Asymptotes

We say a line is an asymptote of a curve if the distance between the line and the curve approaches zero as the curve (specifically the x or y coordinates 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).

Quick Reference

Asymptote Type	Equation	Defining Condition	Typical Occurrence
Vertical	$x = a$	$lim_{x \to a^{-}} f (x) = \pm \infty$ or $lim_{x \to a^{+}} f (x) = \pm \infty$	Denominator equals $0$ with non-zero numerator; logarithmic/trig functions.
Horizontal	$y = b$	$lim_{x \to + \infty} f (x) = b$ or $lim_{x \to - \infty} f (x) = b$	Degree of numerator $\leq$ degree of denominator; exponential functions.
Oblique (Slant)	$y = m x + b$ ( $m \neq 0$ )	$lim_{x \to + \infty} [f (x) - (m x + b)] = 0$ or $lim_{x \to - \infty} [f (x) - (m x + b)] = 0$	Rational functions with $deg (numerator) = deg (denominator) + 1$ .

Vertical Asymptotes

The line $x = a$ is a vertical asymptote of the graph of $f$ if $f (x) \to + \infty$ or $f (x) \to - \infty$ as $x$ approaches $a$ from the left or right.

See the following figures.

The vertical line x = a is a vertical asymptote of a curve — The vertical line $x = a$ is a vertical asymptote of a curve if the $y$ coordinates of the points on the curve approach $+ \infty$ or $- \infty$ as $x$ approaches $a$ (from the left or right or both directions).

When do we have vertical asymptotes? There are a few cases that we need to pay attention to:

There is a fraction: $f (x) = \frac{g (x)}{h (x)} with lim_{x \to a} h (x) = 0.$ If $lim_{x \to a} g (x) \neq 0$ , then: $lim_{x \to a} f (x) = + \infty or - \infty,$ and thus $x = a$ is a vertical asymptote. However, if $lim_{x \to a} h (x) = lim_{x \to a} g (x) = 0$ , then $f (x)$ may or may not approach $+ \infty$ or $- \infty$ .
Logarithmic functions. For example, because $lim_{x \to 0^{+}} \ln x = - \infty$ , the vertical line $x = 0$ is a vertical asymptote of the graph of $y = \ln x$ .
Trigonometric functions like $\tan x = \frac{\sin x}{\cos x}$ , $\cot x = \frac{\cos x}{\sin x}$ , $\sec x = \frac{1}{\cos x}$ , and $\csc x = \frac{1}{\sin x}$ , where the denominator becomes zero.

Horizontal Asymptotes

A line $y = b$ is a horizontal asymptote of the curve $y = f (x)$ if $f (x) \to b$ as $x \to + \infty$ or $x \to - \infty$ .

See the following figures.

A line y = b is an asymptote of a curve — A line $y = b$ is an asymptote of a curve if the distance between the points on the curve and this line gets smaller and smaller as $x$ moves increasingly far away from the origin in the positive or negative direction.

Find horizontal asymptote(s) of the graph of $f (x) = \frac{1}{1 + e^{- x}}$ .

Solution

To this end, we need to find the limits of $f (x)$ as $x \to + \infty$ and $x \to - \infty$ . Since $lim_{x \to + \infty} e^{- x} = 0$ , we have:

\begin{aligned} \lim_{x\to +\infty} \frac{1}{1 + e^{-x}} &= \frac{\displaystyle \lim_{x\to +\infty} 1}{\displaystyle \lim_{x\to +\infty} 1 + \lim_{x\to +\infty} e^{-x}} \\ &= \frac{1}{1 + 0} \\ &= 1. \end{aligned}

Therefore, $y = 1$ is a horizontal asymptote of the graph of f.

Since $lim_{x \to - \infty} e^{- x} = + \infty$ , we have:

\begin{aligned} \lim_{x\to -\infty} \frac{1}{1 + e^{-x}} &= \frac{\displaystyle \lim_{x\to -\infty} 1}{\displaystyle \lim_{x\to -\infty} 1 + \lim_{x\to -\infty} e^{-x}} \\ &= \frac{1}{1 + \infty} \\ &= 0. \end{aligned}

Therefore, $y = 0$ is another horizontal asymptote of the graph of f.

The graph of f(x) = 1/(1+e^(-x)) has two horizontal asymptotes: y=0 and y=1 — The graph of $f (x) = \frac{1}{1 + e^{- x}}$ has two horizontal asymptotes: $y = 0$ and $y = 1$ .

Oblique (Also Known as Inclined or Slant) Asymptotes

A line with equation $y = m x + b$ ( $m \neq 0$ ) is an oblique asymptote (of the graph) of $f$ if either:

lim_{x \to + \infty} [f (x) - (m x + b)] = 0

lim_{x \to - \infty} [f (x) - (m x + b)] = 0.

See the following figures.

A line L is an oblique asymptote of f(x) — 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 $x \to - \infty$ .

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 $degree of P_{1} (x) = degree of P_{2} (x) + 1$ , then $f (x)$ has an oblique asymptote, which is the quotient of dividing $P_{1} (x)$ by $P_{2} (x)$ .

Find the oblique asymptote of the graph of $f (x) = \frac{3 x^{2} + 5 x - 3}{x + 1}$ .

Solution

Let's divide the numerator by the denominator:

\begin{aligned} \frac{3x^2 + 5x - 3}{x + 1} &= \frac{(x + 1)(3x + 2) - 5}{x + 1} \\ &= 3x + 2 - \frac{5}{x + 1}. \end{aligned}

Because:

lim_{x \to \pm \infty} | f (x) - (3 x + 2) | = lim_{x \to \pm \infty} | \frac{- 5}{x + 1} | = 0,

the line $y = 3 x + 2$ is the oblique asymptote of the graph of f, and as shown below, the distance between $f (x)$ and this line approaches zero as $x$ approaches $+ \infty$ or $- \infty$ . Also, notice that the graph of f has a vertical asymptote at $x = - 1$ because the denominator is zero at $x = - 1$ but the numerator is not.

The distance between the graph of f(x) = (3x^2+5x-3)/(x+1) and the line y = 3x+2 goes to zero — The distance between the graph of $f (x) = \frac{3 x^{2} + 5 x - 3}{x + 1}$ and the line $y = 3 x + 2$ goes to zero as $x \to \pm \infty$ .

Exercises

Find the vertical asymptotes (if any) of the following functions:

$f (x) = \frac{4 - 2 x}{2 x^{2} + x - 1}$
$g (x) = \frac{x^{2} + 4 x + 3}{x^{2} + 7 x + 12}$
$h (x) = \frac{x + 7}{x^{2} + 1}$

Answer

(a) $x = - 1$ and $x = \frac{1}{2}$ (b) $x = - 4$ (c) None

Solution

(a) Let's factor the numerator and denominator. The roots of $2 x^{2} + x - 1 = 0$ are $x = - 1$ and $x = \frac{1}{2}$ , so:

\frac{4 - 2 x}{2 x^{2} + x - 1} = \frac{- 2 (x - 2)}{2 (x + 1) (x - \frac{1}{2})} .

This fraction cannot be simplified further. Thus, the zeros of the denominator, $x = - 1$ and $x = \frac{1}{2}$ , are the vertical asymptotes.

Graph of f(x) = (4-2x)/(2x^2+x-1) — Graph of $f (x) = \frac{4 - 2 x}{2 x^{2} + x - 1}$ .

(b) For the numerator, we need two numbers whose sum is 4 and whose product is 3; these are 3 and 1:

x^{2} + 4 x + 3 = (x + 3) (x + 1) .

For the denominator, we need two numbers whose sum is 7 and whose product is 12; these are 4 and 3:

x^{2} + 7 x + 12 = (x + 3) (x + 4) .

Thus:

\frac{x^{2} + 4 x + 3}{x^{2} + 7 x + 12} = \frac{(x + 3) (x + 1)}{(x + 4) (x + 3)} = \frac{x + 1}{x + 4} (if x \neq - 3) .

The zero of the simplified fraction is $x = - 4$ , so $x = - 4$ is the only vertical asymptote. Although $x = - 4$ and $x = - 3$ both make the denominator of the original fraction zero, $x = - 3$ is not a vertical asymptote because:

\begin{aligned} \lim_{x\to -3} g(x) &= \lim_{x\to -3} \frac{x + 1}{x + 4} \\ &= \frac{-3 + 1}{-3 + 4} \\ &= -2, \end{aligned}

while $x = - 4$ is a vertical asymptote because:

\begin{aligned} \lim_{x\to -4^-} g(x) &= \lim_{x\to -4^-} \frac{x + 1}{x + 4} \quad \left[\text{form } \frac{-3}{0^-}\right] \\ &= +\infty, \end{aligned}

and:

\begin{aligned} \lim_{x\to -4^+} g(x) &= \lim_{x\to -4^+} \frac{x + 1}{x + 4} \quad \left[\text{form } \frac{-3}{0^+}\right] \\ &= -\infty. \end{aligned}

Graph of g(x) = (x^2+4x+3)/(x^2+7x+12) — Graph of $g (x) = \frac{x^{2} + 4 x + 3}{x^{2} + 7 x + 12}$ .

(c) We cannot factor the denominator, and the denominator is never zero, because:

x^{2} + 1 \geq 1 > 0.

Therefore, the graph of $h (x)$ has no vertical asymptotes.

Graph of h(x) = (x+7)/(x^2+1) — Graph of $h (x) = \frac{x + 7}{x^{2} + 1}$ .

Find the vertical asymptotes of the function given by:

f (x) = \frac{x - 2}{\sqrt{4 - x^{2}}} .

Answer

$x = - 2$

Solution

The denominator $\sqrt{4 - x^{2}}$ is zero when $x = \pm 2$ , 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 for which the expression under the square root is positive:

\begin{aligned} \text{Dom}(f) &= \{x \mid 4 - x^2 > 0\} \\ &= \{x \mid x^2 < 4\} \\ &= \{x \mid -2 < x < 2\} \\ &= (-2, 2). \end{aligned}

Therefore x can approach $2$ only from the left and $- 2$ only from the right.

\begin{aligned} \lim_{x\to 2^-} f(x) &= \lim_{x\to 2^-} \frac{x - 2}{\sqrt{4 - x^2}} \\ &= \lim_{x\to 2^-} \frac{x - 2}{\sqrt{(2 - x)(2 + x)}} \\ &= \lim_{x\to 2^-} \frac{-(2 - x)}{\sqrt{2 - x}\sqrt{2 + x}} \\ &= \lim_{x\to 2^-} \frac{-\sqrt{2 - x}\sqrt{2 - x}}{\sqrt{2 - x}\sqrt{2 + x}} \\ &= \lim_{x\to 2^-} \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.

On the other hand, as $x \to - 2^{+}$ , the denominator goes to $0$ through positive numbers (the square root is always nonnegative) and the numerator goes to $- 4$ . Therefore, $f (x) \to - \infty$ (symbolically $\frac{- 4}{0^{+}} = - \infty$ ). Therefore, $x = - 2$ is a vertical asymptote of the graph of f.

Graph of f(x) = (x-2)/sqrt(4-x^2) — Graph of $f (x) = \frac{x - 2}{\sqrt{4 - x^{2}}}$ . The domain of f is the open interval $(- 2, 2)$ . Because $lim_{x \to - 2^{+}} f (x) = - \infty$ , $x = - 2$ is a vertical asymptote to the graph of f.

Find the vertical asymptotes of the following functions:

$F (x) = \ln (x - 2)$
$g (x) = \ln (\sqrt{x^{2} - 9})$

Answer

(a) $x = 2$ (b) $x = - 3$ and $x = 3$

Solution

(a) Because $x - 2 = 0$ when $x = 2$ (and $F$ is defined only for $x > 2$ ), the graph of $F$ has a vertical asymptote at $x = 2$ . Alternatively, the graph of $y = \ln (x - 2)$ is obtained by shifting the graph of $y = \ln x$ to the right 2 units. Because 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 $x = 2$ .

Graph of F(x) = ln(x-2) — The graph of $F (x) = \ln (x - 2)$ is obtained by shifting the graph of $y = \ln x$ to the right 2 units, so $x = 2$ is a vertical asymptote of the graph of $F$ .

(b) Because $\sqrt{x^{2} - 9}$ is zero when $x = \pm 3$ , the graph of $g$ has vertical asymptotes at $x = \pm 3$ . Note that the domain of $g$ is ${x ∣ x < - 3 or x > 3} = (- \infty, - 3) \cup (3, \infty)$ , so $x$ can approach $- 3$ only from the left and $3$ only from the right.

Graph of g(x) = ln(sqrt(x^2-9)) — The graph of $g (x) = \ln (\sqrt{x^{2} - 9})$ has two vertical asymptotes at $x = 3$ and $x = - 3$ .

Determine the end behavior of the graph of:

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

Answer

Horizontal asymptote $y = 2$

Solution

Because:

lim_{x \to \pm \infty} \frac{2 x^{2} - 3}{x^{2} + 1} = lim_{x \to \pm \infty} \frac{2 x^{2}}{x^{2}} = 2,

its graph has only one horizontal asymptote, namely the line $y = 2$ .

Graph of y = (2x^2-3)/(x^2+1) — As we can see, $y \to 2$ as $x \to \pm \infty$ .

Determine the end behavior of the function:

f (x) = \frac{1 - x}{| x | + 3} .

Answer

$y = - 1$ as $x \to + \infty$ and $y = 1$ as $x \to - \infty$

Solution

Since $| x | = x$ if $x > 0$ :

\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}

while since $| x | = - x$ if $x < 0$ :

\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}

Hence, f has two horizontal asymptotes, namely $y = - 1$ and $y = 1$ .

Graph of f(x) = (1-x)/(|x|+3) — As we can see, $y \to 1$ as $x \to - \infty$ and $y \to - 1$ as $x \to + \infty$ .

Find the oblique asymptote of:

f (x) = - 3 x + 1 + \frac{3 x + 7}{x^{4} - 2 x + 2} .

Answer

$y = - 3 x + 1$

Solution

Because:

lim_{x \to \pm \infty} \frac{3 x + 7}{x^{4} - 2 x + 2} = lim_{x \to \pm \infty} \frac{3 x}{x^{4}} = 0,

we have:

lim_{x \to \pm \infty} [f (x) - (- 3 x + 1)] = 0,

and therefore the line $y = - 3 x + 1$ is the oblique asymptote of $f (x)$ (both as $x \to + \infty$ and as $x \to - \infty$ ).

Graph of f(x) approaching y = -3x + 1 — The distance between the graph of $f (x)$ and the line $y = - 3 x + 1$ goes to zero as $x \to \pm \infty$ .

Show that if the line $y = m x + b$ is the oblique asymptote of $f (x)$ as $x \to + \infty$ then:

m = lim_{x \to + \infty} \frac{f (x)}{x}

and

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

If the line is the oblique asymptote of $f (x)$ as $x \to - \infty$ , we simply need to replace $x \to + \infty$ by $x \to - \infty$ in the above two equations.

Answer

$m = lim_{x \to + \infty} \frac{f (x)}{x}$ and $b = lim_{x \to + \infty} [f (x) - m x]$

Solution

Suppose $y = m x + b$ is the oblique asymptote of $f (x)$ as $x \to + \infty$ .
By definition:

lim_{x \to + \infty} [f (x) - (m x + b)] = 0, (i)

or, dividing and multiplying by $x$ :

lim_{x \to + \infty} x [\frac{f (x)}{x} - m - \frac{b}{x}] = 0. (ii)

Since $lim_{x \to + \infty} x = + \infty$ , the bracketed factor in (ii) must itself approach $0$ (otherwise the product in (ii) could not approach $0$ ); that is:

lim_{x \to + \infty} [\frac{f (x)}{x} - m - \frac{b}{x}] = 0. (iii)

Because:

lim_{x \to + \infty} [\frac{f (x)}{x} - m - \frac{b}{x}] = lim_{x \to + \infty} \frac{f (x)}{x} - m - lim_{x \to + \infty} \frac{b}{x}

and $b / x \to 0$ as $x \to + \infty$ (with $m$ a constant), (iii) becomes:

lim_{x \to + \infty} \frac{f (x)}{x} - m = 0,

that is:

m = lim_{x \to + \infty} \frac{f (x)}{x} .

After finding $m$ , we use (i) to obtain $b$ :

\begin{aligned} \lim_{x\to +\infty} [f(x) - mx - b] &= 0 \\ \lim_{x\to +\infty} [f(x) - mx] - b &= 0, \end{aligned}

that is:

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

If instead $y = m x + b$ is the oblique asymptote of $f (x)$ as $x \to - \infty$ , the same argument goes through with every $+ \infty$ replaced by $- \infty$ , giving:

m = lim_{x \to - \infty} \frac{f (x)}{x} and b = lim_{x \to - \infty} [f (x) - m x] .

Use Exercise 7 to find the oblique asymptote(s) of the graph of the function:

f (x) = \sqrt{2 x^{2} - 1} - 7 x .

Answer

$y = (\sqrt{2} - 7) x$ as $x \to + \infty$ , and $y = - (\sqrt{2} + 7) x$ as $x \to - \infty$

Solution

By Exercise 7, the slope of the oblique asymptote as $x \to + \infty$ is:

\begin{aligned} 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{2 - \frac{1}{x^2}} - 7x}{x}. \end{aligned}

Since $| x | = x$ when $x > 0$ :

\begin{aligned} m_1 &= \lim_{x\to +\infty} \frac{x\sqrt{2 - \frac{1}{x^2}} - 7x}{x} \\ &= \lim_{x\to +\infty} \left(\sqrt{2 - \frac{1}{x^2}} - 7\right) \\ &= \sqrt{2 - 0} - 7 \\ &= \sqrt{2} - 7. \end{aligned}

The $y$ -intercept of this asymptote is:

\begin{aligned} b_1 &= \lim_{x\to +\infty} \left[f(x) - (\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 $\sqrt{2 x^{2} - 1} + \sqrt{2} x$ :

\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)}{\sqrt{2x^2 - 1} + \sqrt{2}x} \\ &= \lim_{x\to +\infty} \frac{(2x^2 - 1) - 2x^2}{\sqrt{2x^2 - 1} + \sqrt{2}x} \\ &= \lim_{x\to +\infty} \frac{-1}{\sqrt{2x^2 - 1} + \sqrt{2}x} \\ &= 0. \end{aligned}

Hence, $y = (\sqrt{2} - 7) x$ is the oblique asymptote of $f (x)$ as $x \to + \infty$ .

The slope of the oblique asymptote as $x \to - \infty$ is obtained similarly. Since $| x | = - x$ when $x < 0$ :

\begin{aligned} m_2 &= \lim_{x\to -\infty} \frac{f(x)}{x} \\ &= \lim_{x\to -\infty} \frac{|x|\sqrt{2 - \frac{1}{x^2}} - 7x}{x} \\ &= \lim_{x\to -\infty} \frac{-x\sqrt{2 - \frac{1}{x^2}} - 7x}{x} \\ &= \lim_{x\to -\infty} \left(-\sqrt{2 - \frac{1}{x^2}} - 7\right) \\ &= -\sqrt{2 - 0} - 7 \\ &= -\sqrt{2} - 7. \end{aligned}

By an analogous computation, the $y$ -intercept in this case is also $b_{2} = 0$ . Thus $y = - (\sqrt{2} + 7) x$ is the oblique asymptote of $f (x)$ as $x \to - \infty$ .

Graph of f(x) = sqrt(2x^2-1) - 7x with dual oblique asymptotes — The graph of $f (x) = \sqrt{2 x^{2} - 1} - 7 x$ has an oblique asymptote as $x \to + \infty$ and another one as $x \to - \infty$ .

Quick Reference

Vertical Asymptotes

Horizontal Asymptotes

Oblique (Also Known as Inclined or Slant) Asymptotes

Exercises

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