Horizontal asymptotes

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\CancelColor }[0]{} \newcommand {\cancelto }[0]{1} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{\vec {proj}} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$ $% This file was *autogenerated* from digInHorizontalAsymptotes.sagetex.sage with % sagetex.py version 2015/08/26 v3.0-92d9f7a %a2e364af8d80a597d1d39362b8686753% md5sum of corresponding .sage file (minus "goboom","current_tex_line",and pause/unpause lines)$

We explore functions that behave like horizontal lines as the input grows without bound.

Consider the function $f(x) = \frac {6x-9}{x-1}.$

As $x$ approaches infinity, it seems like $f(x)$ approaches a specific value. Such a limit is called a limit at infinity.

If $f(x)$ becomes arbitrarily close to a specific value $L$ by making $x$ sufficiently large, we write $\lim _{x\to \infty } f(x) = L$ and we say, the limit at infinity of $f(x)$ is $L$ .

If $f(x)$ becomes arbitrarily close to a specific value $L$ by making $x$ sufficiently large and negative, we write $\lim _{x\to -\infty } f(x) = L$ and we say, the limit at negative infinity of $f(x)$ is $L$ .

In order to calculate limits of rational functions at positive and negative infinity, we will often multiply the numerator and denominator by $\displaystyle \frac {1}{\phantom {xx}x^{\text {highest power in denominator}}\phantom {xx}}$ .

Why is this the case? Let’s consider the basic rational function

$f(x) = \frac {1}{x}$

A few cards ago, you discovered that as $x$ approaches positive and negative infinity, $f(x)$ approaches $0$ . Therefore, if we can get individual terms in our rational function to ‘‘look like’’ the known basic rational function $f(x)=\frac {1}{x}$ , then we can more easily evaluate the limit as $x$ approaches infinity. Therefore, the common technique applied to $\displaystyle \lim _{x\to \pm \infty }\frac {numerator}{denominator}$ is to multiply the numerator and denominator by $\displaystyle \frac {1}{\phantom {xx}x^{\text {highest power in denominator}}\phantom {xx}}$ . Let’s try an example.

Compute $\lim _{x\to \infty } \frac {6x-9}{x-1}.$

$\begin{align*} \lim_{x\to\infty}\frac{6x-9}{x-1} &= \lim_{x\to\infty}\frac{6x-9}{x-1}\cdot \frac{1/x}{1/x}\\ &=\lim_{x\to\infty}\frac{\frac{6x}{x} - \frac{9}{x}}{\frac{x}{x} - \frac{1}{x}}\\ &=\lim_{x\to\infty}\frac{6 - 9 \cdot \frac{1}{x}}{1 - \frac{1}{x}}\\ &=\frac{\displaystyle\lim_{x\to\infty} \left( 6 - 9 \cdot \frac{1}{x}\right)}{\displaystyle\lim_{x\to\infty} \left( 1 - \frac{1}{x} \right)} \\ &=\frac{\displaystyle\lim_{x\to\infty} 6 - 9 \displaystyle\lim_{x\to\infty} \cdot \frac{1}{x}}{\displaystyle\lim_{x\to\infty} 1 - \displaystyle\lim_{x\to\infty} \frac{1}{x}} \\ &=\frac{6 - 9 \cdot 0}{1 - 0} \\ &= 6 \end{align*}$

Compute $\lim _{x\to \infty } \frac {6x-9}{x^2-1}.$

$\begin{align*} \lim_{x\to\infty}\frac{6x-9}{x^2-1} &= \lim_{x\to\infty}\frac{6x-9}{x^2-1}\cdot \frac{1/x^2}{1/x^2}\\ &=\lim_{x\to\infty}\frac{\frac{6x}{x^2} - \frac{9}{x^2}}{\frac{x}{x^2} - \frac{1}{x^2}}\\ &= \lim_{x\to\infty} \frac{\frac{6}{x}-\frac{9}{x^2}}{1-\frac{1}{x^2}}\\ &=\frac{6 \displaystyle\lim_{x\to\infty} \frac{1}{x} - 9 \displaystyle\lim_{x\to\infty} \frac{1}{x^2}}{\displaystyle\lim_{x\to\infty} 1 - \displaystyle\lim_{x\to\infty} \frac{1}{x^2}} \\ &= \frac{0}{1} \\ &= 0 \end{align*}$

Compute $\lim _{x\to \infty } \frac {3x^4-5x^2+1}{x^2-x+2x^4}.$

Write $\begin{align*} \lim_{x\to\infty}\frac{3x^4-5x^2+1}{x^2-x+2x^4} &= \lim_{x\to\infty}\frac{3x^4-5x^2+1}{x^2-x+2x^4}\cdot \frac{1/x^4}{1/x^4}\\ &=\lim_{x\to\infty}\frac{\frac{3x^4}{x^4} - \frac{5x^2}{x^4}+\frac{1}{x^4}}{\frac{x^2}{x^4}-\frac{x}{x^4} + \frac{2x^4}{x^4}}\\ &= \lim_{x\to\infty} \frac{3-\frac{5}{x^2}+\frac{1}{x^4}}{\frac{1}{x^2}-\frac{1}{x^3}+2}\\ &= \frac{\displaystyle\lim_{x\to\infty} 3 - 5 \displaystyle\lim_{x\to\infty} \frac{1}{x^2} + \displaystyle\lim_{x\to\infty} \frac{1}{x^4}}{\displaystyle\lim_{x\to\infty} \frac{1}{x^2} - \displaystyle\lim_{x\to\infty} \frac{1}{x^3} + \displaystyle\lim_{x\to\infty} 2} \\ &= \frac{3}{2}. \end{align*}$

Sometimes one must be careful, consider this example.

Compute $\lim _{x\to -\infty } \frac {x^3+1}{\sqrt {x^6+5}}$

In this case we multiply the numerator and denominator by $-1/x^3$ , which is a positive number as since $x\to -\infty$ , $x^3$ is a negative number. $\begin{align*} \lim_{x\to -\infty} \frac{x^3+1}{\sqrt{x^6+5}} &= \lim_{x\to -\infty} \frac{x^3+1}{\sqrt{x^6+5}} \cdot \frac{-1/x^3}{-1/x^3}\\ &= \lim_{x\to -\infty} \frac{-1-1/x^3}{\sqrt{x^6/x^6+5/x^6}}\\ &= \frac{-\displaystyle\lim_{x\to -\infty} 1 - \displaystyle\lim_{x\to -\infty} \frac{1}{x^3}}{\sqrt{\displaystyle\lim_{x\to -\infty} 1 + 5\displaystyle\lim_{x\to -\infty} \frac{1}{x^6}}} \\ &= \frac{-1}{\sqrt{1}} \\ &= -1. \end{align*}$

Limits at infinity are used to define what is meant by a horizontal asymptote.

If $\lim _{x\to \infty } f(x) = L \qquad \text {or}\qquad \lim _{x\to -\infty } f(x) = L,$ then the line $y=L$ is a horizontal asymptote of $f(x)$ .

Give the horizontal asymptotes of $f(x) = \frac {6x-9}{x-1}$

From our previous work, we see that $\displaystyle \lim _{x\to \infty } f(x) = 6$ , and upon further inspection, we see that $\displaystyle \lim _{x\to -\infty } f(x) = 6$ . Hence the horizontal asymptote of $f(x)$ is the line $y=6$ .

Note, since $\lim _{x\to \infty } f(x) = \lim _{x\to 0^+} f\left (\frac {1}{x}\right )$ and $\lim _{x\to -\infty } f(x) = \lim _{x\to 0^-} f\left (\frac {1}{x}\right )$

we can also apply the Squeeze Theorem when taking limits at infinity. This is also the reason why the limit laws apply to limits at infinity, by the way.

It is a common misconception that a function cannot cross an asymptote. As the next example shows, a function can, indeed, cross a horizontal asymptote, and in the example this occurs an infinite number of times!

Compute: $\lim _{x\to \infty } \frac {\sin (7x)+4x}{x}$

We can bound our function $\frac {-1+4x}{x} \le \frac {\sin (7x)+4x}{x} \le \frac {1+4x}{x}.$ Now write with me $\begin{align*} \displaystyle\lim_{x\to\infty}\frac{-1+4x}{x} \cdot \frac{1/x}{1/x} &= \lim_{x\to\infty}\frac{-1/x+4}{1}\\ &=4. \end{align*}$ And we also have $\begin{align*} \displaystyle\lim_{x\to\infty}\frac{1+4x}{x} \cdot \frac{1/x}{1/x} &= \lim_{x\to\infty}\frac{1/x+4}{1}\\ &=4. \end{align*}$ Since $\lim _{x\to \infty } \frac {-1+4x}{x} = 4 = \lim _{x\to \infty }\frac {1+4x}{x}$ we conclude by the Squeeze Theorem, $\displaystyle \lim _{x\to \infty }\frac {\sin (7x)}{x}+4 = \answer {4}$ .

Give a horizontal asymptote of $f(x) = \frac {\sin (7x)+4x}{x}.$

Again from previous work, we see that $\displaystyle \lim _{x\to \infty } f(x) = \answer [given]{4}$ . Hence $y=\answer [given]{4}$ is a horizontal asymptote of $f(x)$ .