Distinguish a hole from an asymptote
The same bottom, two different curves
Two functions have the same bottom, x − 3. The bottom is 0 at x = 3, so neither function has a value there.
The first function is \(f(x) = \dfrac{x^2 + 2x - 15}{x - 3}\). Drag the slider to move the orange point toward 3. The height is 7.99 at \(x = 2.99\) and 8.01 at \(x = 3.01\). The curve has a ring at \((3, 8)\), which marks one missing point. A break like this is a hole.
Press the second toggle to show \(g(x) = \dfrac{x + 5}{x - 3}\). The height is \(-799\) at \(x = 2.99\) and 801 at \(x = 3.01\). The curve leaves along the dashed line, which is a vertical asymptote.
The top of f factors as \((x - 3)(x + 5)\), so \(x - 3\) cancels and \(f(x) = x + 5\) for \(x \neq 3\). The top of g has no factor \(x - 3\), so the zero of its bottom stays.
Count the factors
The bottom has the factor \(x - c\) \(n\) times, with \(n\) at least 1, so the bottom is 0 at \(c\). The top has the same factor \(m\) times.
Set \(m\) equal to \(n\). Every factor \(x - c\) in the bottom cancels, so the curve has a ring at \(c\) at a finite height. This is a hole.
Raise \(m\) above \(n\). A factor \(x - c\) remains in the top, so the hole is now at height 0.
Set \(m\) below \(n\). Some factors stay in the bottom after the cancelling, and the curve leaves along the dashed line. This is a vertical asymptote.
Notation
Factor the top and the bottom completely, so that \(A(c)\) and \(B(c)\) are not 0.
\[f(x) = \frac{N(x)}{D(x)} = \frac{(x - c)^m A(x)}{(x - c)^n B(x)}\]Cancel the common factors. This is allowed for \(x \neq c\), and a limit uses only those inputs.
If the simplified bottom is not 0 at \(c\), the curve has a hole. The limit is the value of the simplified fraction at \(c\).
\[\lim_{x \to c} f(x) = \cy{L}\]If the simplified bottom is still 0 at \(c\) and the simplified top is not, the line \(x = c\) is a vertical asymptote.
\[\lim_{x \to c^{\pm}} f(x) = \pm\infty \ \text{ or } \ \mp\infty\]In both cases the original function has no value at c, because its bottom is 0 there.
When it fails
The form \(\tfrac{0}{0}\) does not decide the question. For \(\dfrac{x - c}{(x - c)^3}\), the top and the bottom are both 0 at \(c\), but cancelling leaves \(\dfrac{1}{(x - c)^2}\).
The bottom is still 0 at c, so the curve climbs on both sides of the dashed line. Count the factors that remain after the cancelling.
Example 1: both in one function
Find the hole and the vertical asymptote of \(f(x) = \dfrac{x^2 - 2x - 8}{x^2 - 16}\).
Factor the top and the bottom.
\[f(x) = \frac{(x - 4)(x + 2)}{(x - 4)(x + 4)}\]The bottom is 0 at \(x = 4\) and at \(x = -4\). The factor \(x - 4\) is in the top and in the bottom, so it cancels.
\[f(x) = \frac{x + 2}{x + 4} \quad (x \neq 4)\]At \(x = 4\) the simplified fraction is \(\tfrac{6}{8} = \tfrac{3}{4}\). This value is finite, so the curve has a hole at \((4, \tfrac{3}{4})\).
\[\lim_{x \to 4} f(x) = \cy{\tfrac{3}{4}}\]At \(x = -4\) the simplified bottom is 0 and the top is \(-2\), so the line \(x = -4\) is a vertical asymptote.
Example 2: one factor cancels
Does \(q(x) = \dfrac{x^2 - 4x + 3}{x^2 - 2x + 1}\) have a hole or a vertical asymptote at \(x = 1\)?
Factor the top and the bottom.
\[q(x) = \frac{(x - 1)(x - 3)}{(x - 1)^2}\]A tempting wrong move: The factor \(x - 1\) is in the top and in the bottom, so it cancels completely. The break at 1 is a hole.
\[q(x) = x - 3\]The top has one copy of \(x - 1\), and the bottom has two. At \(x = 1.01\) the height is \(-199\), and at \(x = 0.99\) it is 201. The curve leaves the figure on both sides of \(x = 1\), so it does not approach the red ring.
Cancel one copy of \(x - 1\). The other copy stays in the bottom.
\[q(x) = \frac{x - 3}{x - 1} \quad (x \neq 1)\]The simplified bottom is still 0 at \(x = 1\) and the top is \(-2\), so \(x = 1\) is a vertical asymptote.
Practice
For \(s(x) = \dfrac{x^2 - 4x + 4}{x^2 - 2x}\), find each zero of the bottom. Decide whether it gives a hole or a vertical asymptote.
For each zero c of the bottom, count the factors x − c in the top and in the bottom. Then move the point toward c to see whether the height stays finite.
Solution
Factor the top and the bottom. The bottom is 0 at x = 0 and at x = 2.
\[s(x) = \frac{(x - 2)^2}{x\,(x - 2)}\]The top has two factors \(x - 2\), and the bottom has one. Cancel one copy.
\[s(x) = \frac{x - 2}{x} \quad (x \neq 2)\]At \(x = 2\) the simplified fraction is 0, so the curve has a hole at \((2, 0)\). At \(x = 0\) the simplified bottom is 0 and the top is \(-2\), so \(x = 0\) is a vertical asymptote.
\[\lim_{x \to 2} s(x) = \cy{0}\]
Questions
The graph shows a fraction. Its bottom is 0 at \(x = 2\) and at \(x = -3\).
What does the graph show at \(x = 2\)?
- A hole at height 0.6
- A vertical asymptote, because the bottom is 0
- A jump
Answer
A hole at height 0.6
Which description fits the curve near \(x = -3\)?
- The left part climbs and the right part falls, both without bound
- Both parts approach height \(-2\)
- The curve has a ring at \(x = -3\)
Answer
The left part climbs and the right part falls, both without bound
How many vertical asymptotes does \(z(x) = \dfrac{x^2 - 4}{x^3 - 4x}\) have?
- 1
- 3
- 2
- 0
Answer
1
Summary
You factor the top and the bottom and cancel the common factors.
A zero that cancels leaves a hole with a finite limit, and a zero that stays in the bottom leaves a vertical asymptote.
Do this before you call a zero of the bottom an asymptote, because the zero may cancel.
This is one of four skills in topic 1.14 of Which & Why: AP Calculus.