Horizontal Asymptotes and Vertical Asymptotes

What exactly are asymptotes? They are often mentioned in precalculus. But without a rigorous definition, you may have been left wondering. In this article, I go through, rigorously, exactly what horizontal asymptotes and vertical asymptotes are. I also illustrate them using graphs of functions.

We discuss limits that involve infinity in some way. First we study unbounded growth of functions using infinite limits and then the long term behavior of functions using limits at infinity. We also consider vertical asymptotes and horizontal asymptotes.

Infinite Limits

Infinite limits are used to described unbounded behavior of a function near a given real number which is not necessarily in the domain of the function. They are particularly useful for showing the intentions of the graph of a function by drawing dashed lines representing unbounded growth which are called vertical asymptotes.

Definition. Let $f$ be a function defined on both sides of $c,$ except possible at $c$ it self. Then $$ \lim_{x\to c}f(x)=+\infty $$ means that the values of $f(x)$ can be made arbitrarily large by taking $x$ sufficiently close to $c$ with $x\neq c$.

Definition. Let $f$ be a function defined on both sides of $c,$ except possible at $c$ itself. Then $$ \lim_{x\to c}f(x)=-\infty $$ means that the values of $f(x)$ can be made arbitrarily large (negatively) by taking $x$ sufficiently close to $c$ with $x\neq c$.

Example. Determine $\displaystyle \lim_{x\to 0}\frac{1}{x}.$

Solution. Since $f(x)=1/x$ decreases without bound as $x\to 0^-$ , we notice $\lim_{x\to 0^-}f(x)=-\infty.$ Also since $f$ increases without bound as $x\to 0^+,$ we notice $\lim_{x\to 0^+}f(x)=+\infty.$ Thus $\lim_{x\to0}f(x)$ does not exist.

Compare the previous example with the following.

Example. Determine $\displaystyle \lim_{x\to2} \frac{2x^3+x^2-16x+12}{x^2-4}.$

Solution. By factoring, \begin{align} \lim_{x\to 2}\frac{2x^3+x^2-16x+12}{x^2-4} & =\lim_{x\to 2}\frac{(x-2)\left(2x^2+5x-6\right)}{(x-2)(x+2)} \\ & =\lim_{x\to 2}\frac{\left(2x^2+5x-6\right)}{(x+2)} \\ & = \frac{\left(2(2)^2+5(2)-6\right)}{((2)+2)} \\ & =3. \end{align} as needed.

Theorem. Let $A$ be a positive real number.

$(1) \quad$ If $n$ is a positive even integer, then $\displaystyle\lim_{x\to c}\frac{A}{(x-c)^n}=+\infty.$

$(2) \quad$ If $n$ is a positive odd integer, then \begin{equation} \lim_{x\to c^+}\frac{A}{(x-c)^n}=+\infty \qquad \text{and} \qquad \lim_{x\to c^-}\frac{A}{(x-c)^n}=-\infty.
\end{equation}

Example. Evaluate $\displaystyle \lim_{x\to 2^+}\frac{-3}{\sqrt[3]{x-2}}.$

Solution. Notice that $ \frac{1}{\sqrt[3]{x-2}}$ increases without bound as $x\to 2^+$ and therefore, $$ \lim_{x\to 2^+}\frac{-3}{\sqrt[3]{x-2}}=-\infty. $$

Example. Evaluate $\displaystyle \lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{x^2}\right).$

Solution. Note that \begin{equation} f(x)=\frac{1}{x}-\frac{1}{x^2}=\frac{-1+x}{x^2} \end{equation} and that $\frac{-1+x}{x^2}$ decreases without bound as $x\to 0^+$ and therefore, \begin{equation} \lim_{x\to 0^+}\frac{-1+x}{x^2}=-\infty. \end{equation} This is true because when $x>0$ is close to $0$ we know that $-1+x$ is negative. Similarly, $$ \lim_{x\to 0^-}\frac{-1+x}{x^2}=-\infty $$ because when $x<0$ and close to $0$ we know that $-1+x$ is negative. Therefore it follows \begin{equation} \lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{x^2}\right)=-\infty \end{equation}

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Vertical Asymptotes

Definition. The line $x=c$ is called a vertical asymptote of the curve $y=f(x)$ if at least one of the following statements is true: \begin{alignat}{3} & \lim_{x\to c}f(x)=+\infty &\qquad & \lim_{x\to c^-}f(x)=+\infty &\qquad & \lim_{x\to c^+}f(x)=+\infty \\ & \lim_{x\to c}f(x)=-\infty &\qquad & \lim_{x\to c^-}f(x)=-\infty &\qquad & \lim_{x\to c^+}f(x)=-\infty
\end{alignat}

The following example demonstrates that not all rational functions have vertical asymptotes.

Example. Determine the vertical asymptotes of the function \begin{equation} g(x)=\frac{x^3}{x^2+3x+10}. \end{equation}

Solution. The function $g$ is continuous on its domain which is $\mathbb{R}$ and therefore there are no vertical asymptotes for this function.

Example. Determine the vertical asymptotes of the function \begin{equation} f(x)=\frac{x^3}{x^2+3x-10}. \end{equation}

Solution. Since \begin{equation} f(x)=\frac{x^3}{x^2+3x-10}=\frac{x^3}{(x-2)(x+5)} \end{equation} and therefore the vertical asymptotes are $x=2$ and $x=-5$ because \begin{equation} \lim_{x\to 2^+}\frac{x^3}{(x-2)(x+5)}=+\infty \end{equation} and \begin{equation} \lim_{x\to 5^+}\frac{x^3}{(x-2)(x+5)}=+\infty. \end{equation} as needed.

The following example demonstrates that there can be an unlimited number of vertical asymptotes for a function.

Example. Determine the vertical asymptotes of the function \begin{equation}
h(x)=\tan x-\cot x. \end{equation}

Solution. We can rewrite this function as \begin{align} h(x) &=\tan x-\cot x
\\ & =\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x} \\ & =\frac{\sin ^2x-\cos ^2x}{\sin x \cos x} \\& =\frac{\cos 2x}{\sin x \cos x}. \end{align} Therefore the zeros of the sine and cosine functions yield the vertical asymptotes of $x=\pm \frac{\pi }{2}+k$ for all integers $k,$ since $\cos 2x$ is not zero for these values of $x$ and because, for any $c=\pm \frac{\pi }{2}+k$ , $\lim_{x\to c}h(x)$ is either $\pm \infty.$

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Limits at Infinity

Definition. Let $f$ be a function defined on some interval $(a,+\infty ).$ Then \begin{equation}\lim_{x\to \infty }f(x)=L\end{equation} means that the values of $f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently large; or more precisely, for every $\epsilon>0$ there exists an $N$ such that if $x>N$ then $|f(x)-L|<\epsilon.$

Definition. Let $f$ be a function defined on some interval $(-\infty ,a).$ Then \begin{equation}\lim_{x\to -\infty }f(x)=L\end{equation} means that the values of $f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently large negative; or more precisely, for every$\epsilon >0$ there exists an $N$ such that if $x<N$ then $|f(x)-L|<\epsilon.$

The following limit rules are similar to the limit rules used for when $x\to c$ but instead use $x\to +\infty ;$ and they are also valid for when $x\to +\infty $ is replaced by $x\to -\infty.$

Theorem. If $\lim_{x\to \infty }f(x)$ and $\lim_{x\to \infty }g(x)$ exist, then

$(1)\quad \displaystyle \lim_{x\to \infty } k=k$ for any constant $k$

$(2)\quad \displaystyle \lim_{x\to \infty }k f(x)=k \lim_{x\to \infty }f(x)$

$(3)\quad \displaystyle \lim_{x\to \infty }[f(x)+g(x)]=\lim_{x\to \infty }f(x)+\lim_{x\to \infty }g(x)$

$(4)\quad \displaystyle \lim_{x\to \infty }[f(x)-g(x)]=\lim_{x\to \infty }f(x)-\lim_{x\to \infty }g(x)$

$(5)\quad \displaystyle \lim_{x\to \infty }[f(x)g(x)]=\left( \lim_{x\to \infty }f(x) \right)\left( \lim_{x\to \infty }g(x) \right)$

$(6)\quad \displaystyle \lim_{x\to \infty }[f(x)/g(x)]=\left( \lim_{x\to \infty }f(x) \right)/\left( \lim_{x\to \infty }g(x) \right)$

$(7)\quad \displaystyle \lim_{x\to \infty }[f(x)]^n=\left( \lim_{x\to \infty }f(x) \right){}^n$ where $n$ is a rational number and whenever the limits exist.

Theorem. Let $A$ be a real number.

$(1) \quad $ If $r>0$ is a rational number, then \begin{equation} \lim_{x\to \infty }\frac{A}{x^r}=0. \end{equation}

$(2) \quad $ If $r>0$ is a rational number with $x^r$ defined for all $x,$ then \begin{equation} \lim_{x\to -\infty }\frac{A}{x^r}=0. \end{equation}

Example. Evaluate $\displaystyle \lim_{x\to \infty }\frac{3x^2-x-2}{5x^2+4x+1}.$

Solution. To evaluate this limit we divide both the numerator and the denominator by the highest power of $x$ that occurs. So we have \begin{align}
\lim_{x\to \infty }\frac{3x^2-x-2}{5x^2+4x+1} & =\lim_{x\to \infty }\frac{\frac{3x^2}{x^2}-\frac{x}{x^2}-\frac{2}{x^2}}{\frac{5x^2}{x^2}+\frac{4x}{x^2}+\frac{1}{x^2}} \\ & =\lim_{x\to \infty }\frac{3-\frac{1}{x}-\frac{2}{x^2}}{5+\frac{4}{x}+\frac{1}{x^2}} \\ & =\frac{3-0-0}{5+0+0} \\ & =\frac{3}{5}. \end{align} as desired.

Example. Evaluate $\displaystyle \lim_{x\to \infty }\left(x-\sqrt{x^2+1}\right).$

Solution. We use the conjugate radical as follows \begin{align} \lim_{x\to \infty }\left(x-\sqrt{x^2+1}\right) & =\lim_{x\to \infty }\frac{\left(x-\sqrt{x^2+1}\right)\left(x+\sqrt{x^2+1}\right)}{\left(x+\sqrt{x^2+1}\right)} \\
& =\lim_{x\to \infty }\frac{x^2-\left(x^2+1\right)}{\left(x+\sqrt{x^2+1}\right)} \\ & = \lim_{x\to \infty }\frac{-1}{x+\sqrt{x^2+1}} \\ & =\lim_{x\to \infty }\frac{\frac{-1}{x}}{\frac{x}{x}+\sqrt{\frac{x^2+1}{x^2}}} \\ & =\frac{0}{1+\sqrt{1+0}} \\ & =0. \end{align} as desired.

Horizontal Asymptotes

In mathematics, the symbol $\infty$ is not a number, rather it is used to describe the process of unrestricted growth or the result of such a growth.

Definition. The line $y=L$ is called a horizontal asymptote of the curve $y=f(x)$ if either $\lim_{x\to \infty }f(x)=L$ or $\lim_{x\to -\infty }f(x)=L.$

Theorem. If $f$ is a rational functions of the form: \begin{equation} f(x)=\frac{a_0+a_1x+a_2x^2+\cdots +a_nx^n}{b_0+b_1x+b_2x^2+\cdots +b_mx^m} \end{equation} where $n$ is the degree of the polynomial in the numerator and $m$ is the degree of the polynomial in the denominator, then the horizontal asymptote of the curve $y=f(x)$ is determined by the following. If $n=m,$ then $y=\frac{a_n}{b_m}$ is the horizontal asymptote. If $n > m,$ then there is no horizontal asymptote, but rather a slant (oblique) asymptote and can be found be using long division.

Example. Find the horizontal asymptote of the graph of the function \begin{equation} f(x)=\frac{(1-x)(2+x)}{(1+2x)(2-3x)}. \end{equation}

Solution. The degree of the numerator $(1-x)(2+x)=-x^2-x+2$ is 2 and the degree of the denominator $(1+2x)(2-3x)=-6 x^2+x+2$ is 2 we divide both numerator and denominator by $x^2$ and using the properties of limits, we have \begin{align} \lim_{x\to \infty} f(x) & = \lim_{x\to \pm\infty} \frac{-x^2-x+2}{-6 x^2+x+2} \\ & = \lim_{x\to \pm\infty} \frac{\frac{-x^2-x+2}{x^2}}{\frac{-6 x^2+x+2}{x^2}} \\ & = \lim_{x\to \pm\infty} \frac{\frac{-x^2}{x^2}-\frac{x}{x^2}+\frac{2}{x^2}}{\frac{-6 x^2}{x^2}+\frac{x}{x^2}+\frac{2}{x^2}} \\ & = \frac{1}{6} \end{align} Therefore the only horizontal asymptote of the graph of $f(x)$ is $y=6.$

Example. Find the horizontal asymptote of the graph of the function \begin{equation} f(x)=\frac{\sqrt{2x^2+1}}{3x-5}. \end{equation}

Solution. Dividing both numerator and denominator by $x$ and using the properties of limits, we have \begin{align} \lim_{x\to \infty }\frac{\sqrt{2x^2+1}}{3x-5} & =\lim_{x\to \infty }\frac{\sqrt{2+\frac{1}{x^2}}}{3-\frac{5}{x}} \\ & =\frac{\sqrt{\lim_{x\to \infty }\left(2+\frac{1}{x^2}\right)}}{\lim_{x\to \infty }\left(3-\frac{5}{x}\right)} \\ & =\frac{\sqrt{2+0}}{3-5(0)} \\ & =\frac{\sqrt{2}}{3} \end{align} Therefore the line $y=\left.\sqrt{2}\right/3$ is a horizontal asymptote. It is also important to realize that \begin{align} \lim_{x\to -\infty }\frac{\sqrt{2x^2+1}}{3x-5} & =\lim_{x\to -\infty }\frac{\frac{1}{x}\sqrt{2x^2+1}}{3-\frac{5}{x}} \\ & =\lim_{x\to -\infty }\frac{\left(\frac{-1}{\sqrt{x^2}}\right)\sqrt{2x^2+1}}{3-\frac{5}{x}} \\ & =\lim_{x\to -\infty }\frac{-\sqrt{2+\frac{1}{x^2}}}{3-\frac{5}{x}} \\ & =-\frac{\sqrt{2}}{3} \end{align} Therefore, the line $y=\left.-\sqrt{2}\right/3$ is another horizontal asymptote.

Example. Find all horizontal asymptotes of the graph of the function \begin{equation} f(x)=\frac{x}{\sqrt[4]{x^4+1}}. \end{equation}

Solution. Dividing both numerator and denominator by $x$ and using the properties of limits, we have \begin{align} \lim_{x\to \infty }\frac{x}{\sqrt[4]{x^4+1}} & =\lim_{x\to \infty }\frac{\frac{x}{x}}{\frac{1}{x}\sqrt[4]{x^4+1}}
\\ & =\lim_{x\to \infty }\frac{1}{\sqrt[4]{\frac{x^4+1}{x^4}}} \\ & =\lim_{x\to \infty }\frac{1}{\sqrt[4]{1+\frac{1}{x^4}}} \\ & =\frac{1}{\sqrt[4]{\lim_{x\to \infty }\left(1+\frac{1}{x^4}\right)}} \\ & =\frac{1}{\sqrt[4]{1+0}} \\ & =1. \end{align} Therefore, the line $y=1$ is a horizontal asymptote. In computing the limit $x\to -\infty $ we must remember that for $x<0,$ we have $\sqrt[4]{x^4}=-x,$ so when we divide the numerator by $x,$ when $x<0$ we have, \begin{align} \lim_{x\to -\infty }\frac{x}{\sqrt[4]{x^4+1}} \\ & =\lim_{x\to -\infty }\frac{\frac{x}{x}}{\frac{1}{x}\sqrt[4]{x^4+1}} \\ & =\lim_{x\to \infty }\frac{1}{\frac{1}{\sqrt[4]{x^4}}\sqrt[4]{\frac{x^4+1}{x^4}}} \\ & =\lim_{x\to \infty }\frac{1}{-\text{ }\sqrt[4]{1+\frac{1}{x^4}}} \\ & =-1. \end{align} Therefore, the horizontal asymptotes are $y=\pm 1.$

Exercises on Asymptotes

Exercise. Evaluate the following limits.

$(1) \quad \displaystyle \lim_{x\to 0^-}\frac{5}{2x}$

$(2) \quad \displaystyle \lim_{x\to -8^+}\frac{2x}{x+8}$

$(3) \quad \displaystyle \lim_{x\to 0}\frac{-1}{x^2(x+1)}$

$(4) \quad \displaystyle \lim_{x\to 0}\frac{-1}{x^2(x-1)}$

$(5) \quad \displaystyle \lim_{x\to 0^+}\frac{2}{x^{1/5}}$

$(6) \quad \displaystyle \lim_{x\to \frac{\pi}{2}^-}\tan x$

Exercise. Find the limits for the following functions.

$(1) \quad \displaystyle f(x)=\frac{1}{x^2-4}$ as $x\to 2^+,$ $x\to 2^-,$ $x\to -2^+,$ and $x\to -2^-.$

$(2) \quad \displaystyle f(x)=\frac{x^2-1}{2x+4}$ as $x\to -2^+,$ $x\to -2^-,$ $x\to 1^+,$ and $x\to 0^-.$

$(3) \quad \displaystyle f(t)=2-\frac{3}{t^{1/3}}$ as $t\to 0^+,$ and $t\to 0^-.$

$(4) \quad \displaystyle f(x)=\frac{1}{x^{1/3}}-\frac{1}{(x-1)^{4/3}}$ as $x\to 0^+,$ $x\to 0^-,$ $x\to 1^+,$ and $x\to 1^-.$

Exercise. Graph the following rational functions and include the graphs and equations of their asymptotes.

$(1) \quad \displaystyle f(x)=\frac{-3}{x-3}$

$(2) \quad \displaystyle f(x)=\frac{x^2}{x-1}$

$(3) \quad \displaystyle f(x)=\frac{x^2-1}{2x+4}$

$(4) \quad \displaystyle f(x)=\frac{x^3-x^2+x-1}{(2x+3)(2x-1)}$

Exercise. For each of the following sketch the graph of a function $y=f(x)$ that satisfies the conditions.

$(1) \quad f(0)=0,$ $f(1)=2,$ $f(-1)=-2,$ $\lim_{x\to -\infty }f(x)=-1,$ and $\lim_{x\to \infty }f(x)=1$

$(2) \quad f(2)=1,$ $f(-1)=0,$ $\lim_{x\to \infty }f(x)=0,$ $\lim_{x\to0^+}f(x)=\infty ,$ $\lim_{x\to 0^-}f(x)=-\infty ,$ and $\lim_{x\to -\infty }f(x)=1$

$(3) \quad \lim_{x\to \pm \infty }g(x)=0,$ $\lim_{x\to 3^-}g(x)=-\infty,$ and $\lim_{x\to 3^+}g(x)=+\infty $

Exercise. Find the following one-sided limits algebraically.

$(1) \quad \displaystyle \lim_{x\to 2^+} \frac{x(2x+5)}{(x+1)(x^2+x)}$

$(2) \quad \displaystyle \lim_{h\to 0^+}\frac{\sqrt{h^2+4h+5}-\sqrt{5}}{h}$

$(3) \quad \displaystyle \lim_{x\to 1^+}\frac{\sqrt{2x}(x-1)}{|x-1|}$

Exercise. Find the following two-sided limits.

$(1) \quad \displaystyle \lim_{t\to 0}\left(\frac{\sin k t}{t}\right)$

$(2) \quad \displaystyle \lim_{x\to 0}\left(\frac{\tan 2x}{x}\right)$

$(3) \quad \displaystyle \lim_{x\to 0}\left(6x^2(\cot x)(\csc 2x)\right)$

$(4) \quad \displaystyle \lim_{x\to 0}\left(6x^2(\tan x)(\sec 2x)\right)$

$(5) \quad \displaystyle \lim_{h\to 0}\left(\frac{\sin (\sin h)}{\sin h}\right)$

$(6) \quad \displaystyle \lim_{\theta \to 0}\left(\frac{\sin \theta }{\sin 2\theta }\right)$

Exercise. Find the limits for the following functions for both $x\to +\infty $ and $x\to -\infty $.

$(1) \quad \displaystyle f(x)=\pi -\frac{2}{x^2}$

$(2) \quad \displaystyle f(x)=\frac{-5+\left(\frac{7}{x}\right)}{3-\left(\frac{1}{x}\right)}$

$(3) \quad \displaystyle f(\theta )=\frac{\cos \theta }{3 \theta }$

$(4) \quad \displaystyle f(x)=e^{-x}\sin x$

$(5) \quad \displaystyle f(x)=\frac{2x+3}{5x+7}$

$(6) \quad \displaystyle f(x)=\frac{x+1}{x^2+3}$

$(7) \quad \displaystyle f(x)=\frac{9x^4+x}{2x^4+5x^2-x+6}$

$(8) \quad \displaystyle f(x)=\frac{2\sqrt{x}+x^{-1}}{3x-7}$

$(9) \quad \displaystyle f(x)=\frac{x^{-1}+x^{-4}}{x^{-2}-x^{-3}}$

David A. Smith at Dave4Math

David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.

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