Asymptotes

An asymptote is a line that a curve approaches but does not (necessarily) reach. The three types, vertical, horizontal, and oblique, are all defined using limits and describe the global and local behavior of a function's graph.

Type	Description	How to find
Vertical ( $x = a$ )	Graph blows up as $x \to a$	Find where denominator is $0$ (or $\ln$ domain boundary, etc.)
Horizontal ( $y = b$ )	Graph flattens to $b$ as $x \to \pm \infty$	Compute $lim_{x \to \pm \infty} f (x)$
Oblique ( $y = m x + b$ )	Graph approaches a line with slope $m \neq 0$	Divide numerator by denominator; check remainder $\to 0$

Vertical Asymptotes

The line $x = a$ is a vertical asymptote of the graph of $f$ if $f (x) \to + \infty$ or $f (x) \to - \infty$ as $x$ approaches $a$ from the left or right.

There are three main sources of vertical asymptotes:

Rational functions: $f (x) = g (x) / h (x)$ where $lim_{x \to a} h (x) = 0$ . If additionally $lim_{x \to a} g (x) \neq 0$ , then $x = a$ is definitely a vertical asymptote. If both numerator and denominator vanish at $a$ , the asymptote may or may not exist after simplification.
Logarithmic functions: For example, $lim_{x \to 0^{+}} \ln x = - \infty$ , so $x = 0$ is a vertical asymptote of $y = \ln x$ .
Trigonometric functions: For example, $\tan x$ has vertical asymptotes at every $x = \frac{π}{2} + k π$ .

Horizontal Asymptotes

A line $y = b$ is a horizontal asymptote of the curve $y = f (x)$ if $f (x) \to b$ as $x \to + \infty$ or $x \to - \infty$ .

Diagrams showing a curve approaching horizontal asymptote y=b from both directions

Find the horizontal asymptote(s) of $f (x) = \frac{1}{1 + e^{- x}}$ .

Solution

Since

lim_{x \to + \infty} e^{- x} = 0

lim_{x \to + \infty} \frac{1}{1 + e^{- x}} = \frac{1}{1 + 0} = 1.

y = 1

is a horizontal asymptote. Since

lim_{x \to - \infty} e^{- x} = + \infty

lim_{x \to - \infty} \frac{1}{1 + e^{- x}} = \frac{1}{1 + \infty} = 0.

y = 0

is another horizontal asymptote. The graph has **two** horizontal asymptotes:

y = 0

and

y = 1

Graph of f(x)=1/(1+e^(-x)) with horizontal asymptotes y=0 and y=1

Oblique (Slant) Asymptotes

A line $y = m x + b$ ( $m \neq 0$ ) is an oblique asymptote of $f$ if either

lim_{x \to + \infty} [f (x) - (m x + b)] = 0 or lim_{x \to - \infty} [f (x) - (m x + b)] = 0.

Diagram showing a curve approaching an oblique asymptote as x approaches ±∞

If $f (x) = P_{1} (x) / P_{2} (x)$ and $\deg P_{1} = \deg P_{2} + 1$ (numerator degree is exactly one more than denominator degree), then $f$ has an oblique asymptote equal to the quotient of dividing $P_{1}$ by $P_{2}$ .

Find the oblique asymptote of $f (x) = \frac{3 x^{2} + 5 x - 3}{x + 1}$ .

Solution

The numerator has degree 2 and the denominator has degree 1, so an oblique asymptote exists. Perform polynomial division:

\frac{3 x^{2} + 5 x - 3}{x + 1} = 3 x + 2 - \frac{5}{x + 1} .

Since

| f (x) - (3 x + 2) | = | \frac{- 5}{x + 1} | \to 0

x \to \pm \infty

, the line

y = 3 x + 2

is the oblique asymptote. The graph also has a vertical asymptote at

x = - 1

(denominator is zero there, numerator is not).

Graph of f(x)=(3x^2+5x-3)/(x+1) with oblique asymptote y=3x+2 and vertical asymptote x=-1

Find the horizontal asymptote(s) of $f (x) = \frac{2 x^{2} - 3}{x^{2} + 1}$ .

Solution

Using the leading-term technique:

lim_{x \to \pm \infty} \frac{2 x^{2} - 3}{x^{2} + 1} = lim_{x \to \pm \infty} \frac{2 x^{2}}{x^{2}} = 2.

y = 2

is a horizontal asymptote (the same in both directions). There is no vertical asymptote since

x^{2} + 1 > 0

for all real

x

Find the vertical asymptotes (if any) of:

$f (x) = \frac{4 - 2 x}{2 x^{2} + x - 1}$
$g (x) = \frac{x^{2} + 4 x + 3}{x^{2} + 7 x + 12}$
$h (x) = \frac{x + 7}{x^{2} + 1}$

Solution

**(a)** Factor the denominator:

2 x^{2} + x - 1 = (2 x - 1) (x + 1)

. The denominator is zero at

x = 1 / 2

and

x = - 1

. Check the numerator:

4 - 2 (1 / 2) = 3 \neq 0

and

4 - 2 (- 1) = 6 \neq 0

. So both

x = 1 / 2

and

x = - 1

are vertical asymptotes. **(b)** Factor:

x^{2} + 4 x + 3 = (x + 1) (x + 3)

and

x^{2} + 7 x + 12 = (x + 3) (x + 4)

. So g(x) = \frac{(x+1)\cancel{(x+3)}}{\cancel{(x+3)}(x+4)} = \frac{x+1}{x+4} \quad (x \neq -3). The denominator of the simplified form is zero at

x = - 4

(and

- 4 \neq - 3

), so

x = - 4

is a vertical asymptote. At

x = - 3

there is a **removable discontinuity** (hole), not an asymptote. **(c)** The denominator

x^{2} + 1 \geq 1 > 0

for all real

x

, so there are **no vertical asymptotes**.

Frequently Asked Questions

Can a curve cross its own asymptote?

Yes. A horizontal or oblique asymptote describes end behavior as

x \to \pm \infty

; the curve can cross it at finite values of

x

. A vertical asymptote, however, is never crossed: the function grows without bound there.

How do I find oblique asymptotes when polynomial division is messy?

Use the formulas: if

y = m x + b

is an oblique asymptote as

x \to + \infty

, then

m = lim_{x \to + \infty} f (x) / x

and

b = lim_{x \to + \infty} [f (x) - m x]

. These formulas work for any function, not just rational ones.

A rational function has equal degrees in numerator and denominator. What kind of asymptote does it have?

A horizontal asymptote at

y = a_{n} / b_{m}

(the ratio of leading coefficients). For example,

f (x) = (3 x^{2} + 1) / (x^{2} - 2)

has horizontal asymptote

y = 3

Vertical Asymptotes

Horizontal Asymptotes

Oblique (Slant) Asymptotes

Frequently Asked Questions

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