Vertical Asymptotes and Discontinuity Alternative Format
Lesson Part 1
In This Module
Asymptotes play a key role in understanding the graph of a rational function.
- We will identify the vertical asymptotes and/or point(s) of discontinuity of a rational function.
- We will study the behaviour of the rational function to the left and right side of the discontinuity.
- We will introduce the language of limits to assist us in communicating our understanding of the behaviour of the function close to its vertical asymptotes.
In the subsequent module, we will discuss horizontal asymptotes and other end behaviour asymptotes. Once we have a good understanding of the behaviour of a rational function about its asymptotes, we can then move on to analysing and graphing various rational functions,
Vertical Asymptotes
A rational function \( y = \dfrac{g(x)}{h(x)},\ h(x) \neq 0 \) will have a vertical asymptote at \( x = a \) if \( h(a) = 0 \) and \( g(a) \neq 0 \), when the function is in simplest form.
What does this mean? I will try to clarify with the following two examples.
Example 1
Consider the function \( f(x) = \dfrac{2x}{x - 5} \). We begin by identifying any inadmissible values of \(x\).
The domain of the function is \( \{x \mid x \neq 5,\ x \in \mathbb{R}\} \).
Observe that \( f(5) = \dfrac{2(5)}{5 - 5} = \dfrac{10}{0} \) which is an undefined value. The function is undefined at \(x=5\) and the graph of the function is discontinuous at \( x = 5 \).
What happens to the value of the function (the value of \( y \)) as the value of \( x \rightarrow 5 \)?
This table shows, as \(x\) approaches \(5\) from the left, that is from numbers less than \(5\), \(y\) approaches a large negative value \((y \rightarrow -\infty)\).
| \(x\) |
\(f(x)\) |
| \(4\) |
\(-8\) |
| \(4.5\) |
\(-18\) |
| \(4.9\) |
\(-98\) |
| \(4.99\) |
\(-998\) |
| \(4.999\) |
\(-9998\) |
This table shows, as \(x\) approaches \(5\) from the right, that is from numbers greater than \(5\), \(y\) approaches a large positive value \((y \rightarrow \infty)\).
| \(x\) |
\(f(x)\) |
| \(6\) |
\(12\) |
| \(5.5\) |
\(22\) |
| \(5.1\) |
\(102\) |
| \(5.01\) |
\(1002\) |
| \(5.001\) |
\(10\,002\) |
The function \( f(x) = \dfrac{2x}{x - 5} \) has a vertical asymptote of \( x = 5 \). The graph of the function will approach, but never touch this line.
Lesson Part 2
The Language of Limits
We can communicate this information using the language of limits. You are already familiar with the concept of approaching and the notation used. For example, we use the notation \(x\rightarrow 5\) to indicate “\(x\) approaches \(5\)”.
In our last example, we introduced the concept of “\(x\) approaching \(5\) from the left”. The notation used will have a raised negative sign following the \(5\). That is, \(x\rightarrow 5^-\). Think of this as “\(5\) minus a small amount”. (For example, \(4.9\) is close and to the left of \(5\).)
The concept “\(x\) approaches \(5\) from the right” will have a raised positive sign following the \(5\). That is, \(x\rightarrow 5^+\). Again, think “\(5\) plus a small amount”. (For example, \(5.1\) is to the right of \(5\).)
The following table gives a summary of the use of the three notations:
| Concept |
Notation |
| \(x\) approaches \(5\) |
\(x\rightarrow 5\) |
| \(x\) approaches \(5\) from the left |
\(x\rightarrow 5^-\)
(think \(5\) substract a small amount) |
| \(x\) approaches \(5\) from the right |
\(x\rightarrow 5^+\)
(think \(5\) plus a small amount) |
In our example, when \(x \rightarrow 5^-, y \rightarrow -\infty\). We can denote this by a limit, \(\displaystyle\lim_{x \rightarrow 5^-}f(x) = -\infty \).
This means that \( y \) will continue to decrease in value as \( x \) gets closer and closer to \(5\) from the left of \(5\).
| \(x\) |
\(f(x)\) |
| \(4\) |
\(-8\) |
| \(4.5\) |
\(-18\) |
| \(4.9\) |
\(-98\) |
| \(4.99\) |
\(-998\) |
| \(4.999\) |
\(-9998\) |
Similarly, when \(x \rightarrow 5^+, y \rightarrow \infty\). We can denote this by \(\displaystyle \lim_{x \rightarrow 5^+}f(x) = \infty \).
This means that \( y \) will continue to increase in value as \( x \) gets closer and closer to \(5\) from the right of \(5\), as shown in the table.
| \(x\) |
\(f(x)\) |
| \(6\) |
\(12\) |
| \(5.5\) |
\(22\) |
| \(5.1\) |
\(102\) |
| \(5.01\) |
\(1002\) |
| \(5.001\) |
\(10\,002\) |
The concept of a limit, which is fundamental to calculus, is used when mathematicians are concerned with the behaviour of a function near a particular value of \( x \) (or as \( x \rightarrow \pm \infty \)).
The following is the definition of a limit.
If the limit of a function, \( y = f(x) \), as \( x \) approaches \( a \) is equal to \( L \), then we write
\[ \displaystyle \lim_{x \rightarrow a}f(x) = L \]
This means that \( f(x) \) gets closer and closer to the value \( L \), as \( x \) gets closer and closer to the value \( a \). That is, \( y \rightarrow L \) as \( x \rightarrow a \).
A limit provides information about how a function behaves near, not at, a specific value of \( x \). This is a simplified definition of a limit, but sufficient for our purposes. We will use limits to help us communicate our understanding of the behaviour of the function nearest asymptotes and at its ends. Limits are studied in more depth in calculus.
It should also be mentioned that the \( \displaystyle \lim_{x \rightarrow a}f(x) \) exists only when the left and right side limits exist and are equal.
In the first example, the left and right side limits, as \( x \rightarrow 5 \), do not exist.
For these limits to exist, \( f(x) \) (that is, \( y \)) must approach a specific finite value when \( x \) approaches \(5\) from the left or right.
Since, in this case,
\[\lim_{x \rightarrow 5^-}\dfrac{2x}{x - 5} = -\infty\]
and
\[\lim_{x \rightarrow 5^+} \dfrac{2x}{x - 5} = \infty \]
\( y \) continues to grow larger in a negative or positive direction.
It has unbounded behaviour. This type of behaviour happens at a vertical asymptote.
The \( \displaystyle \lim_{x \rightarrow 5} \dfrac{2x}{x - 5} \) does not exist; however, the left and right side limits have allowed us to communicate the behaviour of the function at the vertical asymptote.
Check Your Understanding A
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Lesson Part 3
Examples
Example 2
Determine the vertical asymptotes, if any, for the function \( f(x) = \dfrac{-2x + 4}{x^2 - x - 2} \) and discuss the behaviour of the function near these asymptotes.
Solution
We start by factoring the denominator of the function to identify any restrictions on the value of \( x \). By factoring the denominator, we can determine the restrictions on \(x\).
\[ f(x) = \dfrac{-2x + 4}{(x + 1)(x - 2)} \]
The domain of the function is \( \{x \mid x \neq -1, 2,\ x \in \mathbb{R} \} \).
We need to determine what the graph of the function is doing at \(x=-1,\ 2\).
\( f(-1) = \dfrac{6}{0} \), an undefined value. This indicates that there is a vertical asymptote at \( x = -1 \): the denominator is \(0\) at \(x=-1\), but the numerator is not.
What happens to the value of \( f(x) \) as \( x \rightarrow -1^{-} \) ?
We can see from this table that \( y \rightarrow \infty \) as \( x \rightarrow -1^{-} \).
Therefore,
\[\lim_{x \rightarrow -1^{-}}f(x) = \infty \]
| \(x\) |
\(f(x)\) |
| \(-2\) |
\(2\) |
| \(-1.5\) |
\(4\) |
| \(-1.1\) |
\(20\) |
| \(-1.01\) |
\(200\) |
| \(-1.001\) |
\(2000\) |
What happens to the value of \( f(x) \) as \( x \rightarrow -1^{+} \)?
Similarly, the this table shows that \( y \rightarrow -\infty \) as \( x \rightarrow -1^{+} \), so
\[ \lim_{x \rightarrow -1^+}f(x) = -\infty \]
| \(x\) |
\(f(x)\) |
| \(0\) |
\(-2\) |
| \(-0.5\) |
\(-4\) |
| \(-0.9\) |
\(-20\) |
| \(-0.99\) |
\(-200\) |
| \(-0.999\) |
\(-2000\) |
Thus,
\[ \lim_{x \rightarrow -1^-}\dfrac{-2x + 4}{x^2 - x - 2} = \infty\]
and
\[\lim_{x \rightarrow -1^+}\dfrac{-2x + 4}{x^2 - x - 2} = -\infty \]
So the limit \( \displaystyle \lim_{x \rightarrow -1}\dfrac{-2x + 4}{x^2 - x - 2} \) does not exist.
This unbounded behaviour of the function, to the left and right of \(-1\), supports the fact that a vertical asymptote occurs at \( x = -1 \).
What happens at \(x=2\)?
Note that \( f(2) = \dfrac{-2(2) + 4}{2^2 - 2 - 2} = \dfrac{0}{0} \). This value \(\dfrac{0}{0}\) is said to be indeterminate, as opposed to undefined. Indeterminate form will be treated in more detail in calculus.
Since the polynomials in both the numerator and denominator are \(0\) at \( x = 2 \), \( x - 2 \) must be a factor of both the numerator and denominator (by factor theorem). Therefore, the equation of the function can be simplified.
Before simplifying, let's first study the behaviour of the function \( f(x) = \dfrac{-2x + 4}{x^2 - x - 2} \) as \( x \rightarrow 2 \).
We can see from this table that \( y \rightarrow -0.\overline{6} \) or \( -\dfrac{2}{3} \) as \( x \rightarrow 2^{-}\), that is,
\[ \displaystyle \lim_{x \rightarrow 2^{-}}f(x) = -\frac{2}{3} \]
| \(x\) |
\(f(x)\) |
| \(1\) |
\(-1\) |
| \(1.5\) |
\(-0.8\) |
| \(1.9\) |
\(-0.688965\) |
| \(1.99\) |
\(-0.6688963\) |
| \(1.999\) |
\(-0.666889\) |
Similarly, \( y \rightarrow -0.\overline{6} \) or \( -\dfrac{2}{3} \) as \( x \rightarrow 2^{+}\), that is,
\[ \lim_{x \rightarrow 2^{+}}f(x) = -\frac{2}{3} \]
| \(x\) |
\(f(x)\) |
| \(3\) |
\(-0.5\) |
| \(2.5\) |
\(-0.5714286\) |
| \(2.1\) |
\(-0.6451613\) |
| \(2.01\) |
\(-0.6644518\) |
| \(2.001\) |
\(-0.6664445\) |
Thus, the left and right hand limits exist and are equal. So,
\[ \lim_{x \rightarrow 2^{-}}~f(x) = \lim_{x \rightarrow 2^{+}}f(x) = -\dfrac{2}{3} \]
So the limit of \( f(x) \) as \( x \rightarrow 2 \) exists and is \( -\dfrac{2}{3} \). Therefore,
\[ \lim_{x \rightarrow 2} \dfrac{-2x + 4}{(x + 1)(x - 2)} = -\dfrac{2}{3} \]
This indicates that there is a point of discontinuity (a hole) at \( x = 2 \), and not a vertical asymptote. The curve will approach \( \left(2, -\frac{2}{3} \right) \) as the value of \( x \) approaches \(2\). However, the function is not defined at \( x = 2 \).
An open point on the graph is used to indicate the discontinuity at \( x = 2 \).
Algebraically, the equation of the function simplifies. We can remove the common factor \(x-2\) from the numerator and denominator:
\begin{align*}f(x) &= \dfrac{-2x + 4}{x^2 - x - 2} \\&= \dfrac{-2(x - 2)}{(x + 1)(x - 2)}\\f(x) & = \dfrac{-2}{x + 1}, \ x \neq -1, 2\end{align*}
The graph of \( f(x)=\dfrac{-2x + 4}{x^2 - x - 2} \) is the same as the graph \( y = \dfrac{-2}{x + 1} \) everywhere except at the point \( \left(2, -\frac{2}{3} \right) \), where \(f(x)\) has a hole, or point of discontinuity, since the function maintains the restriction \(x \neq 2\) after simplification. Note that both \(f(x)\) and \(y\) have the vertical asymptote at \(x=-1\).
Note also that using the simplified equation and ignoring the restriction \( x \neq 2 \), \( f(2) = \dfrac{-2}{2 + 1}=-\dfrac{2}{3} \), the \( y \)-coordinate of the point of discontinuity. This rational function has a horizontal asymptote of \(y=0\). We will discuss horizontal asymptotes and the behavior of a function near its horizontal asymptote in the next module on end behavior asymptotes.
Check Your Understanding B and C
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Lesson Part 4
Examples
Example 3 — Part A
Determine, with support, an equation for each rational function of the form \( y = \dfrac{g(x)}{h(x)} \) that satisfy the given conditions:
- A hole exists at \( (4, -2) \) and
- a vertical asymptote occurs at \( x = 0 \).
Solution
For a hole to exist at \( x = 4 \), we need \( g(4) = 0 \) and \( h(4) = 0 \), so \( x - 4 \) is a factor of both the numerator and the denominator.
For a vertical asymptote to exist at \( x = 0 \), then \( h(0) = 0 \) and \( g(0) \neq 0 \), so \( x \) is a factor of the denominator, but not the numerator.
A function which satisfies these conditions is of the form
\[ f(x) = \dfrac{k(x - 4)}{x(x - 4)},\ \text{where } k\neq 0,\ k \in \mathbb{R}, \text{ and } x \neq 0,\ 4\]
We must determine the value of \( k \) such that the hole is located at \( (4, -2) \). The equation we have for \(f(x)\) can be simplified:
\[ f(x) = \dfrac{k(x - 4)}{x(x - 4)} = \dfrac{k}{x},\ x \neq 0, 4,\ k \in \mathbb{R} \]
The graph of the function, \(g(x) = \dfrac{k}{x},\ x \neq 0 \) is identical to the graph of \( f(x) \), with the exception of the hole at \( (4, -2) \).
Using \( g(4) = -2 \), we can determine the value of \( k \) that will place the hole at the correct location.
Here, \( \dfrac{k}{4} = -2 \) and \( k = -8 \). Thus,
\[ f(x) = \dfrac{-8(x - 4)}{x(x - 4)} ,\ x\neq 0, 4 \]
is a function that has a hole at \((4,-2)\) and a vertical asymptote at \(x=0\).
Therefore, a function with a hole at \((4,-2)\) and vertical asymptote of \(x=0\) is \(f(x)=\dfrac{-8x+32}{x^2-4x}\).
Example 3 — Part B
Determine, with support, an equation for each rational function of the form \( y = \dfrac{g(x)}{h(x)} \) that satisfy the given conditions:
- \( g(-3) = 0 \) and \( h(-3) = 0 \), but
- a vertical asymptote occurs at \( x = -3 \).
Solution
Given \( g(-3) = 0 \) and \( h(-3) = 0 \), then \( x + 3 \) must be a factor of the numerator and denominator.
However, a vertical asymptote exists at \( x = -3 \).
This means that \( h(-3) = 0 \) and \( g(-3) \neq 0 \) when the function is in simplest form.
For this to happen, \( x + 3 \) must be a factor of multiplicity \(2\) or greater in the denominator (at least two factors of \( x + 3 \)), and of lesser multiplicity (but at least \(1\)) in the numerator.
One such function is given by \( f(x) = \dfrac{x + 3}{(x + 3)^2} = \dfrac{x + 3}{x^2 + 6x + 9},\ x \neq -3 \).
When this equation is simplified to \( y = \dfrac{1}{x + 3},\ x \neq -3 \), the indeterminate form \( \left( \dfrac{0}{0} \right) \) of the equation at \( x = -3 \) is lost, but the graph of the function remains the same.
This simplified equation is not a valid solution to this problem as it does not satisfy the first condition. Therefore, the solution is \(f(x)\) in unsimplified form.
Therefore, the function \( f(x) = \dfrac{x + 3}{x^2 + 6x + 9},\ x \neq 3 \) is indeterminate in form at \( x = -3 \), but has a vertical asymptote at \( x = -3 \).
In both situations, other solutions can be generated by using \( y = kf(x),\ k \in \mathbb{R},\ k \neq 0 \).
As well, there are other, more complicated, rational functions that would satisfy the given conditions.
Summary
For a rational function \( y = \dfrac{g(x)}{h(x)}, h(x) \neq 0 \):
- The function will be discontinuous at \( x = a \) if \( h(a) = 0 \).
- The function has a vertical asymptote at \( x = a \) if \( h(a) = 0 \) and \( g(a) \neq 0 \), when the function is in simplest form.
- If \( h(a) = 0 \) and \( g(a) = 0 \) for some value of \( a \in \mathbb{R} \), then \( x - a \) is a factor of the numerator and denominator of the function, and a point of discontinuity (a hole) may occur at \( x = a \). To verify this, express the function in simplified form and then determine if it generates a single point of discontinuity, or a vertical asymptote.
- Since the value of a limit provides information near, but not at, a specific value of \( x \), limits are often used to analyze a function near its asymptotes.
We will continue our study of asymptotes in the next module with discussion of end-behavior asymptotes, horizontal, and oblique.