frac{6}{x^2}}=\lim_{x\to -\infty}\frac{2x + 0+0}{-2+0 + 0}=\lim_{x\to -\infty}\frac{2x}{-2}=\lim_{x\to -\infty}(-x)
\]

As \(x\to -\infty\), \(-x\to \infty\) (because if \(x\) is a large negative number, then \(-x\) is a large positive number). So the limit does not exist (it goes to \(-\infty\) or \(\infty\), depending on the sign, but in this case, as we saw, it goes to \(\infty\) in the sense that the magnitude goes to infinity, but since the question says to enter "DNE" if the limit does not exist, we can also analyze the leading terms directly.

Alternative approach: The degree of the numerator (\(n = 3\)) is greater than the degree of the denominator (\(m = 2\)). For a rational function \(\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}\) where \(a_n
eq 0\) and \(b_m
eq 0\), if \(n>m\), then \(\lim_{x\to\pm\infty}\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}\) does not exist (it goes to \(\pm\infty\) depending on the sign of \(a_n/b_m\) and the parity of \(n - m\) and the sign of \(x\)).

In our case, the leading term of the numerator is \(2x^3\) and the leading term of the denominator is \(-2x^2\). So we can write the limit as:

\[
\lim_{x\to -\infty}\frac{2x^3}{-2x^2}=\lim_{x\to -\infty}(-x)
\]

As \(x\to -\infty\), \(-x\to \infty\), so the limit does not exist.

Answer:

DNE