![[Picture]](digInSlantAsymptotes.online-d6e2884fb1cd59898572627b1630ed8f.svg)
\[ \l (x) = \answer {x/2 -1} \]
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:
Vertical 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
\[ x = a. \]
Horizontal asymptotes 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?
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 \(y=\l (x)\): \[ \graph [xmin=-40,xmax=40,ymin=-50,ymax=50]{\frac {3x^2+x+2}{x+2}, 3x-5} \]