Horizontal asymptotes

Functions: Limits and asymptotes

Horizontal asymptotes

Asymptotes and limits have a lot in common. We will investigate how we can use the definition of a limit to find horizontal asymptotes.

The function $\blue{f(x)}$ has a horizontal asymptote $y=\green a$ to infinity if $\lim_{x \to \infty}\blue{f(x)}=\green a$ .

This intuitively means: when $x$ gets really big, $\blue{f(x)}$ gets arbitrarily close to value $\green a$ .

Similarily $\blue{f(x)}$ has a horizontal asymptote $y=\green a$ to minus infinity if $\lim_{x \to -\infty}\blue{f(x)}=\green a$ .

This intuitively means: when $x$ gets really small, $\blue{f(x)}$ gets arbitrarily close to value $\green a$ .

Example

$\blue{f(x)}=\blue{\frac{1}{x+1}}$

has a horizontal asymptote $y=\green0$

to infinity, because

$\lim_{x \to \infty} \blue{\frac{1}{x+1}}=\green0$

Example in a graph

In this figure we have plotted the graph $\blue{f(x)}=\blue{\frac{1}{x+1}}$ . We can see that the line $y=\green0$ is also the horizontal asymptote to minus infinity of $\blue{f(x)}$ . This is true, since $\lim_{x \to -\infty} \blue{\frac{1}{x+1}}=\green0$ .

Calculating asymptotes

Therefore, in order to determine whether a function has a horizontal asymptote to infinity, we calculate $\lim_{x \to \infty}\blue{f(x)}$ and in order to determine whether a function has a horizontal asymptote to minus infinity, we calculate $\lim_{x \to -\infty}\blue{f(x)}$ .

For this calculation, we can use the theory about The definition of a limit.

Given the function $f(x)={{-5 x^2+x+7}\over{6 x^2+6 x-3}}$ . Find the horizontal asymptotes in infinity and minus infinity of this function.

Write your answer in the form $y=a$ for a suitable number $a$ . If there is no horizontal asymptote, write "none".

This function has a horizontal asymptote $y=-{{5}\over{6}}$ in infinity and $y=-{{5}\over{6}}$ in minus infinity.

To find the horizontal asymptote of $f(x)$ in minus infinity, we calculate the following limit:

$\begin{array}{rcl} \displaystyle \lim_{x\rightarrow \, -\infty } f(x) &=& \displaystyle \lim_{x\rightarrow \, -\infty } {{-5 x^2+x+7}\over{6 x^2+6 x-3}}\\ && \qquad \blue{\text{function } f(x)=\frac{p(x)}{q(x)} \text{ substituted}}\\ &=& \displaystyle \lim_{x\rightarrow \, -\infty } \frac{x^2\cdot\left({{1}\over{x}}+{{7}\over{x^2}}-5\right)}{x^2\cdot\left({{6}\over{x}}-{{3}\over{x^2}}+6\right)}\\ && \qquad \blue{\text{the degree of } q(x) \text{ equals }2 \text{, so } x^2 \text{ is factored out}}\\ &=& \displaystyle \lim_{x\rightarrow \, -\infty } \frac{{{1}\over{x}}+{{7}\over{x^2}}-5}{{{6}\over{x}}-{{3}\over{x^2}}+6}\\ && \qquad \blue{\text{ simplified}}\\ &=& \displaystyle\frac{-5}{6}\\ && \qquad \blue{ \text{calculated}}\\ &=& \displaystyle -{{5}\over{6}}\\ && \qquad \blue{ \text{simplified}}\\ \end{array}$
The limit is $-{{5}\over{6}}$ , so the horizontal asymptote in minus infinity is $y=-{{5}\over{6}}$ .

Similarly, we find

$\displaystyle \lim_{x\rightarrow \, +\infty } f(x)= -{{5}\over{6}},$
so the horizontal asymptote in infinity is $y=-{{5}\over{6}}$ .

New example