We say a line is the asymptote of a curve if the distance between the line and curve approaches zero as the curve (specifically the \(x\) or \(y\) coordinate of the points on the curve) goes to \(+\infty\) or \(-\infty\).
We study three types of asymptotes: (1) vertical, (2) horizontal, and (3) oblique (or inclined or slant).
In this section, we learn how to find them.
Vertical asymptotes
The curve has a vertical asymptote \(x=a\) when:
as \(x\) approaches \(a\) (from the right or left or both directions), the distance between the points on the graph of \(y=f(x)\) and the vertical line \(x=a\) gets smaller and smaller but they never reach the line \(x=a\) (Figure 2).
More precisely:
Definition 1. We say \(x=a\) is a vertical asymptote of (the graph of) the function \(y=f(x)\), if one of the following conditions holds:
(a) \(f(x)\to+\infty\) as \(x\to a\) or \(x\to a^{+}\) or \(x\to a^{-}\)
(b) \(f(x)\to-\infty\) as \(x\to a\) or \(x\to a^{+}\) or \(x\to a^{-}\)
When do we have vertical asymptotes? There are two cases that we need to pay attention
- There is a fraction \[f(x)=\frac{g(x)}{h(x)}\quad\text{and}\quad\lim_{x\to a}h(x)=0\] If \[\lim_{x\to a}g(x)\neq0\] then \(x=a\) is a vertical asymptote, because it follows from Theorem 1 in Section 4.6 that \[\lim_{x\to a}f(x)=+\infty\text{ or }-\infty.\] However, if \({\displaystyle \lim_{x\to a}h(x)=\lim_{x\to a}}g(x)=0\), then \(f(x)\) may or may not approach \(+\infty\) or \(-\infty\).
- Logarithmic functions. For example, because \[\lim_{x\to0^{+}}\ln x=-\infty\] the vertical line \(x=0\) is a vertical asymptote of the graph of \(y=\ln x\). See Figure 3.
Figure 3. Graph of \(y=\ln x\).
The function is a fraction f(x) = g(x)/h(x)
Let’s start with simple situations:
Rational functions
If \(f(x)=\frac{P(x)}{Q(x)}\) where \(P(x)\) and \(Q(x)\) are polynomials, you should follow these steps:
(a) Let’s factor out the numerator and denominator: \[4-2x=2(2-x)\] To factor out the denominator, we can find the zeros: \(\underbrace{2}_{a}x^{2}+\underbrace{1}_{b}x+\underbrace{(-1)}_{c}=0\) \[x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-1\pm\sqrt{1^{2}-4\cdot2\cdot(-1)}}{2\cdot2}\] (b) To factor out the numerator and denominator, we can use the quadratic formula (\(-b\pm\sqrt{b^{2}-4ac})/(2a)\), or in this case to simply guess them. For the numerator, we need two numbers that their sum is 4 and their product is 3. It is easy to guess that they are 3 and thus \[x^{2}+4x+3=(x+3)(x+1)\] For the denominator, we are looking for two numbers that their sum is 7 and their product is 12. Again, it is easy to guess that they are 4 and 3. Thus \[x^{2}+7x+12=(x+3)(x+4)\] and \[\frac{x^{2}+4x+3}{x^{2}+7x+12}=\frac{(x+3)(x+1)}{(x+4)(x+3)}=\frac{x+1}{x+4}\quad(\text{if }x\neq-3)\] The zero of the simplified fraction is \(x=-4\). Therefore, \(x=-4\) is the only vertical asymptote. Note that although \(x=-4\) and \(x=-3\) both make the denominator of the original zero, but \(x=-3\) is not a vertical asymptote because \[\lim_{x\to-3}g(x)=\lim_{x\to-3}\frac{x+1}{x+4}=\frac{-3+1}{-3+4}=-2,\] and \(x=-4\) is a vertical asymptote because \[\lim_{x\to-4^{-}}g(x)=\lim_{x\to-4^{-}}\frac{x+1}{x+4}\overset{\left[\frac{-3}{0^{-}}\right]}{=}+\infty\] and \[\lim_{x\to-4^{+}}g(x)=\lim_{x\to-4^{+}}\frac{x+1}{x+4}\overset{\left[\frac{-3}{0^{+}}\right]}{=}-\infty\] The graph of \(g\) is shown in Figure 5. (c) We cannot factor the denominator and numerator and the denominator is never zero, because \[x^{2}+1\geq1>0\] Therefore, the graph of \(h(x)\) does not have any vertical asymptotes. See Figure 6.
\[x=-1\qquad x=-\frac{1}{2}\] [In fact, because \(a+c=b=1\), we already knew that \(x=-1\) is one of the zeros. See Section on Solutions and Roots. Therefore, \[\frac{4-2x}{2x^{2}+x-1}=\frac{-2(x-2)}{2(x+1)(x+\frac{1}{2})}\] We cannot simplify this fraction further. Thus, the zeros of the denominator \(x=-1\)) and \(x=-1/2\) are the vertical asymptotes (see Figure 4).
[Whenever in doubt about the sign of the denominator as \(x\) approaches \(a\) from the left or right, test it for a number on the correct side of \(a\). For example, as \(x\to-4^{-}\), we test for a number on the left of \(-4\), say \(x=-4.5\). In this case \(x+4=-4.5+4=-0.5<0\). And as \(x\to-4^{+}\), we test for a number on the right of \(-4\), say \(x=-3.9\). In this case \(x+4=-3.9+4=0.1>0\)]
Fractions in General
We learned how to find the vertical asymptotes of rational functions. More generally, if \(f(x)\) is a fraction, every zero (= every root) of the denominator can be potentially where a vertical asymptote occurs.
In other words, if \(f(x)=\frac{g(x)}{h(x)}\) and \(h(a)=0\), then we should test \(x=a\) and see if \(f(x)\) goes to \(+\infty\) or \(-\infty\) as \(x\) approaches \(a\) (from either side).
If \(f(x)=\dfrac{g(x)}{h(x)}\), to find the vertical asymptotes of the graph of \(f(x)\):
- In above, we assumed \(g(x)\) and \(h(x)\) are continuous functions on their domains. Almost all functions that we know are continuous except the greatest integer function \(y=\left\lfloor x\right\rfloor\) (also denoted by \(y=[\![ x]\!]\)) and some piecewise-defined functions.
- Note that if \(f(x)=g(x)/h(x)\) and \(h(a)=0\) but \(g(a)\neq0\), then \(x=a\) is a vertical asymptote because as \(x\) approaches \(a\) (from right or left) by Theorem 1 in Section 4.6 \[f(x)\overset{\left[\frac{g(a)(\neq0)}{0}\right]}{\longrightarrow}+\infty\text{ or }-\infty\] [Again, here we assumed \(h(x)\) and \(g(x)\) are continuous functions on their domains. Therefore, as \(x\) approaches \(a\) (from right or left), \(g(x)\to g(a)\) and \(h(x)\to h(a)\)]
The denominator \(\sqrt{4-x^{2}}\) is zero when \(x=\pm2\), so we need to find the limit of \(f\) as \(x\) approaches \(-2\) and 2. The domain of \(f\) is the set of all \(x\) such that the expression under the square root is nonnegative \(4-x^{2}\geq0\) and the denominator is not zero \(4-x^{2}\neq0\). That is \[\begin{aligned} On the other hand, as \(x\to-2^{+}\), the denominator goes to 0 through positive numbers [recall that the square root is always nonnegative] and the numerator goes to \(-4\). Therefore by part (c) of Theorem 1 in Section 4.6 \(f(x)\) goes to \(-\infty\) [Symbolically \(\frac{-4}{0^{+}}=-\infty\) or \(\frac{(-)}{(+)}=(-)\)]. Therefore, \(x=-2\) is a vertical asymptote of the graph of \(f\). The graph of \(f\) is shown in Figure 7.
Dom(f) & =\{x|\ 4-x^{2}>0\}\\
& =\{x|\ x^{2}<4\}\\
& =\{x|\ -2<x<2\}\\
& =(-2,2).\end{aligned}\] [Recall that \(x^{2}<a^{2}\) is equivalent to \(-|a|<x<|a|\)] Therefore \(x\) can approach \(2\) only from the left and \(-2\) only from the right. \[\begin{aligned}
\lim_{x\to2^{-}}f(x) & =\lim_{x\to2^{-}}\frac{x-2}{\sqrt{4-x^{2}}}\\
& =\lim_{x\to2^{-}}\frac{x-2}{\sqrt{(2-x)(2+x)}}\\
& =\lim_{x\to2^{-}}\frac{-(2-x)}{\sqrt{2-x}\sqrt{2+x}}\\
& =\lim_{x\to2^{-}}\frac{-\cancel{\sqrt{2-x}}\sqrt{2-x}}{\cancel{\sqrt{2-x}}\sqrt{2+x}}\\
& =\lim_{x\to2^{-}}\frac{-\sqrt{2-x}}{\sqrt{2+x}}\\
& =-\frac{\sqrt{2-2}}{\sqrt{2+2}}\\
& =0.\end{aligned}\] Therefore \(x=2\) is not a vertical asymptote of the graph of \(f\).
Trigonometric functions
- Recall that the graph of cosine is obtained by shifting the graph of sine to the left \(\pi/2\), so if \(\cos\alpha=0\), we know \(\sin\alpha\neq0\) and if \(\sin\beta=0\), we know \(\cos\beta\neq0\).
Because \[\tan x=\frac{\sin x}{\cos x},\quad\text{and}\quad\sec x=\frac{1}{\cos x}\] their graphs have vertical asymptotes wherever \(\cos x=0\) (see Figure 8); that is, at \[x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2},\dots\]
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| (a) | (b) |
| Figure 8. Wherever cosine is zero, tangent and secant have a vertical asymptote. | |
Similarly because \[\cot x=\frac{\cos x}{\sin x},\quad\text{and}\quad\csc x=\frac{1}{\sin x}\] their graphs have vertical asymptotes wherever \(\sin x=0\); that is, at \[x=0,\pm\pi,\pm2\pi,\pm3\pi,\dots\]
Logarithmic Functions
If \(f(x)=\ln(g(x))\) and if \(\lim_{x\to a}g(x)=0\), then \(x=a\) is a vertical asymptote to the graph of \(f\). This is also true if \(g\to0\) as \(x\to a^{+}\) or \(x\to a^{-}\).
- Because the input of the logarithm has to be positive, in the above, we have to assume \(g(x)\) approaches 0 through positive values as \(x\to a^{+}\) or \(x\to a^{-}\).
(a) Method 1: Because \(x-2=0\) for \(x=2\) [If \(x>2\), \(x-2>0\). Therefore \(F(x)\) is defined for \(x>2\)], the graph of \(F\) has a vertical asymptote at \(x=2\). (b) Because \(\sqrt{x^{2}-9}\) is zero when \(x=\pm3\), the graph of \(g\) has vertical asymptotes at \(x=\pm3\). Note that because the domain of \(g\) is \(\{x|\ x<-3\text{ or }x>3\}=(-\infty,-3)\cup(3,\infty)\), \(x\) can approach \(-3\) only from left and 3 only from the right. The graph of \(g\) is shown in Figure 10.
Method 2: Recall that if the graph of \(y=f(x)\) is given, the graph of \(y=f(x-a)\) (\(a>0\)) and is obtained by shifting the graph of \(f\) to the right \(a\) units. Because the graph of \(y=\ln(x-2)\) is obtained by shifting the graph of \(y=\ln x\) to the right 2 units, and the graph of \(y=\ln x\) has a vertical asymptote at \(x=0\), the graph of \(y=\ln(x-2)\) has a vertical asymptote at \(x=2\). See Figure 9.
- A function may have any number of vertical asymptotes or none at all. For example, \(f(x)=\tan x\) has infinitely many vertical asymptotes at \(x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2},\cdots\); \(g(x)=\frac{1}{x-1}\) has only one asymptote \(x=1\), and \(F(x)=x^{3}\) has zero vertical asymptotes.
Horizontal asymptotes
We say a horizontal line is an asymptote of a curve if the distance between the points on the curve and this line gets smaller and smaller as \(x\) goes to \(+\infty\) or \(-\infty\) (see Figure 11)
More precisely
Definition 2. We say the horizontal line \(y=b\) is a horizontal asymptote of (the graph of) \(f(x)\), if one of the following conditions holds: (a) \(\lim_{x\to+\infty}f(x)=b\) (b) \(\lim_{x\to-\infty}f(x)=b\)
- A function has, at most, two horizontal asymptotes, but a function can have infinitely many vertical asymptotes. For example, \(f(x)=\tan x\) has infinitely many vertical asymptotes.
Because \[\lim_{x\to\pm\infty}\frac{2x^{2}-3}{x^{2}+1}=\lim_{x\to\pm\infty}\frac{2x^{2}}{x^{2}}=2,\] its graph has only one asymptote, namely the line \(y=2\), as shown in Figure 12. 
\[\begin{aligned}
\lim_{x\to+\infty}f(x) & =\lim_{x\to+\infty}\frac{1-x}{|x|+3}\\
& =\lim_{x\to+\infty}\frac{1-x}{x+3}\\
& =\lim_{x\to+\infty}\frac{-x}{x}\\
& =-1\end{aligned}\] since \(|x|=x\) if \(x>0\), while \[\begin{aligned}
\lim_{x\to+\infty}f(x) & =\lim_{x\to+\infty}\frac{1-x}{|x|+3}\\
& =\lim_{x\to+\infty}\frac{1-x}{-x+3}\\
& =\lim_{x\to+\infty}\frac{-x}{-x}\\
& =1\end{aligned}\] since \(|x|=-x\) if \(x<0\). Hence, \(f\) has two horizontal asymptotes, namely \(y=-1\) and \(y=1\), as shown in Figure 13.
Recall that \(\lim_{x\to+\infty}e^{-x}=0\) (Important Limits at Infinity in Section 4.7)
\[\begin{aligned}
\lim_{x\to+\infty}\frac{1}{e^{-x}+1} &
=\frac{\lim_{x\to+\infty}1}{\lim_{x\to+\infty}e^{-x}+\lim_{x\to+\infty}1}\\
& =\frac{1}{0+1}\\
& =1.\end{aligned}\] Since \(\lim_{x\to-\infty}e^{-x}=+\infty\), \(\lim_{x\to-\infty}(e^{-x}+1)\overset{[\infty+1=\infty]}{=}\infty\) \[\begin{aligned}
\lim_{x\to+\infty}\frac{1}{e^{-x}+1} & =\frac{\lim_{x\to+\infty}1}{\lim_{x\to+\infty}(e^{-x}+1)}\\
& \overset{\left[\frac{1}{\infty}=0\right]}{=}0.\end{aligned}\] Hence, the graph of \(g\) has two asymptotes, namely \(y=1\) and \(y=0\), as shown in Figure 14.
Oblique (also known as inclined or slant) asymptotes
We say a line \(L\) is an oblique asymptote of \(f(x)\) if the graph of \(f(x)\) gets closer and closer to this line as \(x\to+\infty\) or \(x\to-\infty\) (Figure 15).
More precisely
Definition 3. We say a line with equation \(y=mx+b\) (\(m\neq0\)) is the oblique asymptote (of the graph) of \(f\) if
(a) \({\displaystyle \lim_{x\to+\infty}\left[f(x)-(mx+b)\right]=0}\) or
(b) \({\displaystyle \lim_{x\to-\infty}\left[f(x)-(mx+b)\right]=0}\)
- The distance between a point on the curve \(P\left(x,f(x)\right)\) and the line \(y=mx+b\) is \[d=\frac{|f(x)-(mx+b)|}{\sqrt{m^{2}+1}}.\] But \(d\to0\) as \(x\to+\infty\) (or \(-\infty\)) if and only if \[f(x)-(mx+b)\to0\] as \(x\to+\infty\) (or \(-\infty\)).
- Consider a function of the form \[f(x)=mx+b+\frac{P(x)}{Q(x)},\] where \(P(x)\) and \(Q(x)\) are two polynomials such that \[\text{degree of }P(x)<\text{degree of }Q(x).\] As we explained in Section on the Other Indeterminate Forms \[\frac{P(x)}{Q(x)}\to0\quad\text{as }x\to\pm\infty.\] Therefore \[f(x)-(mx+b)=\frac{P(x)}{Q(x)}\to0\quad\text{as }x\to\pm\infty,\] which shows the line \(y=mx+b\) is the oblique asymptote of \(f(x)\).
Because \[\lim_{x\to\pm\infty}\frac{3x+7}{x^{4}-2x+2}=\lim_{x\to\pm\infty}\frac{3x}{x^{4}}=0,\]
\[\lim_{x\to\pm\infty}\left[f(x)-(-3x+1)\right]=0\] and therefore, the line \(y=-3x+1\) is the oblique asymptote of \(f(x)\), as shown in Figure 16.
- If \[f(x)=\frac{P_{1}(x)}{P_{2}(x)}\] where \(P_{1}(x)\) and \(P_{2}(x)\) are two polynomials without any factors in common such that \[\text{degree of }P_{1}(x)=\text{degree of }P_{2}(x)+1\] then \(f(x)\) has an oblique asymptote which is the quotient of dividing \(P_{1}(x)\) by \(P_{2}(x)\).
Let’s divide the numerator by the denominator: Thus \[\frac{3x^{2}+5x-3}{x+1}=3x+2+\frac{-5}{x+1}\] Because \[\frac{-5}{x+1}\to0\quad\text{as}\quad x\to\pm\infty\] the line \(y=3x+2\) is the oblique asymptote of the graph of \(f\), as shown in Figure 17. Also notice that the graph of \(f\) has a vertical asymptote at \(x=-1\). 
- More generally, when \[f(x)=mx+b+g(x)\] where \(g(x)\to0\) as \(x\to+\infty\) (or \(-\infty)\), the line \(y=mx+b\) is an oblique asymptote of \(f(x)\).
How to determine the oblique asymptotes
If the line \(y=mx+b\) is the oblique asymptote of \(f(x)\) as \(x\to+\infty\) then \[\bbox[#F2F2F2,5px,border:2px solid black]{m=\lim_{x\to+\infty}\frac{f(x)}{x}}\] and \[\bbox[#F2F2F2,5px,border:2px solid black]{b=\lim_{x\to+\infty}\left[f(x)-mx\right].}\] If this line is the oblique asymptote of \(f(x)\) as \(x\to-\infty\), it is sufficient to replace \(+\infty\) by \(-\infty\) in the above equations.
\begin{align}\lim_{x\to+\infty}\left(f(x)-(mx+b)\right)=0\tag{i}\end{align} or \[\begin{align}\lim_{x\to+\infty}x\left[\frac{f(x)}{x}-m-\frac{b}{x}\right]=0.\tag{ii}\end{align}\]
\[\begin{align}\lim_{x\to+\infty}\left[\frac{f(x)}{x}-m-\frac{b} {x}\right]=0.\tag{iii}\end{align}\]
otherwise, by Theorem the limit in (ii) would be infinite. Because \[\lim_{x\to+\infty}\left[\frac{f(x)}{x}-m-\frac{b}{x}\right]=\lim_{x\to+\infty}\frac{f(x)}{x}-\lim_{x\to+\infty}m-\lim_{x\to+\infty}\frac{b}{x}\]
The slope of the oblique asymptote as \(x\to+\infty\) is \[\begin{aligned} The slope of the oblique asymptote as \(x\to-\infty\) is obtained similarly: \[\begin{aligned} Read why the above formulas work
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and \(b/x\to0\) as \(x\to+\infty\) and \(m\) is a constant, we have \[\lim_{x\to+\infty}\frac{f(x)}{x}-m=0.\] Hence (iii) implies \[\lim_{x\to+\infty}\frac{f(x)}{x}=m.\] After finding \(m\), we may use (i) to obtain \(b\): \[\begin{aligned}
\lim_{x\to+\infty}[f(x)-mx-b] & =0\\
\lim_{x\to+\infty}[f(x)-mx]-\lim_{x\to+\infty}b & =0\\
\lim_{x\to+\infty}[f(x)-mx]-b & =0\end{aligned}\] \[\Rightarrow\lim_{x\to+\infty}[f(x)-mx]=b.\]
m_{1} & =\lim_{x\to+\infty}\frac{f(x)}{x}\\
& =\lim_{x\to+\infty}\frac{\sqrt{2x^{2}-1}-7x}{x}\\
& =\lim_{x\to+\infty}\frac{\sqrt{x^{2}\left(2-\frac{1}{x^{2}}\right)}-7x}{x}\\
& =\lim_{x\to+\infty}\frac{|x|\sqrt{\left(2-\frac{1}{x^{2}}\right)}-7x}{x}\end{aligned}\] Since \(|x|=x\) when \(x>0\), we have \[\begin{aligned}
m_{1} & =\lim_{x\to+\infty}\frac{\bcancel{x}\sqrt{\left(2-\frac{1}{x^{2}}\right)}-7\bcancel{x}}{\bcancel{x}}\\
& =\lim_{x\to+\infty}\left(\sqrt{2-\frac{1}{x^{2}}}-7\right)\\
& =\sqrt{2-\lim_{x\to+\infty}\frac{1}{x^{2}}}-7\\
& =\sqrt{2-0}-7\\
& =\sqrt{2}-7\end{aligned}\] The \(y\)-intercept of the asymptote is
\[\begin{aligned}
b_{1} & =\lim_{x\to+\infty}[f(x)-(\sqrt{2}-7)x]\\
& =\lim_{x\to+\infty}\left[\left(\sqrt{2x^{2}-1}-7x\right)-(\sqrt{2}-7)x\right]\\
& =\lim_{x\to+\infty}\left[\sqrt{2x^{2}-1}-\sqrt{2}x\right]\end{aligned}\] Multiplying and dividing by
the conjugate of \(\sqrt{2x^{2}-1}-\sqrt{2}\) \[\begin{aligned}
b_{1} &
=\lim_{x\to+\infty}\frac{\left(\sqrt{2x^{2}-1}-\sqrt{2}x\right)\left(\sqrt{2x^{2}-1}+\sqrt{2}x\right)}{\left(\sqrt{2x^{2}-1}+\sqrt{2}x\right)}\\
& =\lim_{x\to+\infty}\frac{(2x^{2}-1)-\left(\sqrt{2}x\right)^{2}}{\sqrt{2x^{2}-1}+\sqrt{2}x}\\
& =\lim_{x\to+\infty}\frac{-1}{\sqrt{2x^{2}-1}+\sqrt{2}x}\\
& \overset{\left[\frac{1}{\infty}\right]=0}{=}0.\end{aligned}\] Hence, \(y=(\sqrt{2}-7)x+0\) is the oblique asymptote of \(f(x)\) as \(x\to+\infty\).
m_{2} & =\lim_{x\to-\infty}\frac{f(x)}{x}\\
& =\lim_{x\to-\infty}\frac{\sqrt{2x^{2}-1}-7x}{x}\\
& =\lim_{x\to-\infty}\frac{\sqrt{x^{2}\left(2-\frac{1}{x^{2}}\right)}-7x}{x}\\
& =\lim_{x\to-\infty}\frac{|x|\sqrt{\left(2-\frac{1}{x^{2}}\right)}-7x}{x}\end{aligned}\] Since \(|x|=-x\) when \(x<0\), we have \[\begin{aligned}
m_{2} &
=\lim_{x\to-\infty}\frac{-\bcancel{x}\sqrt{\left(2-\frac{1}{x^{2}}\right)}-7\bcancel{x}}{\bcancel{x}}\\
& =\lim_{x\to-\infty}\left(-\sqrt{2-\frac{1}{x^{2}}}-7\right)\\
& =-\sqrt{2-\lim_{x\to-\infty}\frac{1}{x^{2}}}-7\\
& =\sqrt{2-0}-7\\
& =-\sqrt{2}-7\end{aligned}\] In this case, we can show the \(y\)-intercept is also zero. That is, \(y=-(\sqrt{2}+7)x\) is the oblique asymptote of \(f(x)\) as \(x\to-\infty\). The two oblique asymptotes of \(f(x)\) are shown in Figure 18.