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)={{2 x^2+5 x+9}\over{4 x^2-6 x}}#. 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={{1}\over{2}}# in infinity and #y={{1}\over{2}}# in minus infinity.

To find the horizontal asymptote of #f(x)# in infinity, we calculate the following limit:
\[\begin{array}{rcl}
\displaystyle \lim_{x\rightarrow \, \infty } f(x)
&=& \displaystyle \lim_{x\rightarrow \, \infty } {{2 x^2+5 x+9}\over{4 x^2-6 x}}\\
&& \qquad \blue{\text{function } f(x)=\frac{p(x)}{q(x)} \text{ substituted}}\\
&=& \displaystyle \lim_{x\rightarrow \, \infty } \frac{x^2\cdot\left({{5}\over{x}}+{{9}\over{x^2}}+2\right)}{x^2\cdot\left(4-{{6}\over{x}}\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{{{5}\over{x}}+{{9}\over{x^2}}+2}{4-{{6}\over{x}}}\\
&& \qquad \blue{\text{ simplified}}\\
&=& \displaystyle\frac{2}{4}\\
&& \qquad \blue{ \text{calculated}}\\

&=& \displaystyle {{1}\over{2}}\\
&& \qquad \blue{ \text{simplified}}\\
\end{array}\]
The limit is #{{1}\over{2}}# , so the horizontal asymptote in infinity is #y={{1}\over{2}}#.

Similarly, we find
\[\displaystyle \lim_{x\rightarrow \, -\infty } f(x)= {{1}\over{2}},\]
so the horizontal asymptote in minus infinity is #y={{1}\over{2}}#.

New example