Asymptote

The line x = a is a vertical asymptote of the graph of the function y = ƒ(x) if at least one of the following statements is true:

1. ${\displaystyle \lim _{x\to a^{-}}f(x)=\pm \infty ,}$
2. ${\displaystyle \lim _{x\to a^{+}}f(x)=\pm \infty ,}$

where ${\displaystyle \lim _{x\to a^{-}}}$ is the limit as x approaches the value a from the left (from lesser values), and ${\displaystyle \lim _{x\to a^{+}}}$ is the limit as x approaches a from the right.

For example, if ƒ(x) = x/(x–1), the numerator approaches 1 and the denominator approaches 0 as x approaches 1. So

${\displaystyle \lim _{x\to 1^{+}}{\frac {x}{x-1}}=+\infty }$

${\displaystyle \lim _{x\to 1^{-}}{\frac {x}{x-1}}=-\infty }$

and the curve has a vertical asymptote x = 1.

The function ƒ(x) may or may not be defined at a, and its precise value at the point x = a does not affect the asymptote. For example, for the function

${\displaystyle f(x)={\begin{cases}{\frac {1}{x}}&{\text{if }}x>0,\\5&{\text{if }}x\leq 0.\end{cases}}}$

has a limit of +∞ as x → 0⁺, ƒ(x) has the vertical asymptote x = 0, even though ƒ(0) = 5. The graph of this function does intersect the vertical asymptote once, at (0, 5). It is impossible for the graph of a function to intersect a vertical asymptote (or a vertical line in general) in more than one point. Moreover, if a function is continuous at each point where it is defined, it is impossible that its graph does intersect any vertical asymptote.

A common example of a vertical asymptote is the case of a rational function at a point x such that the denominator is zero and the numerator is non-zero.

If a function has a vertical asymptote, then it isn't necessarily true that the derivative of the function has a vertical asymptote at the same place. An example is

${\displaystyle f(x)={\tfrac {1}{x}}+\sin({\tfrac {1}{x}})\quad }$ at ${\displaystyle \quad x=0}$.

This function has a vertical asymptote at ${\displaystyle x=0,}$ because

${\displaystyle \lim _{x\to 0^{+}}f(x)=\lim _{x\to 0^{+}}\left({\tfrac {1}{x}}+\sin \left({\tfrac {1}{x}}\right)\right)=+\infty ,}$

and

${\displaystyle \lim _{x\to 0^{-}}f(x)=\lim _{x\to 0^{-}}\left({\tfrac {1}{x}}+\sin \left({\tfrac {1}{x}}\right)\right)=-\infty }$.

The derivative of ${\displaystyle f}$ is the function

${\displaystyle f'(x)={\frac {-(\cos({\tfrac {1}{x}})+1)}{x^{2}}}}$.

For the sequence of points

${\displaystyle x_{n}={\frac {(-1)^{n}}{(2n+1)\pi }},\quad }$ for ${\displaystyle \quad n=0,1,2,\ldots }$

that approaches ${\displaystyle x=0}$ both from the left and from the right, the values ${\displaystyle f'(x_{n})}$ are constantly ${\displaystyle 0}$. Therefore, both one-sided limits of ${\displaystyle f'}$ at ${\displaystyle 0}$ can be neither ${\displaystyle +\infty }$ nor ${\displaystyle -\infty }$. Hence ${\displaystyle f'(x)}$ doesn't have a vertical asymptote at ${\displaystyle x=0}$.

The arctangent function has two different asymptotes

Horizontal asymptotes are horizontal lines that the graph of the function approaches as x → ±∞. The horizontal line y = c is a horizontal asymptote of the function y = ƒ(x) if

${\displaystyle \lim _{x\rightarrow -\infty }f(x)=c}$ or ${\displaystyle \lim _{x\rightarrow +\infty }f(x)=c}$.

In the first case, ƒ(x) has y = c as asymptote when x tends to −∞, and in the second ƒ(x) has y = c as an asymptote as x tends to +∞.

For example, the arctangent function satisfies

${\displaystyle \lim _{x\rightarrow -\infty }\arctan(x)=-{\frac {\pi }{2}}}$ and ${\displaystyle \lim _{x\rightarrow +\infty }\arctan(x)={\frac {\pi }{2}}.}$

So the line y = –π/2 is a horizontal asymptote for the arctangent when x tends to –∞, and y = π/2 is a horizontal asymptote for the arctangent when x tends to +∞.

Functions may lack horizontal asymptotes on either or both sides, or may have one horizontal asymptote that is the same in both directions. For example, the function ƒ(x) = 1/(x²+1) has a horizontal asymptote at y = 0 when x tends both to −∞ and +∞ because, respectively,

${\displaystyle \lim _{x\to -\infty }{\frac {1}{x^{2}+1}}=\lim _{x\to +\infty }{\frac {1}{x^{2}+1}}=0.}$

Other common functions that have one or two horizontal asymptotes include x ↦ 1/x (that has an hyperbola as it graph), the Gaussian function ${\displaystyle x\mapsto \exp(-x^{2}),}$ the error function, and the logistic function.

In the graph of ${\displaystyle f(x)=x+{\tfrac {1}{x}}}$, the y-axis (x = 0) and the line y = x are both asymptotes.

When a linear asymptote is not parallel to the x- or y-axis, it is called an oblique asymptote or slant asymptote. A function ƒ(x) is asymptotic to the straight line y = mx + n (m ≠ 0) if

${\displaystyle \lim _{x\to +\infty }\left[f(x)-(mx+n)\right]=0\,{\mbox{ or }}\lim _{x\to -\infty }\left[f(x)-(mx+n)\right]=0.}$

In the first case the line y = mx + n is an oblique asymptote of ƒ(x) when x tends to +∞, and in the second case the line y = mx + n is an oblique asymptote of ƒ(x) when x tends to −∞.

An example is ƒ(x) = x + 1/x, which has the oblique asymptote y = x (that is m = 1, n = 0) as seen in the limits

${\displaystyle \lim _{x\to \pm \infty }\left[f(x)-x\right]}$

${\displaystyle =\lim _{x\to \pm \infty }\left[\left(x+{\frac {1}{x}}\right)-x\right]}$

${\displaystyle =\lim _{x\to \pm \infty }{\frac {1}{x}}=0.}$

The oblique asymptote, for the function f(x), will be given by the equation y = mx + n. The value for m is computed first and is given by

${\displaystyle m\;{\stackrel {\text{def}}{=}}\,\lim _{x\rightarrow a}f(x)/x}$

where a is either ${\displaystyle -\infty }$ or ${\displaystyle +\infty }$ depending on the case being studied. It is good practice to treat the two cases separately. If this limit doesn't exist then there is no oblique asymptote in that direction.

Having m then the value for n can be computed by

${\displaystyle n\;{\stackrel {\text{def}}{=}}\,\lim _{x\rightarrow a}(f(x)-mx)}$

where a should be the same value used before. If this limit fails to exist then there is no oblique asymptote in that direction, even should the limit defining m exist. Otherwise y = mx + n is the oblique asymptote of ƒ(x) as x tends to a.

For example, the function ƒ(x) = (2x² + 3x + 1)/x has

${\displaystyle m=\lim _{x\rightarrow +\infty }f(x)/x=\lim _{x\rightarrow +\infty }{\frac {2x^{2}+3x+1}{x^{2}}}=2}$ and then

${\displaystyle n=\lim _{x\rightarrow +\infty }(f(x)-mx)=\lim _{x\rightarrow +\infty }\left({\frac {2x^{2}+3x+1}{x}}-2x\right)=3}$

so that y = 2x + 3 is the asymptote of ƒ(x) when x tends to +∞.

The function ƒ(x) = ln x has

${\displaystyle m=\lim _{x\rightarrow +\infty }f(x)/x=\lim _{x\rightarrow +\infty }{\frac {\ln x}{x}}=0}$ and then

${\displaystyle n=\lim _{x\rightarrow +\infty }(f(x)-mx)=\lim _{x\rightarrow +\infty }\ln x}$, which does not exist.

So y = ln x does not have an asymptote when x tends to +∞.

A rational function has at most one horizontal asymptote or oblique (slant) asymptote, and possibly many vertical asymptotes.

The degree of the numerator and degree of the denominator determine whether or not there are any horizontal or oblique asymptotes. The cases are tabulated below, where deg(numerator) is the degree of the numerator, and deg(denominator) is the degree of the denominator.

The cases of horizontal and oblique asymptotes for rational functions

deg(numerator)−deg(denominator)	Asymptotes in general	Example	Asymptote for example
< 0	${\displaystyle y=0}$	${\displaystyle f(x)={\frac {1}{x^{2}+1}}}$	${\displaystyle y=0}$
= 0	y = the ratio of leading coefficients	${\displaystyle f(x)={\frac {2x^{2}+7}{3x^{2}+x+12}}}$	${\displaystyle y={\frac {2}{3}}}$
= 1	y = the quotient of the Euclidean division of the numerator by the denominator	${\displaystyle f(x)={\frac {2x^{2}+3x+5}{x}}=2x+3+{\frac {5}{x}}}$	${\displaystyle y=2x+3}$
> 1	none	${\displaystyle f(x)={\frac {2x^{4}}{3x^{2}+1}}}$	no linear asymptote, but a curvilinear asymptote exists

The vertical asymptotes occur only when the denominator is zero (If both the numerator and denominator are zero, the multiplicities of the zero are compared). For example, the following function has vertical asymptotes at x = 0, and x = 1, but not at x = 2.

${\displaystyle f(x)={\frac {x^{2}-5x+6}{x^{3}-3x^{2}+2x}}={\frac {(x-2)(x-3)}{x(x-1)(x-2)}}}$

Black: the graph of ${\displaystyle f(x)=(x^{2}+x+1)/(x+1)}$. Red: the asymptote ${\displaystyle y=x}$. Green: difference between the graph and its asymptote for ${\displaystyle x=1,2,3,4,5,6}$

When the numerator of a rational function has degree exactly one greater than the denominator, the function has an oblique (slant) asymptote. The asymptote is the polynomial term after dividing the numerator and denominator. This phenomenon occurs because when dividing the fraction, there will be a linear term, and a remainder. For example, consider the function

${\displaystyle f(x)={\frac {x^{2}+x+1}{x+1}}=x+{\frac {1}{x+1}}}$

shown to the right. As the value of x increases, f approaches the asymptote y = x. This is because the other term, 1/(x+1), approaches 0.

If the degree of the numerator is more than 1 larger than the degree of the denominator, and the denominator does not divide the numerator, there will be a nonzero remainder that goes to zero as x increases, but the quotient will not be linear, and the function does not have an oblique asymptote.

If a known function has an asymptote (such as y=0 for f(x)=e^x), then the translations of it also have an asymptote.

• If x=a is a vertical asymptote of f(x), then x=a+h is a vertical asymptote of f(x-h)
• If y=c is a horizontal asymptote of f(x), then y=c+k is a horizontal asymptote of f(x)+k

If a known function has an asymptote, then the scaling of the function also have an asymptote.

• If y=ax+b is an asymptote of f(x), then y=cax+cb is an asymptote of cf(x)

For example, f(x)=e^x-1+2 has horizontal asymptote y=0+2=2, and no vertical or oblique asymptotes.