Limits to Determine Horizontal Asymptotes

Limits to Determine Horizontal Asymptotes

One important uses of limits at infinity and negative infinity is that we can determine horizontal asymptotes with them.

Definition: A function $f$ has a Horizontal Asymptote at $y = L$ if either $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$ where $L$ is a finite real number.

Remark: Note that a function $f$ can have infinitely many vertical asymptotes, however, $f$ can have at most 2 horizontal asymptotes by the definition of a horizontal asymptote. That is, if $\lim_{x \to \infty} f(x) = L$ and $\lim_{x \to -\infty} f(x) = M$, then the two horizontal asymptotes of $f$ are denoted by the lines $y = L$ and $y = M$.

Let's look at some examples of computing horizontal asymptotes.

Example 1

Show that the function $f(x) = \frac{1}{x^2 + x}$ has a horizontal asymptote at $y = 0$.


We note that as $x \to \infty$, $x^2 + x \to \infty$. Since the denominator gets arbitrarily large as $x$ gets arbitrarily large, $f(x) \to 0$. Therefore, $\lim_{x \to \infty} f(x) = 0$, which by our definition states that the line $y = 0$ is a horizontal asymptote of $f$.

Alternatively, we can factor $f(x)$ in the following manner, $f(x) = \frac{1}{x(x+1)}$. Clearly, as $x \to \infty$, $x(x+1) \to \infty$ and so $f(x) \to 0$.

Example 2

Show that the function $f(x) = \frac{x^3 + 9x - 4}{4x^3 + 7x^2 + 6}$ has a horizontal asymptote as $x \to \infty$.

We will first divide every term of $f(x)$ by $x^3$ to get that:

\begin{align} f(x) = \frac{1 + \frac{9}{x^2} - \frac{4}{x^3}}{4 + \frac{7}{x} + \frac{6}{x^3}} \end{align}


\begin{align} \lim_{x \to \infty} \frac{1 + \frac{9}{x^2} - \frac{4}{x^3}}{4 + \frac{7}{x} + \frac{6}{x^3}} = \frac{1 + \lim_{x \to \infty} \frac{9}{x^2} - \lim_{x \to \infty} \frac{4}{x^3}}{4 + \lim_{x \to \infty} \frac{7}{x} + \lim_{x \to \infty} \frac{6}{x^3}} = \frac{1}{4} \end{align}

So there exists a horizontal asymptote, namely the line $y = \frac{1}{4}$.

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