AP®MathsCollege BoardCalculus ABRevision NotesUnit 1: Limits & ContinuityLimitsAsymptotes

Asymptotes (College Board AP® Calculus AB): Revision Note

Written by: Roger B

Reviewed by: Dan Finlay

Updated on 1 May 2026

Horizontal asymptote

What is a horizontal asymptote?

A horizontal asymptote is a horizontal line
- that the graph of a function gets closer and closer to (but never touches or intersects)
- as x becomes unbounded in the positive or negative direction
On the following diagram, the horizontal asymptote is indicated by a dashed line

A graph of a function f with a horizontal asymptote at y=c — **An example of a function with a horizontal asymptote at y=c**

How can I identify horizontal asymptotes using limits?

A function will have a horizontal asymptote if it has a finite limit at infinity
- I.e. the line $y = c$ will be a horizontal asymptote for the graph of a function $f$ if
  - $\lim_{x \to \infty} f (x) = c$ , or
  - $\lim_{x \to - \infty} f (x) = c$
Horizontal asymptotes (if any) may therefore be determined by evaluating the limits at infinity

Examiner Tips and Tricks

By graphing a function on your graphing calculator you can:

spot any asymptotic behavior by a function at plus or minus infinity
check limits that you have determined analytically

How can I identify horizontal asymptotes numerically?

If a function has a horizontal asymptote, then its values will tend to a limit for large positive or negative values of $x$
Consider, the function $f (x) = \frac{12 + 5 e^{- \frac{x}{1000}}}{1 + e^{- \frac{x}{1000}}}$

Consider the values for large positive $x$

$x =$	$\frac{12 + 5 e^{- \frac{x}{1000}}}{1 + e^{- \frac{x}{1000}}} =$
100	8.674...
1,000	10.117...
10,000	11.999...
100,000	12

The table suggests that $\lim_{x \to \infty} f (x) = 12$
- Note $f (100 000) \neq 12$
- It starts as 11.999... with a large number of 9s
- Most calculators cannot differentiate between this value and 12
Consider the values for large negative $x$

$x =$	$\frac{12 + 5 e^{- \frac{x}{100}}}{1 + e^{- \frac{x}{100}}} =$
-100	8.325...
-1,000	6.882...
-10,000	5.000...
-100,000	5

The table suggests that $\lim_{x \to - \infty} f (x) = 5$
- Note $f (- 100 000) \neq 5$
- It starts as 5.000... with a large number of 0s
- Most calculators cannot differentiate between this value and 5

Vertical asymptote

What is a vertical asymptote?

A vertical asymptote is a vertical line
- that the graph of a function gets closer and closer to (but never touches or intersects)
- as x gets closer and closer to the x-value of the vertical line
On the following diagram, the vertical asymptote is indicated by a dashed line

Graph of a function f with a vertical asymptote at x=c — **An example of a function with a vertical asymptote at x=c**

How can I identify vertical asymptotes using limits?

A function will have a vertical asymptote at any x-value where the function becomes unbounded
- I.e. the line $x = c$ will be a vertical asymptote for the graph of a function $f$ if
  - $\lim_{x \to c^{-}} f (x) = \pm \infty$ , or
  - $\lim_{x \to c^{+}} f (x) = \pm \infty$
Vertical asymptotes (if any) may therefore be determined by identifying points where the function becomes unbounded
- Usually this will involve a function in the form of a quotient
  - at points where the denominator becomes zero

How can I identify vertical asymptotes numerically?

If a function has a vertical asymptote, then its values tend to positive or negative infinity as the values of $x$ get closer to the vertical asymptote
Consider, the function $f (x) = \csc x$
Consider the values for values of $x$ just smaller than 0

$x =$	$\csc x =$
-0.1	-10.016...
-0.01	-100.001...
-0.001	-1,000.000...
-0.0001	-10,000.000...

The table suggests that $\lim_{x \to 0^{-}} f (x) = - \infty$
- The values of $f$ keep getting further from 0 as $x$ gets closer to 0
Consider the values for values of $x$ just bigger than 0

$x =$	$\csc x =$
0.1	10.016...
0.01	100.001...
0.001	1,000.000...
0.0001	10,000.000...

The table suggests that $\lim_{x \to 0^{+}} f (x) = \infty$
- The values of $f$ keep getting further from 0 as $x$ gets closer to 0

Worked Example

Let $f$ be the function defined by $f (x) = \frac{3 x - 11}{x - 2}$ .

Using limits, identify the vertical and horizontal asymptotes (if any) on the graph of $f$ .

Answer:

The denominator becomes 0 when $x = 2$ , so start by considering the limits there

At 2 the numerator is equal to -5, so zero only occurs in the denominator

Just 'to the left' of 2, $3 x - 11 < 0$ and $x - 2 < 0$ so

$\lim_{x \to 2^{-}} f (x) = \infty$

Just 'to the right' of 2, $3 x - 11 < 0$ and $x - 2 > 0$ so

$\lim_{x \to 2^{+}} f (x) = - \infty$

This confirms that the graph of $f$ has a vertical asymptote at $x = 2$

To identify horizontal asymptotes, start by rearranging to make the behavior of the function more obvious

$\frac{3 x - 11}{x - 2} = \frac{3 (x - 2) - 5}{x - 2} = \frac{3 (x - 2)}{x - 2} - \frac{5}{x - 2} = 3 - \frac{5}{x - 2}$

$\frac{5}{x - 2}$ becomes closer and closer to zero as x increases in the positive or negative directions, so

$\lim_{x \to - \infty} f (x) = \lim_{x \to \infty} f (x) = 3 - 0 = 3$

Alternatively, divide the top and bottom of the function by $x$

$\lim_{x \to \infty} f (x) = \lim_{x \to \infty} \frac{3 - \frac{11}{x}}{1 - \frac{2}{x}} = \frac{3}{1} = 3$

$\lim_{x \to - \infty} f (x) = \lim_{x \to - \infty} \frac{3 - \frac{11}{x}}{1 - \frac{2}{x}} = \frac{3}{1} = 3$

This means that the graph of $f$ has a horizontal asymptote at $y = 3$

The graph of $f$ has a vertical asymptote at $x = 2$ , and a horizontal asymptote at $y = 3$

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