Slant asymptotes

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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

\[ \ell (x) = L. \]

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

If there is a nonhorizontal line $\ell (x) = m\cdot x+b$ such that

\[ \lim _{x\to \infty }\left (f(x) - \ell (x)\right ) = 0 \qquad \text {or}\qquad \lim _{x\to -\infty } \left ( f(x) - \ell (x)\right ) = 0, \]

then $\ell $ is a slant asymptote for $f$.

Consider the graph of the following function.

What is the slant asymptote of this function?

\[ \ell (x) = \answer {x/2 -1} \]

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 $\ell $ such that

\[ \lim _{x\to \infty }\left (f(x) - \ell (x)\right ) = 0\qquad \text {or}\qquad \lim _{x\to -\infty } \left ( f(x) - \ell (x)\right ) = 0. \]

First, let’s consider the limit as $x$ approaches positive infinity. We will imagine that we have such a line

\[ \ell (x) = m\cdot x + b \]

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 $\ell (x)$ 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, $\ell (x) = \answer [given]{3x-5}$ is a slant asymptote for our function $f$. You should check that we get the same slant asymptote $\ell (x) = \answer [given]{3x-5}$ 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 $y=\ell (x)$:

\[ \graph [xmin=-40,xmax=40,ymin=-50,ymax=50]{\frac {3x^2+x+2}{x+2}, 3x-5} \]

2025-01-06 19:32:06