Asymptotes, calculus lesson

Far from the origin, the graph of

y = \frac{2x^2 + 3x - 1}{x - 1}

looks almost exactly like a straight line, and near $x = 1$ it shoots off vertically. Asymptotes are the lines that describe this large scale shape. Each one is defined by a limit from the last few lessons, so finding them is mostly a matter of computing the right limits.

Key terms

Vertical asymptote

Definition The line $x = a$ is a vertical asymptote of the graph of $f$ if at least one of the following holds.

\lim_{x \to a^-} f(x) = \infty, \quad \lim_{x \to a^-} f(x) = -\infty, \quad \lim_{x \to a^+} f(x) = \infty, \quad \lim_{x \to a^+} f(x) = -\infty.

In plain words Near $x = a$ the graph runs up or down without bound, on at least one side.

Example For $f(x) = (x + 1)/(x - 2)$ the numerator tends to $3$ as $x \to 2$ while the denominator tends to $0$ , so $f(x) \to \infty$ as $x \to 2^+$ and $f(x) \to -\infty$ as $x \to 2^-$ . The line $x = 2$ is a vertical asymptote.

Horizontal asymptote

Definition The line $y = L$ is a horizontal asymptote of the graph of $f$ if

\lim_{x \to \infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L.

A graph has at most one horizontal asymptote in each direction, so at most two in total.

In plain words Far to the right, or far to the left, the graph levels off toward the height $L$ .

Example For $f(x) = x/\sqrt{x^2 + 1}$ we found $\lim_{x \to \infty} f(x) = 1$ and $\lim_{x \to -\infty} f(x) = -1$ in the lesson on limits at infinity. So the graph has two horizontal asymptotes, $y = 1$ on the right and $y = -1$ on the left.

Oblique asymptote

Definition The line $y = mx + b$ with $m \neq 0$ is an oblique (or slant) asymptote of the graph of $f$ if

\lim_{x \to \infty} \bigl(f(x) - (mx + b)\bigr) = 0 \quad \text{or} \quad \lim_{x \to -\infty} \bigl(f(x) - (mx + b)\bigr) = 0.

In plain words Far out, the graph hugs a tilted line.

Example For $x \neq 0$ the function

f(x) = \frac{x^2 + 1}{x} = x + \frac{1}{x}

satisfies $f(x) - x = 1/x \to 0$ as $x \to \pm\infty$ , so $y = x$ is an oblique asymptote in both directions.

Difference tending to zero

Definition Horizontal and oblique asymptotes are both cases of one idea. The graph of $f$ approaches the line $\ell(x) = mx + b$ as $x \to \infty$ exactly when $f(x) - \ell(x) \to 0$ . If such a line exists, then

m = \lim_{x \to \infty} \frac{f(x)}{x} \qquad \text{and} \qquad b = \lim_{x \to \infty} \bigl(f(x) - mx\bigr),

and conversely, if both limits exist as real numbers, then $y = mx + b$ is an asymptote as $x \to \infty$ . The case $m = 0$ gives a horizontal asymptote. The same holds with $x \to -\infty$ .

In plain words Find the slope by comparing $f(x)$ with $x$ , then find the intercept from what is left over. If the left over part settles down, you have the line.

Example Let $f(x) = \sqrt{x^2 + 2x}$ as $x \to \infty$ . For $x > 0$ the slope quotient is

\frac{f(x)}{x} = \sqrt{1 + \frac{2}{x}} \to 1,

so $m = 1$ . Then

f(x) - x = \frac{(x^2 + 2x) - x^2}{\sqrt{x^2 + 2x} + x} = \frac{2}{\sqrt{1 + \frac{2}{x}} + 1} \to 1,

so $b = 1$ and $y = x + 1$ is an asymptote. At $x = 100$ the gap $f(x) - (x + 1)$ is about $-0.00495$ .

Rational functions

Definition Let $r = p/q$ with $p$ and $q$ polynomials having no common factor of positive degree. Then

the line $x = a$ is a vertical asymptote exactly when $q(a) = 0$ ,
if $\deg p < \deg q$ , then $y = 0$ is the horizontal asymptote in both directions,
if $\deg p = \deg q$ , then $y = a_p / a_q$ is the horizontal asymptote in both directions, where $a_p$ and $a_q$ are the leading coefficients of $p$ and $q$ ,
if $\deg p = \deg q + 1$ , there is an oblique asymptote, found by polynomial division,
if $\deg p \ge \deg q + 2$ , there is no horizontal or oblique asymptote.

In plain words Cancel common factors first. Zeros of the denominator that survive give vertical asymptotes, and a cancelled zero only leaves a hole in the graph.

Example Factor the function

f(x) = \frac{x^2 - 1}{x^2 - x - 2} = \frac{(x - 1)(x + 1)}{(x - 2)(x + 1)}.

The factor $x + 1$ cancels, and $f(x) = (x - 1)/(x - 2)$ for $x \neq -1$ . So $x = 2$ is a vertical asymptote, while at $x = -1$ the graph has a hole at height $(-2)/(-3) = 2/3$ . The degrees are equal and both leading coefficients are $1$ , so $y = 1$ is the horizontal asymptote.

Polynomial division

Definition For polynomials $p$ and $q$ with $q$ not the zero polynomial, there are unique polynomials $s$ and $R$ with

p(x) = s(x) q(x) + R(x), \qquad \deg R < \deg q \text{ or } R = 0.

Then wherever $q(x) \neq 0$

\frac{p(x)}{q(x)} = s(x) + \frac{R(x)}{q(x)},

and the last term tends to $0$ as $x \to \pm\infty$ . When $s$ has degree $1$ , the line $y = s(x)$ is the oblique asymptote.

In plain words Divide until the remainder has lower degree than the divisor. The quotient is the line, and the remainder part fades away.

Example Dividing $2x^2 + 3x - 1$ by $x - 1$ gives

2x^2 + 3x - 1 = (x - 1)(2x + 5) + 4,

so for $x \neq 1$

\frac{2x^2 + 3x - 1}{x - 1} = 2x + 5 + \frac{4}{x - 1}.

The oblique asymptote is $y = 2x + 5$ , and $x = 1$ is a vertical asymptote.

Crossing an asymptote

Definition Horizontal and oblique asymptotes describe behavior only as $x \to \infty$ or $x \to -\infty$ , so the graph may cross them, even infinitely many times. A vertical asymptote $x = a$ is different. Since $f$ is a function, its graph meets the line $x = a$ in at most one point, namely $(a, f(a))$ when $a$ is in the domain, and otherwise not at all.

In plain words A graph can cut through a horizontal or slanted asymptote and still approach it later. Being an asymptote is about the far ends, not about avoiding the line. A vertical asymptote can be touched at most once.

Example The function

f(x) = \frac{x^2 + x}{x^2 + 1}

has the horizontal asymptote $y = 1$ . Setting $f(x) = 1$ gives $x^2 + x = x^2 + 1$ , so $x = 1$ , and the graph crosses its asymptote at $(1, 1)$ . The function $(\sin x)/x$ has the horizontal asymptote $y = 0$ and crosses it at every $x = k\pi$ with $k$ a nonzero integer. For a vertical asymptote, let $g(x) = 1/x$ for $x \neq 0$ and $g(0) = 0$ . The line $x = 0$ is still a vertical asymptote, and the graph meets it at the single point $(0, 0)$ .

One-sided conditions

Definition Each kind of asymptote needs only one side or one direction. A vertical asymptote at $x = a$ needs just one of the four one-sided infinite limits above, and a horizontal or oblique asymptote needs the limit as $x \to \infty$ or as $x \to -\infty$ , not both.

In plain words Check each side and each direction separately. The graph can behave quite differently on the other side.

Example For $f(x) = e^{1/x}$ , as $x \to 0^+$ the exponent $1/x \to \infty$ , so $f(x) \to \infty$ , while as $x \to 0^-$ the exponent tends to $-\infty$ and $f(x) \to 0$ . The line $x = 0$ is still a vertical asymptote. Similarly $e^x$ has the horizontal asymptote $y = 0$ only as $x \to -\infty$ .

Check yourself

Find all asymptotes of the graph of

f(x) = \frac{3x^2}{x^2 - 9}.

Show the answer

The denominator is $0$ at $x = \pm 3$ , where the numerator is $27 \neq 0$ , so $x = 3$ and $x = -3$ are vertical asymptotes. The degrees are equal, so $y = 3$ is the horizontal asymptote.

Find the oblique asymptote of the graph of

f(x) = \frac{x^3 + 1}{x^2 + 1}.

Show the answer

Polynomial division gives $x^3 + 1 = x(x^2 + 1) + (1 - x)$ , so

f(x) = x + \frac{1 - x}{x^2 + 1}.

The fraction tends to $0$ as $x \to \pm\infty$ , so the asymptote is $y = x$ in both directions.

Where does the graph of the function below cross its horizontal asymptote?

f(x) = \frac{2x^2 + x}{x^2 + 1}

Show the answer

The horizontal asymptote is $y = 2$ . Setting $2x^2 + x = 2(x^2 + 1)$ gives $x = 2$ , so the crossing point is $(2, 2)$ .

Asymptotes

Theory, part 25

Practice, part 26

Key terms

Vertical asymptote

Horizontal asymptote

Oblique asymptote

Difference tending to zero

Rational functions

Polynomial division

Crossing an asymptote

One-sided conditions

Check yourself

How this lesson connects

Builds on

Used later in