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Section 1.3 End Behavior

Consider the function \(\dfrac{1}{x}\text{,}\)

Figure 1.3.1.

  • what does the graph of \(\dfrac{1}{x}\) do as \(x\) increases without bound? \(\displaystyle\lim_{x \to \infty} \dfrac{1}{x} = 0\)

  • similarly, \(\displaystyle\lim_{x \to -\infty} \dfrac{1}{x} = 0\)

  • contrast with the limits at a finite point such as, \(\displaystyle\lim_{x \to 0} \dfrac{1}{x} = DNE\text{,}\) or \(\displaystyle\lim_{x \to 1} \dfrac{1}{x} = 1\)

Limits at infinity are related to horizontal asymptotes; this graph has a horizontal aspymtote at zero.

Table 1.3.2. Consider the following graphs:
Figure 1.3.3. \(y = x\)
\begin{equation*} \lim_{x \to \infty} x = \infty \end{equation*}
\begin{equation*} \lim_{x \to -\infty} x = -\infty \end{equation*}
Figure 1.3.4. \(y = x^2\)
\begin{equation*} \lim_{x \to \infty} x^2 = \infty \end{equation*}
\begin{equation*} \lim_{x \to -\infty} x^2 = \infty \end{equation*}
Figure 1.3.5. \(y = x^3\)
\begin{equation*} \lim_{x \to \infty} x^3 = \infty \end{equation*}
\begin{equation*} \lim_{x \to -\infty} x^3 = -\infty \end{equation*}
Figure 1.3.6. \(y = x^4\)
\begin{equation*} \lim_{x \to \infty} x^4 = \infty \end{equation*}
\begin{equation*} \lim_{x \to -\infty} x^4 = \infty \end{equation*}

The general pattern that we can get from these graphs is:

  • \(\displaystyle\lim_{x \to \infty} x^{n} = \infty\) for \(n = 1 \mbox{, } 2 \mbox{, } \dots\)

  • \(\displaystyle\lim_{x \to - \infty} x^{n}\) \(\begin{cases} -\infty \amp \text{ for } n = 1 \mbox{, } 2 \mbox{, } \dots \\ \infty \amp \text{ for } n = 1 \mbox{, } 2 \mbox{, } \dots \end{cases}\)

Note 1.3.7.

Multiplying by a positive constant doesn't affect end behavior, but a nehative will reverse it.

Subsection 1.3.1 Rules for More Complicated Limits

Firstly, if \(C_{n} \neq 0 \) then, \(\displaystyle\lim_{x \to \infty} (C_{0} + C_{1}x^{1} + \dots + C_{n}x^{n})\) is equal to \(\displaystyle\lim_{x \to \infty} C_{n}x^{n}\)

\begin{align*} \lim_{x \to \infty} (C_{0} + C_{1}x^{1} + \dots + C_{n}x^{n}) \amp = \lim_{x \to \infty} \left( x^{n} (\dfrac{C_{0}}{x^{n}} + \dfrac{C_{1}}{x^{n-1}} + \dots + C_{n}) \right)\\ \amp = (\lim_{x \to \infty} x^{n}) \left( \lim_{x \to \infty} (\dfrac{C_{0}}{x^{n}} + \dfrac{C_{1}}{x^{n-1}} + \dots + C_{n}) \right)\\ \amp = (\lim_{x \to \infty} x^{n}) ( 0 + 0 + \dots + C_{n})\\ \amp = C_{n} \cdot \lim_{x \to \infty} x^{n}\\ \amp = \lim_{x \to \infty} C_{n} x^{n} \end{align*}
Note 1.3.8.

We can't take the limit of each of the terms as orginally written because it can't be calculated as the limit of a sum since all of the x's have limits of \(\infty\) and we can't use the individual limits of \(\infty\) in calculations.

Example 1.3.1.

Example:

\begin{equation*} \lim_{x \to - \infty} -6x^6 + 5x^4 + 2x^3 + 200x^2 + 9 = \lim_{x \to - \infty} -6x^6 = - \infty \end{equation*}

What is the point? As complicated as the formula and graph may be, we can build up info about end behavior. The goal is to skip the steps and focus on the leading term.

Example 1.3.2.

Now let's try a rational function:

\begin{align*} \lim_{x \to - \infty} \dfrac{-6x^6 + 5x^4 + 2x^3 + 200x^2 + 9}{x^7 + 3x^4 + x^2 + 17} \amp = \lim_{x \to - \infty} \dfrac{\frac{-6x^6}{x^7} + \frac{5x^4}{x^7} + \frac{2x^3}{x^7} + \frac{200x^2}{x^7} + \frac{9}{x^7}}{\frac{x^7}{x^7} + \frac{3x^4}{x^7} + \frac{x^2}{x^7} + \frac{17}{x^7}}\\ \amp = \dfrac{\lim_{x \to - \infty}\frac{-6}{x} + \frac{5}{x^3} + \frac{2}{x^4} + \frac{200}{x^5} + \frac{9}{x^7}}{\lim_{x \to - \infty}1 + \frac{3}{x^3} + \frac{1}{x^5} + \frac{17}{x^7}}\\ \amp = \dfrac{0 + 0 + 0 + 0 + 0}{1 + 0 + 0 + 0}\\ \amp = \dfrac{0}{1}\\ \amp = 0 \end{align*}

So what? This graph has a horizontal asymptote at zero for bigger and bigger values of x.

Now for the fast way:

\begin{align*} \lim_{x \to - \infty} \dfrac{-6x^6 + 5x^4 + 2x^3 + 200x^2 + 9}{x^7 + 3x^4 + x^2 + 17} \amp = \lim_{x \to - \infty} \dfrac{6x^6}{x^7} \\ \amp = \lim_{x \to - \infty} \dfrac{-6}{x}\\ \amp = 0 \end{align*}

Note 1.3.9.
The functions themselves are not equal but in the limits they are.

For limits at infinity, the limit of a rational function can be taken as the limits of the ratio of its leading terms. We can't do this at a finite point because those lower-order terms matter.

Example 1.3.3.

Let's find the limit of \(\displaystyle\lim_{x \to - \infty} \dfrac{\sqrt{x^2 + 6}}{4x+2}\text{:}\)

\begin{align*} \lim_{x \to - \infty} \dfrac{\sqrt{x^2 + 6}}{4x+2} \amp = \lim_{x \to - \infty} \dfrac{ \left( \frac{\sqrt{x^{2} + 6}}{\left| x \right|} \right) }{ \left( \frac{4x+2}{\left| x \right|} \right) } \\ \amp = \lim_{x \to - \infty} \dfrac{ \left( \frac{\sqrt{x^{2} + 6}}{\sqrt{x^{2}}} \right) }{ \left( \frac{4x+2}{-x} \right) } \\ \amp = \lim_{x \to - \infty} \dfrac{\sqrt{\frac{x^{2} + 6}{x^{2}}}}{-4 - \frac{2}{x}} \\ \amp = \lim_{x \to - \infty} \dfrac{\sqrt{1 + \frac{6}{x^{2}}}}{- 4 - \frac{2}{x}}\\ \amp = \dfrac{\sqrt{1 + \displaystyle\lim_{x \to - \infty} \frac{6}{x^{2}}}}{- 4 - \displaystyle\lim_{x \to - \infty} \frac{2}{x}}\\ \amp = \dfrac{\sqrt{1 + 0}}{-4 - 0}\\ \amp = - \dfrac{1}{4} \end{align*}