Connecting Limits at Infinity and Horizontal Asymptotes

If a function has a limit at infinity, when we get farther and farther from the origin along the \(x\)- axis, it will appear to straighten out into a line.

As \(x\) approaches infinity, we can find the equation of this line by considering the limit of our equation. This line is called an asymptote.

Related Topics

• How to Find Infinite Limits and Vertical Asymptotes

A step-by-step guide to connecting limits at infinity and horizontal asymptotes

It is common for functions to display asymptote behavior. Graphically, this means that their graph approaches another geometric object (typically a line) as the function’s graph moves away from its origin.

We use limits in mathematics to illustrate circumstances in which a function approaches a certain value, which is why asymptotic behavior requires limits.

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You may recall learning about asymptotes in algebra class. Limits in calculus allow us to prove the existence of these places.

Horizontal Asymptotes:

In order for \(f(x)\) to have the horizontal asymptote \(y=L\):

\(\color{blue}{lim_{x\to ∞} f(x)=L }\) \(or\) \(\color{blue}{lim_{x\to -∞} f(x)= L}\)

Must be true.

Therefore, to find horizontal asymptotes, we simply test the function’s limit as it approaches infinity and again as it approaches negative infinity. Asymptotes may only be horizontal in one direction at a time.

Connecting Limits at Infinity and Horizontal Asymptotes – Example 1:

Find the horizontal asymptotes of the function. \(f(x)=\frac{1-2x^3}{x^3-4x^2+2x}\)

To determine whether there are horizontal asymptotes we must evaluate the limits at infinity:

\(lim_{x\to ∞}\frac{1-2x^3}{x^3-4x^2+2x}\) \(=lim_{x\to ∞}\frac{\frac{1}{x^3}-2}{1-\frac{4}{x}+\frac{2}{x^2}}\)

\(=\frac{0-2}{1-0-0}=\frac{-2}{1}=-2\)

The function \(f(x)\) has a horizontal asymptote \(y=-2\).

We can calculate \(lim_{x\to – ∞ }f(x)=-2\), so \(y =- 2\) is the only horizontal asymptotes.

Connecting Limits at Infinity and Horizontal Asymptotes – Example 2:

Find the horizontal asymptotes of the function. \(f(x)=\frac{x^2-1}{x^2-x-2}\)

To determine whether there are horizontal asymptotes we must evaluate the limits at infinity:

\(lim_{x\to ∞} \frac{x^2-1}{x^2-x-2}=lim_{x\to ∞} \frac{1-\frac{1}{x^2}}{1-\frac{1}{x}-\frac{2}{x^2}}\)

\(=\frac{1-0}{1-0-0}=1\)

The function \(f(x)\) has a horizontal asymptote \(y=1\).

We can calculate \(lim_{x\to – ∞ }f(x)=1\), so \(y = 1\) is the only horizontal asymptotes.

Exercises for Connecting Limits at Infinity and Horizontal Asymptotes

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Find the horizontal asymptotes of the functions.

1. \(\color{blue}{f(x)=\frac{2x^2-4x+8}{3x^2-27}}\)
2. \(\color{blue}{f\left(x\right)=\frac{x}{1+x^2}}\)
3. \(\color{blue}{f\left(x\right)=e^{\frac{1}{x}}}\)
4. \(\color{blue}{f\left(x\right)=\frac{x^2+7x+12}{-2x^2-2x+12}}\)
5. \(\color{blue}{f\left(x\right)=\frac{x}{\sqrt{x^2+1}}}\)

1. \(\color{blue}{y=\frac{2}{3}}\)
2. \(\color{blue}{y=0}\)
3. \(\color{blue}{y=1}\)
4. \(\color{blue}{y=-\frac{1}{2}}\)
5. \(\color{blue}{y=1, y=-1}\)