Slant asymptotes

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \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 {\sagemath }[0]{\textsf {SageMath}} \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}$ $% This file was *autogenerated* from digInSlantAsymptotes.sagetex.sage with % sagetex.py version 2015/08/26 v3.0-92d9f7a %3dc91be5378ce0a90c54003aefd933c1% md5sum of corresponding .sage file (minus "goboom","current_tex_line",and pause/unpause lines)$

We explore functions that ‘‘shoot to infinity’’ at certain points in their domain.

If we think of an asymptote as a ‘‘line that a function resembles when the input or output is large,’’ then there are three types of asymptotes, just as there are three types of lines: $\begin{align*} \text{Vertical Asymptotes} \qquad&\leftrightarrow\qquad \text{Vertical Lines}\\ \text{Horizontal Aymptotes}\qquad&\leftrightarrow\qquad \text{Horizontal Lines} \\ \text{Slant Asymptotes}\qquad&\leftrightarrow\qquad \text{Slant Lines} \end{align*}$

Here we’ve made up a new term ‘‘slant’’ line, meaning a line whose slope is neither zero, nor is it undefined. Let’s do a quick review of the different types of asymptotes:

Recall, a function $f$ has a vertical asymptote at $x=a$ if at least one of the following hold:

$\lim _{x\to a} f(x) = \pm \infty$ ,
$\lim _{x\to a^+} f(x) = \pm \infty$ ,
$\lim _{x\to a^-} f(x) = \pm \infty$ .

In this case, the asymptote is the vertical line $x = a.$

We have also seen that a function $f$ has a horizontal asymptote if $\lim _{x\to \infty } f(x) = L \qquad \text {or}\qquad \lim _{x\to -\infty } f(x) = L,$ and in this case, the asymptote is the horizontal line $\l (x) = L.$

Slant asymptotes On the other hand, a slant asymptote is a somewhat different beast.

If there is a nonhorizontal line $\l (x) = m\cdot x+b$ such that $\lim _{x\to \infty }\left (f(x) - \l (x)\right ) = 0 \qquad \text {or}\qquad \lim _{x\to -\infty } \left ( f(x) - \l (x)\right ) = 0,$ then $\l$ is a slant asymptote for $f$ .

Consider the graph of the following function.

What is the slant asymptote of this function? $\l (x) = \answer {x/2 -1}$

To analytically find slant asymptotes, one must find the required information to determine a line:

The slope.
The $y$ -intercept.

While there are several ways to do this, we will give a method that is fairly general.

Find the slant asymptote of $f(x) = \frac {3x^2+x+2}{x+2}.$

We are looking to see if there is a line $\l$ such that $\lim _{x\to \infty }\left (f(x) - \l (x)\right ) = 0\qquad \text {or}\qquad \lim _{x\to -\infty } \left ( f(x) - \l (x)\right ) = 0.$ First, let’s consider the limit as $x$ approaches positive infinity. We will imagine that we have such a line $\l (x) = m\cdot x + b$ and attempt to find the correct values for $m$ and $b$ . Let’s look again at our limit. We are assuming: $\lim _{x \to \infty } \left (\frac {3x^2 + x + 2}{x+2} - (mx + b) \right ) = 0.$ We know that $f(x)$ is continuous everywhere except at $x = -2$ and $\l (x)$ is continuous everywhere, so we can apply our limit laws away from $x = -2$ . We’re looking at large values of $x$ , so this is no problem. We use the fact that the sum of the limits is the limit of the sums. $\begin{align*} \lim_{x \to \infty} \left ( \frac{3x^2 + x + 2}{x+2} \right ) - &\lim_{x\to\infty} (mx+b) = 0 \\ \lim_{x \to \infty} \frac{3x^2 + x + 2}{x+2} = &\lim_{x \to \infty} (mx+b) \end{align*}$

We are assuming these two limits are equal. Dividing by $x$ on the right hand side makes the limit equal to $m$ :

$\begin{align*} m = \lim_{x \to \infty} \left(\frac{mx}{x} + \frac{b}{x}\right) = \lim_{x \to \infty} \left(\frac{mx+b}{x}\right). \end{align*}$

To find the value of $m$ , then, we can divide the left hand side by $x$ and evaluate the limit. We see the following.

$\begin{align*} m &=\lim_{x\to\infty}\frac{\frac{3x^2+x+2}{x+2}}{x}\\ &= \lim_{x\to\infty}\frac{\frac{3x^2+x+2}{x+2}}{x}\\ &= \lim_{x\to\infty}\frac{3x^2+x+2}{x^2+2x}\\ &= \lim_{x\to\infty}\left(\frac{3x^2+x+2}{x^2+2x}\cdot\frac{1/x^2}{1/x^2}\right)\\ &= \lim_{x\to\infty}\frac{3+1/x+2/x^2}{1+2/x}\\ &= \answer[given]{3}. \end{align*}$

So $m=3$ . We now know that $\lim _{x \to \infty }\frac {3x^2 +x+2}{x+2} = \lim _{x \to \infty } (3x + b)$ for some value of $b$ . To find the $y$ -intercept $b$ , we use a similar method. Notice that $\lim _{x \to \infty }( 3x + b - 3x )= \lim _{x \to \infty } b = b,$ so if we subtract $3x$ from the right hand side, we are left with just $b$ . Since the two sides are equal, subtracting $3x$ from the left hand side and evaluating the limit will give us the value for $b$ . We write the following.

$\begin{align*} b &=\lim_{x\to\infty} \left(\frac{3x^2+x+2}{x+2} - 3x\right)\\ &=\lim_{x\to\infty} \left( \frac{3x^2+x+2}{x+2} - \frac{3x^2 + 6x}{x+2}\right) \\ &=\lim_{x\to\infty} \frac{3x^2+x+2-3x^2-6x}{x+2}\\ &=\lim_{x\to\infty} \frac{-5x+2}{x+2}\\ &=\lim_{x\to\infty} \left(\frac{-5x+2}{x+2}\cdot\frac{1/x}{1/x}\right)\\ &=\lim_{x\to\infty} \frac{-5+2/x}{1+2/x}\\ &= \answer[given]{-5}. \end{align*}$

By this method, we have determined that $\lim _{x\to \infty }\left (\frac {3x^2+x+2}{x+2} - (3x-5) \right ) = 0.$ In other words, $\l (x) = \answer [given]{3x-5}$ is a slant asymptote for our function $f$ . You should check that we get the same slant asymptote $\l (x) = \answer [given]{3x-5}$ when we take the limit to negative infinity as well. We can confirm our results by looking at the graph of $y=f(x)$ and $y=\l (x)$ : $\graph [xmin=-40,xmax=40,ymin=-50,ymax=50]{\frac {3x^2+x+2}{x+2}, 3x-5}$