• We will study the end behaviour of the graph of a rational function and identify any horizontal asymptote, if it exists.
• We will identify the conditions when a rational function does not have a horizontal asymptote. In this case, we will determine a slant (or oblique) asymptote for the function, if one exists.
• We will continue to use the language of limits to assist us in communicating our understanding of the end behaviour of the function.

A horizontal asymptote is a guideline for the end behaviour of the function.

To identify a horizontal asymptote of a rational function, if it exists, we must study the end behaviours of the function.

Using the language of limits, this means that we must determine

\(\displaystyle \lim_{x \rightarrow \infty}f(x)\) and \(\displaystyle \lim_{x \rightarrow -\infty}f(x)\)

What is the horizontal asymptote of the function \( y = \dfrac{1}{x^2 + 1} \)? How does the graph of the function behave near the horizontal asymptote?

Solution

This function is the reciprocal of the quadratic \( y = x^2 + 1 \).

From previous studies of functions of the form \( y = \dfrac{1}{f(x)} \), where \( f(x) \) is a polynomial function (of degree one or greater), we know that \( y = 0 \) is the horizontal asymptote.

As \( x \rightarrow \infty \), \( x^2 + 1 \rightarrow \infty \) and thus \( \dfrac{1}{x^2 + 1} \rightarrow 0 \).

Since \( y = \dfrac{1}{x^2 + 1} \gt 0 \) for all \( x \in \mathbb{R} \), the curve lies above the \( x \)-axis and will approach \( y = 0 \) from above as \( x \) increases in value.

Similarly, as \( x \rightarrow -\infty,\ x^2 + 1 \rightarrow \infty \) and \( \dfrac{1}{x^2 + 1} \rightarrow 0 \).

So, again, the graph of the function will approach \( y = 0 \) from above as \(x\) decreases in value.

What is the horizontal asymptote of the function \( y = \dfrac{1}{x^2 + 1} \)? How does the graph of the function behave near the horizontal asymptote?

Solution

This function is the reciprocal of the quadratic \( y = x^2 + 1 \).

From previous studies of functions of the form \( y = \dfrac{1}{f(x)} \), where \( f(x) \) is a polynomial function (of degree one or greater), we know that \( y = 0 \) is the horizontal asymptote.

We can say that

\( \displaystyle \lim_{x \rightarrow \infty}\frac{1}{x^2 + 1} = 0\) and \(\displaystyle \lim_{x \rightarrow -\infty}\frac{1}{x^2 + 1} = 0 \)

This function has a horizontal asymptote of \( y = 0 \).

There is no vertical asymptote since \( x^2 + 1 \neq 0 \) for \( x \in \mathbb{R} \).