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Last modified on December 20th, 2023

Horizontal asymptotes, or HA, are horizontal dashed lines on a graph that help determine the end behavior of a function. They show how the input influences the graph’s curve as it extends toward infinity.

Mathematically, they can be represented as the equation of a line y = b when either ${\lim _{x\rightarrow \infty }=b}$ or ${\lim _{x\rightarrow -\infty }=b}$. Here, the value of x tends to infinity or –infinity, and the curve approaches a constant value b. Thus, for very large or very small values of x, the curve would approach, but not touch, a horizontal line y = b, as shown below.

Horizontal asymptotes can be negative and approach infinity. A function may not always have a horizontal asymptote.

Unlike vertical asymptotes, even though these lines do not touch the curve of the rational function, they can cross over in some cases. A slant or oblique asymptote is similar, as it shows the end behavior of a function, but it is a slanted line, as the name suggests.

We follow the steps below to determine the horizontal asymptote of any function y = f(x), where ${x\rightarrow \pm \infty}$.

- We find the value of ${ \lim _{x\rightarrow \infty }f\left( x\right)}$
- We do the same for ${\lim _{x\rightarrow -\infty }f\left( x\right)}$
- If one (or both) values is a real number b, then the horizontal asymptote is given as y = b.

While this method holds for most functions of the form y = f(x), there is an easier way of finding out the horizontal asymptotes of a rational function using three basic rules.

A rational function of the form ${f\left( x\right) =\dfrac{P\left( x\right) }{Q\left( x\right) }}$, where P(x) and Q(x) are polynomials, can have only one horizontal asymptote. We can find it by analyzing the ratio of the leading term coefficients of the numerator and denominator.

There are three such possibilities:

**Case 1: Degree of the Numerator > Degree of the Denominator**

In such a case, there is no horizontal asymptote.

For ${f\left( x\right) =\dfrac{x^{3}-1}{3+x^{2}}}$, the ratio of the leading terms will be ${\dfrac{x^{3}}{x^{2}}}$, which gives x as the result. Since this value is larger than any non-existent denominator, there is no horizontal asymptote for the given function.

When the degree of the numerator is greater by one, we get a slant or oblique asymptote that follows the form y = mx + b.

**Case 2: Degree of the Numerator < Degree of the Denominator**

Here, the horizontal asymptote is y = 0, i.e., the x-axis.

If we take the function ${f\left( x\right) =\dfrac{2x}{3x^{2}+2}}$, the ratio of the leading terms will give us a function that behaves similarly to ${g\left( x\right) =\dfrac{2}{3x}}$. As x approaches infinity, the function will approach 0. Hence, the horizontal asymptote will be y = 0.

As we can see, the graph crosses the horizontal asymptote at point A.

**Case 3: Degree of the Numerator = Degree of the Denominator**

The horizontal asymptote is calculated by finding the coefficient ratio of the leading terms.

For example, for the function ${ f\left( x\right) =\dfrac{2x^{2}-1}{x^{2}+3}}$, the degrees of the numerator and the denominator are equal. Hence, the ratio of the leading terms gives us ${\dfrac{2x^{2}}{x^{2}}=2}$. The horizontal asymptote is thus y = 2.

For exponential functions of the form ${f\left( x\right) =ab^{kx}+c}$, the horizontal asymptote is always y = c. If c = 0, then y = 0, or the x-axis.

Using the above rule, for the function ${f\left( x\right) =4^{x}+1}$, the horizontal asymptote will be y = 1 as c = 1.

- Square root functions have two horizontal asymptotes. For example, ${f\left( x\right) =\dfrac{x+1}{\sqrt{x^{2}-2}}}$ has horizontal asymptotes at y =1 and y = -1, respectively.
- Rectangular hyperbolas of the form ${ f\left( x\right) =\dfrac{a}{x+p}+q}$ have a horizontal asymptote of y = q.

- Polynomial functions of the form ${f\left( x\right) =a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots +a_{1}x^{1}+a_{0}}$ do not have any horizontal asymptote as x does not give any real values at limits of infinity and –infinity.
- Basic trigonometric functions do not have horizontal asymptotes.
- Basic logarithmic functions do not have horizontal asymptotes.

**Give the horizontal asymptote of the function ****${f_{\left( x\right) }=\dfrac{4x^{2}+1}{5x^{2}-2x+1}}$.**

Solution:

Here, the highest powers of the numerator and the denominator are equal, i.e., 2. So we can find the horizontal asymptote from the ratio of the leading term coefficients, 4 and 5, respectively. Therefore, the horizontal asymptote of f(x) is ${y=\dfrac{4}{5}}$.

**Calculate the horizontal asymptote for ${f\left( x\right) =\dfrac{x}{x-1}}$.**

Solution:

To get the horizontal asymptote, we need to solve for limits.

${\lim _{x\rightarrow \infty }f\left( x\right) =\lim _{x\rightarrow \infty }\dfrac{x}{x-1}}$

${=\lim _{x\rightarrow \infty }\dfrac{x}{\left[ x\left( 1-\dfrac{1}{x}\right) \right] }}$

${=\lim _{x\rightarrow \infty }\dfrac{1}{\left( 1-\dfrac{1}{x}\right) }}$

${=\dfrac{1}{1-0}=1}$

So y = 1 is the HA of the function. Taking the other limit,

${\lim _{x\rightarrow -\infty }f\left( x\right) =\lim _{x\rightarrow -\infty }\dfrac{x}{x-1}}$

${=\lim _{x\rightarrow -\infty }\dfrac{x}{\left[ x\left( 1-\dfrac{1}{x}\right) \right] }}$

${=\lim _{x\rightarrow -\infty }\dfrac{1}{\left( 1-\dfrac{1}{x}\right) }}$

${=\dfrac{1}{1+0}=1}$

Again, we got the same value. Therefore there is only one horizontal asymptote, which is y = 1.

**Determine whether ${f\left( x\right) =\dfrac{1-x^{2}}{x^{4}-x^{2}+2}}$ has horizontal asymptotes. Calculate the horizontal asymptotes, if any.**

Solution:

In the given function, the degree of the numerator < the degree of the denominator. Therefore, it will have one horizontal asymptote, y = 0.

**Identify the horizontal asymptote for each graph.**

Solution:

**A. **The horizontal asymptote is x = 0.**B. **The horizontal asymptote is x = 3.

Last modified on December 20th, 2023