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.

Table of Contents

## 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.

### 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:

- Factor \(P(x)\) and \(Q(x)\) if you can.
- Cancel any common factor to simplify the fraction.
- Take the denominator of the simplified fraction and set it equal to zero. All the zeros of the denominator are where there are vertical asymptotes.

##### 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 \(f(x)\):

- Find the zeros of \(h(x)\).
- If \(h(a)=0\) and \(g(a)\neq0\) then \(x=a\) is a vertical asymptote
- If \(h(a)=0\) and \(g(a)=0\), then find \[\lim_{x\to a^{+}}f(x),\quad\text{or}\quad\lim_{x\to a^{-}}f(x).\] If one of the above limits is \(+\infty\) or \(-\infty\), then \(x=a\) is a vertical asymptote.

- 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 \(h(x)\to h(a)\)]

##### 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 \[x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2},\dots\]

(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 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 the 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.

## Oblique (also known as inclined or slant) asymptotes

We say a line \(L\) is the 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 \(f(x)\).

- If \[f(x)=\frac{P_{1}(x)}{P_{2}(x)}\] where \(P_{1}(x)\) and \(P_{2}(x)\) are two polynomials 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 \(P_{2}(x)\).

- 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.

#### Read why the above formulas work

#### Hide the explanation

- If \(y=mx+b\) is the oblique asymptote of \(f(x)\) as \(x\to+\infty\), we have

\begin{align}\lim_{x\to+\infty}\left(f(x)-(mx+b)\right)=0\tag{i}\end{align}

or

\[\begin{align}\lim_{x\to+\infty}x\left[\frac{f(x)}{x}-m-\frac{b}{x}\right]=0.\tag{ii}\end{align}\]

- Since \(\lim_{x\to+\infty}x=+\infty\), we must have

\[\begin{align}\lim_{x\to+\infty}\left[\frac{f(x)}{x}-m-\frac{b} {x}\right]=0.\tag{iii}\end{align}\]

otherwise, by Theorem the limit in (ii) would be infinite. Because \[\lim_{x\to+\infty}\left[\frac{f(x)}{x}-m-\frac{b}{x}\right]=\lim_{x\to+\infty}\frac{f(x)}{x}-\lim_{x\to+\infty}m-\lim_{x\to+\infty}\frac{b}{x}\]

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}\] \[\Rightarrow\lim_{x\to+\infty}[f(x)-mx]=b.\]