Limits at Infinity; Asymptotes of Graphs

Learning Objectives

Calculate the limit of a function as $x$ increases or decreases without bound.
Using correct notation, describe an infinite limit.
Define a vertical asymptote.
Recognize a horizontal asymptote on the graph of a function.
Estimate the end behavior of a function as $x$ increases or decreases without bound.
Recognize an oblique asymptote on the graph of a function.

Infinite Limits

Evaluating the limit of a function at a point or evaluating the limit of a function from the right and left at a point helps us to characterize the behavior of a function around a given value. As we shall see, we can also describe the behavior of functions that do not have finite limits.

We now turn our attention to $h(x)=1/(x-2)^2$ . From its graph we see that as the values of $x$ approach 2, the values of $h(x)=1/(x-2)^2$ become larger and larger and, in fact, become infinite. Mathematically, we say that the limit of $h(x)$ as $x$ approaches 2 is positive infinity. Symbolically, we express this idea as

\underset{x\to 2}{\lim}h(x)=+\infty

More generally, we define infinite limits as follows:

Definition

We define three types of infinite limits.

Infinite limits from the left: Let $f(x)$ be a function defined at all values in an open interval of the form $(b,a)$ .

If the values of $f(x)$ increase without bound as the values of $x$ (where $x<a$ ) approach the number $a$ , then we say that the limit as $x$ approaches $a$ from the left is positive infinity and we write
$\underset{x\to a^-}{\lim}f(x)=+\infty$ .
If the values of $f(x)$ decrease without bound as the values of $x$ (where $x<a$ ) approach the number $a$ , then we say that the limit as $x$ approaches $a$ from the left is negative infinity and we write
$\underset{x\to a^-}{\lim}f(x)=−\infty$ .

Infinite limits from the right: Let $f(x)$ be a function defined at all values in an open interval of the form $(a,c)$ .

If the values of $f(x)$ increase without bound as the values of $x$ (where $x>a$ ) approach the number $a$ , then we say that the limit as $x$ approaches $a$ from the left is positive infinity and we write
$\underset{x\to a^+}{\lim}f(x)=+\infty$ .
If the values of $f(x)$ decrease without bound as the values of $x$ (where $x>a$ ) approach the number $a$ , then we say that the limit as $x$ approaches $a$ from the left is negative infinity and we write
$\underset{x\to a^+}{\lim}f(x)=−\infty$ .

Two-sided infinite limit: Let $f(x)$ be defined for all $x\ne a$ in an open interval containing $a$ .

If the values of $f(x)$ increase without bound as the values of $x$ (where $x\ne a$ ) approach the number $a$ , then we say that the limit as $x$ approaches $a$ is positive infinity and we write
$\underset{x\to a}{\lim}f(x)=+\infty$ .
If the values of $f(x)$ decrease without bound as the values of $x$ (where $x\ne a$ ) approach the number $a$ , then we say that the limit as $x$ approaches $a$ is negative infinity and we write
$\underset{x\to a}{\lim}f(x)=−\infty$ .

It is important to understand that when we write statements such as $\underset{x\to a}{\lim}f(x)=+\infty$ or $\underset{x\to a}{\lim}f(x)=−\infty$ we are describing the behavior of the function, as we have just defined it. We are not asserting that a limit exists. For the limit of a function $f(x)$ to exist at $a$ , it must approach a real number $L$ as $x$ approaches $a$ . That said, if, for example, $\underset{x\to a}{\lim}f(x)=+\infty$ , we always write $\underset{x\to a}{\lim}f(x)=+\infty$ rather than $\underset{x\to a}{\lim}f(x)$ DNE.

Recognizing an Infinite Limit

Evaluate each of the following limits, if possible. Use a table of functional values and graph $f(x)=1/x$ to confirm your conclusion.

$\underset{x\to 0^-}{\lim}\frac{1}{x}$
$\underset{x\to 0^+}{\lim}\frac{1}{x}$
$\underset{x\to 0}{\lim}\frac{1}{x}$

Answer:

Begin by constructing a table of functional values.

Table of Functional Values for $f(x)=\frac{1}{x}$
$x$	$\frac{1}{x}$	$x$	$\frac{1}{x}$
−0.1	−10	0.1	10
−0.01	−100	0.01	100
−0.001	−1000	0.001	1000
−0.0001	−10,000	0.0001	10,000
−0.00001	−100,000	0.00001	100,000
−0.000001	−1,000,000	0.000001	1,000,000

The values of $1/x$ decrease without bound as $x$ approaches 0 from the left. We conclude that
$\underset{x\to 0^-}{\lim}\frac{1}{x}=−\infty$ .
The values of $1/x$ increase without bound as $x$ approaches 0 from the right. We conclude that
$\underset{x\to 0^+}{\lim}\frac{1}{x}=+\infty$ .
Since $\underset{x\to 0^-}{\lim}\frac{1}{x}=−\infty$ and $\underset{x\to 0^+}{\lim}\frac{1}{x}=+\infty$ have different values, we conclude that
$\underset{x\to 0}{\lim}\frac{1}{x}$ DNE.

The graph of $f(x)=1/x$ in (Figure) confirms these conclusions.

The graph of the function f(x) = 1/x. The function curves asymptotically towards x=0 and y=0 in quadrants one and three.

Figure 8. The graph of

f(x)=1/x

confirms that the limit as

x

approaches 0 does not exist.

Evaluate each of the following limits, if possible. Use a table of functional values and graph $f(x)=1/x^2$ to confirm your conclusion.

$\underset{x\to 0^-}{\lim}\frac{1}{x^2}$
$\underset{x\to 0^+}{\lim}\frac{1}{x^2}$
$\underset{x\to 0}{\lim}\frac{1}{x^2}$

Answer:

a. $\underset{x\to 0^-}{\lim}\frac{1}{x^2}=+\infty$ ;

\underset{x\to 0^+}{\lim}\frac{1}{x^2}=+\infty

; c.

\underset{x\to 0}{\lim}\frac{1}{x^2}=+\infty

Hint

Follow the procedures from (Figure).

Evaluating a Limit of the Form $K/0, \, K\ne 0$ Using the Limit Laws

Evaluate $\underset{x\to 2^-}{\lim}\frac{x-3}{x^2-2x}$ .

Answer:

Step 1. After substituting in $x=2$ , we see that this limit has the form $-1/0$ . That is, as $x$ approaches 2 from the left, the numerator approaches −1 and the denominator approaches 0. Consequently, the magnitude of $\frac{x-3}{x(x-2)}$ becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:

\underset{x\to 2^-}{\lim}\frac{x-3}{x^2-2x}=\underset{x\to 2^-}{\lim}\frac{x-3}{x(x-2)}

Step 2. Since $x-2$ is the only part of the denominator that is zero when 2 is substituted, we then separate $1/(x-2)$ from the rest of the function:

=\underset{x\to 2^-}{\lim}\frac{x-3}{x} \cdot \frac{1}{x-2}

Step 3. $\underset{x\to 2^-}{\lim}\frac{x-3}{x}=-\frac{1}{2}$ and $\underset{x\to 2^-}{\lim}\frac{1}{x-2}=−\infty$ . Therefore, the product of $(x-3)/x$ and $1/(x-2)$ has a limit of $+\infty$ :

\underset{x\to 2^-}{\lim}\frac{x-3}{x^2-2x}=+\infty

Evaluate $\underset{x\to 1}{\lim}\frac{x+2}{(x-1)^2}$ .

Answer:

$+\infty$

Hint

Use the methods from (Figure).

It is useful to point out that functions of the form $f(x)=1/(x-a)^n$ , where $n$ is a positive integer, have infinite limits as $x$ approaches $a$ from either the left or right ((Figure)). These limits are summarized in (Figure).

Two graphs side by side of f(x) = 1 / (x-a)^n. The first graph shows the case where n is an odd positive integer, and the second shows the case where n is an even positive integer. In the first, the graph has two segments. Each curve asymptotically towards the x axis, also known as y=0, and x=a. The segment to the left of x=a is below the x axis, and the segment to the right of x=a is above the x axis. In the second graph, both segments are above the x axis.

Figure 9. The function

f(x)=1/(x-a)^n

has infinite limits at

a

Infinite Limits from Positive Integers

If $n$ is a positive even integer, then

\underset{x\to a}{\lim}\frac{1}{(x-a)^n}=+\infty

If $n$ is a positive odd integer, then

\underset{x\to a^+}{\lim}\frac{1}{(x-a)^n}=+\infty

and

\underset{x\to a^-}{\lim}\frac{1}{(x-a)^n}=−\infty

We should also point out that in the graphs of $f(x)=1/(x-a)^n$ , points on the graph having $x$ -coordinates very near to $a$ are very close to the vertical line $x=a$ . That is, as $x$ approaches $a$ , the points on the graph of $f(x)$ are closer to the line $x=a$ . The line $x=a$ is called a vertical asymptote of the graph. We formally define a vertical asymptote as follows:

Definition

Let $f(x)$ be a function. If any of the following conditions hold, then the line $x=a$ is a vertical asymptote of $f(x)$ :

\begin{array}{ccc}\hfill \underset{x\to a^-}{\lim}f(x)& =\hfill & +\infty \, \text{or} \, -\infty \hfill \\ \hfill \underset{x\to a^+}{\lim}f(x)& =\hfill & +\infty \, \text{or} \, −\infty \hfill \\ & \text{or}\hfill & \\ \hfill \underset{x\to a}{\lim}f(x)& =\hfill & +\infty \, \text{or} \, −\infty \hfill \end{array}

Finding a Vertical Asymptote

Evaluate each of the following limits using (Figure). Identify any vertical asymptotes of the function $f(x)=1/(x+3)^4$ .

$\underset{x\to -3^-}{\lim}\frac{1}{(x+3)^4}$
$\underset{x\to -3^+}{\lim}\frac{1}{(x+3)^4}$
$\underset{x\to -3}{\lim}\frac{1}{(x+3)^4}$

Answer:

We can use (Figure) directly.

$\underset{x\to -3^-}{\lim}\frac{1}{(x+3)^4}=+\infty$
$\underset{x\to -3^+}{\lim}\frac{1}{(x+3)^4}=+\infty$
$\underset{x\to -3}{\lim}\frac{1}{(x+3)^4}=+\infty$

The function $f(x)=1/(x+3)^4$ has a vertical asymptote of $x=-3$ .

Evaluate each of the following limits. Identify any vertical asymptotes of the function $f(x)=\frac{1}{(x-2)^3}$ .

$\underset{x\to 2^-}{\lim}\frac{1}{(x-2)^3}$
$\underset{x\to 2^+}{\lim}\frac{1}{(x-2)^3}$
$\underset{x\to 2}{\lim}\frac{1}{(x-2)^3}$

Answer:

a. $\underset{x\to 2^-}{\lim}\frac{1}{(x-2)^3}=−\infty$ ;

\underset{x\to 2^+}{\lim}\frac{1}{(x-2)^3}=+\infty

; c.

\underset{x\to 2}{\lim}\frac{1}{(x-2)^3}

DNE. The line

x=2

is the vertical asymptote of

f(x)=1/(x-2)^3

Hint

Use (Figure).

In the next example we put our knowledge of various types of limits to use to analyze the behavior of a function at several different points.

Behavior of a Function at Different Points

Use the graph of $f(x)$ in (Figure) to determine each of the following values:

$\underset{x\to 3^-}{\lim}f(x); \, \underset{x\to 3^+}{\lim}f(x); \, \underset{x\to 3}{\lim}f(x); \, f(3)$
$\underset{x\to \infty}{\lim}f(x)$
$\underset{x\to -\infty}{\lim}f(x)$

The graph of a function f(x) described by the above limits and values. There is a smooth curve for values below x=-2; at (-2, 3), there is an open circle. There is a smooth curve between (-2, 1] with a closed circle at (1,6). There is an open circle at (1,3), and a smooth curve stretching from there down asymptotically to negative infinity along x=3. The function also curves asymptotically along x=3 on the other side, also stretching to negative infinity. The function then changes concavity in the first quadrant around y=4.5 and continues up.

Figure 10. The graph shows

f(x)

Answer:

Using (Figure) and the graph for reference, we arrive at the following values:

$\underset{x\to 3^-}{\lim}f(x)=−\infty; \, \underset{x\to 3^+}{\lim}f(x)=−\infty; \, \underset{x\to 3}{\lim}f(x)=-\infty; \, f(3)$ is undefined
$\underset{x\to \infty}{\lim}f(x)=\infty$
$\underset{x\to -\infty}{\lim}f(x)=-\infty$

Evaluate $\underset{x\to 1}{\lim}f(x)$ for $f(x)$ shown here:

A graph of a piecewise function. The first segment curves from the third quadrant to the first, crossing through the second quadrant. Where the endpoint would be in the first quadrant is an open circle. The second segment starts at a closed circle a few units below the open circle. It curves down from quadrant one to quadrant four.

Answer:

Does not exist.

Hint

Compare the limit from the right with the limit from the left.

Limits at Infinity and Horizontal Asymptotes

Recall that $\underset{x \to a}{\lim}f(x)=L$ means $f(x)$ becomes arbitrarily close to $L$ as long as $x$ is sufficiently close to $a$ . We can extend this idea to limits at infinity. For example, consider the function $f(x)=2+\frac{1}{x}$ . As can be seen graphically in (Figure) and numerically in (Figure), as the values of $x$ get larger, the values of $f(x)$ approach 2. We say the limit as $x$ approaches $\infty$ of $f(x)$ is 2 and write $\underset{x\to \infty }{\lim}f(x)=2$ . Similarly, for $x<0$ , as the values $|x|$ get larger, the values of $f(x)$ approaches 2. We say the limit as $x$ approaches $−\infty$ of $f(x)$ is 2 and write $\underset{x\to a}{\lim}f(x)=2$ .

The function f(x) 2 + 1/x is graphed. The function starts negative near y = 2 but then decreases to −∞ near x = 0. The function then decreases from ∞ near x = 0 and gets nearer to y = 2 as x increases. There is a horizontal line denoting the asymptote y = 2.

Figure 1. The function approaches the asymptote

y=2

x

approaches

\pm \infty

Values of a function $f$ as $x \to \pm \infty$
$x$	10	100	1,000	10,000
$2+\frac{1}{x}$	2.1	2.01	2.001	2.0001
$x$	-10	-100	-1000	-10,000
$2+\frac{1}{x}$	1.9	1.99	1.999	1.9999

More generally, for any function $f$ , we say the limit as $x \to \infty$ of $f(x)$ is $L$ if $f(x)$ becomes arbitrarily close to $L$ as long as $x$ is sufficiently large. In that case, we write $\underset{x\to a}{\lim}f(x)=L$ . Similarly, we say the limit as $x\to −\infty$ of $f(x)$ is $L$ if $f(x)$ becomes arbitrarily close to $L$ as long as $x<0$ and $|x|$ is sufficiently large. In that case, we write $\underset{x\to −\infty }{\lim}f(x)=L$ . We now look at the definition of a function having a limit at infinity.

Definition

(Informal) If the values of $f(x)$ become arbitrarily close to $L$ as $x$ becomes sufficiently large, we say the function $f$ has a limit at infinity and write

\underset{x\to \infty }{\lim}f(x)=L

If the values of $f(x)$ becomes arbitrarily close to $L$ for $x<0$ as $|x|$ becomes sufficiently large, we say that the function $f$ has a limit at negative infinity and write

\underset{x\to -\infty }{\lim}f(x)=L

If the values $f(x)$ are getting arbitrarily close to some finite value $L$ as $x\to \infty$ or $x\to −\infty$ , the graph of $f$ approaches the line $y=L$ . In that case, the line $y=L$ is a horizontal asymptote of $f$ ((Figure)). For example, for the function $f(x)=\frac{1}{x}$ , since $\underset{x\to \infty }{\lim}f(x)=0$ , the line $y=0$ is a horizontal asymptote of $f(x)=\frac{1}{x}$ .

Definition

If $\underset{x\to \infty }{\lim}f(x)=L$ or $\underset{x \to −\infty}{\lim}f(x)=L$ , we say the line $y=L$ is a horizontal asymptote of $f$ .

The figure is broken up into two figures labeled a and b. Figure a shows a function f(x) approaching but never touching a horizontal dashed line labeled L from above. Figure b shows a function f(x) approaching but never a horizontal dashed line labeled M from below.

Figure 2. (a) As

x\to \infty

, the values of

f

are getting arbitrarily close to

L

. The line

y=L

is a horizontal asymptote of

f

. (b) As

x\to −\infty

, the values of

f

are getting arbitrarily close to

M

. The line

y=M

is a horizontal asymptote of

f

A function cannot cross a vertical asymptote because the graph must approach infinity (or negative infinity) from at least one direction as $x$ approaches the vertical asymptote. However, a function may cross a horizontal asymptote. In fact, a function may cross a horizontal asymptote an unlimited number of times. For example, the function $f(x)=\frac{ \cos x}{x}+1$ shown in (Figure) intersects the horizontal asymptote $y=1$ an infinite number of times as it oscillates around the asymptote with ever-decreasing amplitude.

The function f(x) = (cos x)/x + 1 is shown. It decreases from (0, ∞) and then proceeds to oscillate around y = 1 with decreasing amplitude.

Figure 3. The graph of

f(x)=\cos x/x+1

crosses its horizontal asymptote

y=1

an infinite number of times.

The algebraic limit laws and squeeze theorem we introduced in Introduction to Limits also apply to limits at infinity. We illustrate how to use these laws to compute several limits at infinity.

Computing Limits at Infinity

For each of the following functions $f$ , evaluate $\underset{x\to \infty }{\lim}f(x)$ and $\underset{x\to −\infty }{\lim}f(x)$ . Determine the horizontal asymptote(s) for $f$ .

$f(x)=5-\frac{2}{x^2}$
$f(x)=\frac{\sin x}{x}$
$f(x)= \tan^{-1} (x)$

Answer:

Using the algebraic limit laws, we have $\underset{x\to \infty }{\lim}(5-\frac{2}{x^2})=\underset{x\to \infty }{\lim}5-2(\underset{x\to \infty }{\lim}\frac{1}{x})(\underset{x\to \infty }{\lim}\frac{1}{x})=5-2 \cdot 0=5$ . Similarly, $\underset{x\to \infty }{\lim}f(x)=5$ . Therefore, $f(x)=5-\frac{2}{x^2}$ has a horizontal asymptote of $y=5$ and $f$ approaches this horizontal asymptote as $x\to \pm \infty$ as shown in the following graph.
Figure 4. This function approaches a horizontal asymptote as $x\to \pm \infty$ .
Since $-1\le \sin x\le 1$ for all $x$ , we have
$\frac{-1}{x}\le \frac{\sin x}{x}\le \frac{1}{x}$
for all $x \ne 0$ . Also, since
$\underset{x\to \infty }{\lim}\frac{-1}{x}=0=\underset{x\to \infty }{\lim}\frac{1}{x}$ ,
we can apply the squeeze theorem to conclude that
$\underset{x\to \infty }{\lim}\frac{\sin x}{x}=0$ .
Similarly,
$\underset{x\to −\infty}{\lim}\frac{\sin x}{x}=0$ .
Thus, $f(x)=\frac{\sin x}{x}$ has a horizontal asymptote of $y=0$ and $f(x)$ approaches this horizontal asymptote as $x\to \pm \infty$ as shown in the following graph.

Figure 5. This function crosses its horizontal asymptote multiple times.
To evaluate $\underset{x\to \infty }{\lim} \tan^{-1} (x)$ and $\underset{x\to −\infty}{\lim} \tan^{-1} (x)$ , we first consider the graph of $y= \tan (x)$ over the interval $(−\pi /2,\pi /2)$ as shown in the following graph.

The graph of $\tan x$ has vertical asymptotes at $x=\pm \frac{\pi }{2}$

Since

\underset{x\to (\pi/2)^-}{\lim} \tan x=\infty

it follows that

\underset{x\to \infty }{\lim} \tan^{-1} (x)=\frac{\pi }{2}

Similarly, since

\underset{x\to (\pi/2)^+}{\lim} \tan x=−\infty

it follows that

\underset{x\to −\infty}{\lim} \tan^{-1} (x)=-\frac{\pi }{2}

As a result, $y=\frac{\pi }{2}$ and $y=-\frac{\pi }{2}$ are horizontal asymptotes of $f(x)= \tan^{-1} (x)$ as shown in the following graph.

The function f(x) = tan−1 x is shown. It increases from (−∞, −π/2), passes through the origin, and then increases toward (∞, π/2). There are horizontal dashed lines marking y = ±π/2.

Figure 7. This function has two horizontal asymptotes.

Evaluate $\underset{x\to −\infty}{\lim}(3+\frac{4}{x})$ and $\underset{x\to \infty }{\lim}(3+\frac{4}{x})$ . Determine the horizontal asymptotes of $f(x)=3+\frac{4}{x}$ , if any.

Answer:

Both limits are 3. The line $y=3$ is a horizontal asymptote.

Hint

$\underset{x\to \pm \infty }{\lim}1/x=0$

Special Limits with Sine and Cosine

Definition

$\underset{x\to \infty }{\lim}\sin x=DNE$ $\underset{x\to \infty }{\lim}\cos x=DNE$ $\underset{x\to \infty }{\lim}\frac{\sin x}{x}=\infty$ $\underset{x\to \infty }{\lim}\frac{\cos x}{x}=\infty$

Special limits with sine and cosine

For each of the following functions $f$ , evaluate $\underset{x\to \infty }{\lim}f(x)$

$f(x)=\frac{2-\sin x}{x}$
$f(x)=\sin\left(\frac{1}{x}\right)$
$f(x)= \frac{6}{5x-\cos x}$

Answer:

Using the algebraic limit laws, we have $\underset{x\to \infty }{\lim}\frac{2-\sin x}{x}=\underset{x\to \infty }{\lim}\frac{2x}{x}-\frac{\cos x}{x}=\underset{x\to \infty }{\lim}2-\frac{\cos x}{x}=\underset{x\to \infty }{\lim}2-\underset{x\to \infty }{\lim}\frac{\cos x}{x}=2-0=2$ .
Since $\underset{x\to \infty }{\lim}\frac{1}{x}=0$ , then $\underset{x\to \infty }{\lim}\sin \left(\frac{1}{x} \right)=\sin(0)=0$ .
Using the algebraic limit laws, we have $\underset{x\to \infty }{\lim}\frac{6}{5x-\cos x}=\underset{x\to \infty }{\lim}\frac{\frac{6}{x}}{\frac{5x}{x}-\frac{\cos x}{x}}=\frac{\underset{x\to \infty }{\lim}\frac{6}{x}}{\underset{x\to \infty }{\lim}5-\underset{x\to \infty }{\lim}\frac{\cos x}{x}}=\frac{0}{5-0}=0$ .

Infinite Limits at Infinity

Sometimes the values of a function $f$ become arbitrarily large as $x\to \infty$ (or as $x\to −\infty )$ . In this case, we write $\underset{x\to \infty }{\lim}f(x)=\infty$ (or $\underset{x\to −\infty }{\lim}f(x)=\infty )$ . On the other hand, if the values of $f$ are negative but become arbitrarily large in magnitude as $x\to \infty$ (or as $x\to −\infty )$ , we write $\underset{x\to \infty }{\lim}f(x)=−\infty$ (or $\underset{x\to −\infty }{\lim}f(x)=−\infty )$ .

For example, consider the function $f(x)=x^3$ . As seen in (Figure) and (Figure), as $x\to \infty$ the values $f(x)$ become arbitrarily large. Therefore, $\underset{x\to \infty }{\lim}x^3=\infty$ . On the other hand, as $x\to −\infty$ , the values of $f(x)=x^3$ are negative but become arbitrarily large in magnitude. Consequently, $\underset{x\to −\infty }{\lim}x^3=−\infty$ .

Values of a power function as $x\to \pm \infty$
$x$	10	20	50	100	1000
$x^3$	1000	8000	125,000	1,000,000	1,000,000,000
$x$	-10	-20	-50	-100	-1000
$x^3$	-1000	-8000	-125,000	-1,000,000	-1,000,000,000

The function f(x) = x3 is graphed. It is apparent that this function rapidly approaches infinity as x approaches infinity.

Figure 8. For this function, the functional values approach infinity as

x\to \pm \infty

Definition

(Informal) We say a function $f$ has an infinite limit at infinity and write

\underset{x\to \infty }{\lim}f(x)=\infty

if $f(x)$ becomes arbitrarily large for $x$ sufficiently large. We say a function has a negative infinite limit at infinity and write

\underset{x\to \infty }{\lim}f(x)=−\infty

if $f(x)<0$ and $|f(x)|$ becomes arbitrarily large for $x$ sufficiently large. Similarly, we can define infinite limits as $x\to −\infty$ .

Formal Definitions

Earlier, we used the terms arbitrarily close, arbitrarily large, and sufficiently large to define limits at infinity informally. Although these terms provide accurate descriptions of limits at infinity, they are not precise mathematically. Here are more formal definitions of limits at infinity. We then look at how to use these definitions to prove results involving limits at infinity.

Definition

(Formal) We say a function $f$ has a limit at infinity, if there exists a real number $L$ such that for all $\epsilon >0$ , there exists $N>0$ such that

|f(x)-L|<\epsilon

for all $x>N$ . In that case, we write

\underset{x\to \infty }{\lim}f(x)=L

(see (Figure)).

We say a function $f$ has a limit at negative infinity if there exists a real number $L$ such that for all $\epsilon >0$ , there exists $N<0$ such that

|f(x)-L|<\epsilon

for all $x<N$ . In that case, we write

\underset{x\to −\infty }{\lim}f(x)=L

The function f(x) is graphed, and it has a horizontal asymptote at L. L is marked on the y axis, as is L + ॉ and L – ॉ. On the x axis, N is marked as the value of x such that f(x) = L + ॉ.

Figure 9. For a function with a limit at infinity, for all

x>N

|f(x)-L|<\epsilon

Earlier in this section, we used graphical evidence in (Figure) and numerical evidence in (Figure) to conclude that $\underset{x\to \infty }{\lim}(\frac{2+1}{x})=2$ . Here we use the formal definition of limit at infinity to prove this result rigorously.

A Finite Limit at Infinity Example

Use the formal definition of limit at infinity to prove that $\underset{x\to \infty }{\lim}(2+\frac{1}{x})=2$ .

Answer:

Let $\epsilon >0$ . Let $N=\frac{1}{\epsilon }$ . Therefore, for all $x>N$ , we have

|2+\frac{1}{x}-2|=|\frac{1}{x}|=\frac{1}{x}<\frac{1}{N}=\epsilon

Use the formal definition of limit at infinity to prove that $\underset{x\to \infty }{\lim}(3 - \frac{1}{x^2})=3$ .

Answer:

Let $\epsilon >0$ . Let $N=\frac{1}{\sqrt{\epsilon }}$ . Therefore, for all $x>N$ , we have

$|3-\frac{1}{x^2}-3|=\frac{1}{x^2}<\frac{1}{N^2}=\epsilon$

Therefore, $\underset{x\to \infty }{\lim}(3-1/x^2)=3$ .

Hint

Let $N=\frac{1}{\sqrt{\epsilon }}$ .

We now turn our attention to a more precise definition for an infinite limit at infinity.

Definition

(Formal) We say a function $f$ has an infinite limit at infinity and write

\underset{x\to \infty }{\lim}f(x)=\infty

if for all $M>0$ , there exists an $N>0$ such that

f(x)>M

for all $x>N$ (see (Figure)).

We say a function has a negative infinite limit at infinity and write

\underset{x\to \infty }{\lim}f(x)=−\infty

if for all $M<0$ , there exists an $N>0$ such that

f(x)<M

for all $x>N$ .

Similarly we can define limits as $x\to −\infty$ .

The function f(x) is graphed. It continues to increase rapidly after x = N, and f(N) = M.

Figure 10. For a function with an infinite limit at infinity, for all

x>N

f(x)>M

Earlier, we used graphical evidence ((Figure)) and numerical evidence ((Figure)) to conclude that $\underset{x\to \infty }{\lim}x^3=\infty$ . Here we use the formal definition of infinite limit at infinity to prove that result.

An Infinite Limit at Infinity

Use the formal definition of infinite limit at infinity to prove that $\underset{x\to \infty }{\lim}x^3=\infty$ .

Answer:

Let $M>0$ . Let $N=\sqrt[3]{M}$ . Then, for all $x>N$ , we have

x^3>N^3=(\sqrt[3]{M})^3=M

Therefore, $\underset{x\to \infty }{\lim}x^3=\infty$ .

Use the formal definition of infinite limit at infinity to prove that $\underset{x\to \infty }{\lim}3x^2=\infty$ .

Answer:

Let $M>0$ . Let $N=\sqrt{\frac{M}{3}}$ . Then, for all $x>N$ , we have

$3x^2>3N^2=3(\sqrt{\frac{M}{3}})^2=\frac{3M}{3}=M$

Hint

Let $N=\sqrt{\frac{M}{3}}$ .

Key Equations

Infinite Limits from the Left $\underset{x\to a^-}{\lim}f(x)=+\infty$ $\underset{x\to a^-}{\lim}f(x)=−\infty$
Infinite Limits from the Right $\underset{x\to a^+}{\lim}f(x)=+\infty$ $\underset{x\to a^+}{\lim}f(x)=−\infty$
Two-Sided Infinite Limits $\underset{x\to a}{\lim}f(x)=+\infty: \underset{x\to a^-}{\lim}f(x)=+\infty[/latex] and [latex]\underset{x\to a^+}{\lim}f(x)=+\infty$ $\underset{x\to a}{\lim}f(x)=−\infty: \underset{x\to a^-}{\lim}f(x)=−\infty$ and $\underset{x\to a^+}{\lim}f(x)=−\infty$

In the following exercises, use direct substitution to obtain an undefined expression. Then, use the method of (Figure) to simplify the function to help determine the limit.

1. $\underset{x\to -2^-}{\lim}\frac{2x^2+7x-4}{x^2+x-2}$

Answer:

$-\infty$

2. $\underset{x\to -2^+}{\lim}\frac{2x^2+7x-4}{x^2+x-2}$

3. $\underset{x\to 1^-}{\lim}\frac{2x^2+7x-4}{x^2+x-2}$

Answer:

$-\infty$

4. $\underset{x\to 1^+}{\lim}\frac{2x^2+7x-4}{x^2+x-2}$

For the following exercises, evaluate the limit.

5. $\underset{x\to \infty }{\lim}\frac{1}{3x+6}$

Answer:

6. $\underset{x\to \infty }{\lim}\frac{2x-5}{4x}$

7. $\underset{x\to \infty }{\lim}\frac{x^2-2x+5}{x+2}$

Answer:

$\infty$

8. $\underset{x\to −\infty }{\lim}\frac{3x^3-2x}{x^2+2x+8}$

9. $\underset{x\to −\infty }{\lim}\frac{x^4-4x^3+1}{2-2x^2-7x^4}$

Answer:

$-\frac{1}{7}$

10. $\underset{x\to \infty }{\lim}\frac{3x}{\sqrt{x^2+1}}$

11. $\underset{x\to −\infty }{\lim}\frac{\sqrt{4x^2-1}}{x+2}$

Answer:

-2

12. $\underset{x\to \infty }{\lim}\frac{4x}{\sqrt{x^2-1}}$

13. $\underset{x\to −\infty }{\lim}\frac{4x}{\sqrt{x^2-1}}$

Answer:

-4

14. $\underset{x\to \infty }{\lim}\frac{2\sqrt{x}}{x-\sqrt{x}+1}$

For the following exercises, find the horizontal and vertical asymptotes.

15. $f(x)=x-\frac{9}{x}$

Answer:

Horizontal: none, vertical: $x=0$

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

17. $f(x)=\frac{x^3}{4-x^2}$

Answer:

Horizontal: none, vertical: $x=\pm 2$

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

19. $f(x)= \sin (x) \sin (2x)$

Answer:

Horizontal: none, vertical: none

20. $f(x)= \cos x+ \cos (3x)+ \cos (5x)$

21. $f(x)=\frac{x \sin (x)}{x^2-1}$

Answer:

Horizontal: $y=0$ , vertical: $x=\pm 1$

$f(x)=\frac{x}{ \sin (x)}$

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

Answer:

Horizontal: $y=0$ , vertical: $x=0$ and $x=-1$

23. $f(x)=\frac{1}{x-1}-2x$

24. $f(x)=\frac{x^3+1}{x^3-1}$

Answer:

Horizontal: $y=1$ , vertical: $x=1$

25. $f(x)=\frac{ \sin x+ \cos x}{ \sin x- \cos x}$

26. $f(x)=x- \sin x$

Answer:

Horizontal: none, vertical: none

27. $f(x)=\frac{1}{x}-\sqrt{x}$

In the following exercises, sketch the graph of a function with the given properties.

28. $\underset{x\to -\infty }{\lim}f(x)=0, \, \underset{x\to -1^-}{\lim}f(x)=−\infty$ , $\underset{x\to -1^+}{\lim}f(x)=\infty, \, \underset{x\to 0}{\lim}f(x)=f(0), \, f(0)=1, \, \underset{x\to \infty }{\lim}f(x)=−\infty$

Answer:

Answers may vary.

A graph of a piecewise function with two segments. The first segment is in quadrant three and asymptotically goes to negative infinity along the y axis and 0 along the x axis. The second segment consists of two curves. The first appears to be the left half of an upward opening parabola with vertex at (0,1). The second appears to be the right half of a downward opening parabola with vertex at (0,1) as well.

29. $\underset{x\to -\infty}{\lim}f(x)=2, \, \underset{x\to 3^-}{\lim}f(x)=−\infty$ , $\underset{x\to 3^+}{\lim}f(x)=\infty, \, \underset{x\to \infty }{\lim}f(x)=2, \, f(0)=\frac{-1}{3}$

30. $\underset{x\to -\infty }{\lim}f(x)=2, \, \underset{x\to -2}{\lim}f(x)=−\infty$ , $\underset{x\to \infty }{\lim}f(x)=2, \, f(0)=0$

Answer:

Answers may vary.

A graph containing two curves. The first goes to 2 asymptotically along y=2 and to negative infinity along x = -2. The second goes to negative infinity along x=-2 and to 2 along y=2.

31. $\underset{x\to -\infty }{\lim}f(x)=0, \, \underset{x\to -1^-}{\lim}f(x)=\infty, \, \underset{x\to -1^+}{\lim}f(x)=−\infty, \, f(0)=-1, \, \underset{x\to 1^-}{\lim}f(x)=−\infty, \, \underset{x\to 1^+}{\lim}f(x)=\infty, \, \underset{x\to \infty }{\lim}f(x)=0$

32. [T] In physics, the magnitude of an electric field generated by a point charge at a distance $r$ in vacuum is governed by Coulomb’s law: $E(r)=\large \frac{q}{4\pi \epsilon_0 r^2}$ , where $E$ represents the magnitude of the electric field, $q$ is the charge of the particle, $r$ is the distance between the particle and where the strength of the field is measured, and $\large \frac{1}{4\pi \epsilon_0}$ is Coulomb’s constant: $8.988 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$ .

Use a graphing calculator to graph $E(r)$ given that the charge of the particle is $q=10^{-10}$ .
Evaluate $\underset{r\to 0^+}{\lim}E(r)$ . What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?

Answer:

A graph of a function with two curves. The first is in quadrant two and curves asymptotically to infinity along the y axis and to 0 along the x axis as x goes to negative infinity. The second is in quadrant one and curves asymptotically to infinity along the y axis and to 0 along the x axis as x goes to infinity.

\underset{r\to 0^+}{\lim}E(r)=\infty

. The magnitude of the electric field as you approach the particle

q

becomes infinite. It does not make physical sense to evaluate negative distance.

33. [T] The density of an object is given by its mass divided by its volume: $\rho =m/V$ .

Use a calculator to plot the volume as a function of density $(V=m/\rho)$ , assuming you are examining something of mass 8 kg ( $m=8$ ).
Evaluate $\underset{\rho \to 0^+}{\lim}V(\rho)$ and explain the physical meaning.

Limits at Infinity; Asymptotes of Graphs

Learning Objectives

Definition

Recognizing an Infinite Limit

Hint

Evaluating a Limit of the Form $K/0, \, K\ne 0$ Using the Limit Laws

Hint

Infinite Limits from Positive Integers

Definition

Finding a Vertical Asymptote

Hint

Behavior of a Function at Different Points

Hint

Limits at Infinity and Horizontal Asymptotes

Definition

Definition

Computing Limits at Infinity

Hint

Special Limits with Sine and Cosine

Definition

Special limits with sine and cosine

Infinite Limits at Infinity

Definition

Formal Definitions

Definition

A Finite Limit at Infinity Example

Hint

Definition

An Infinite Limit at Infinity

Hint

Key Equations

Licenses & Attributions

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Study Guides > Calculus Volume 1

Limits at Infinity; Asymptotes of Graphs

Learning Objectives

Definition

Recognizing an Infinite Limit

Hint

Evaluating a Limit of the Form K/0, K≠0K/0, \, K\ne 0K/0,K=0 Using the Limit Laws

Hint

Infinite Limits from Positive Integers

Definition

Finding a Vertical Asymptote

Hint

Behavior of a Function at Different Points

Hint

Limits at Infinity and Horizontal Asymptotes

Definition

Definition

Computing Limits at Infinity

Hint

Special Limits with Sine and Cosine

Definition

Special limits with sine and cosine

Infinite Limits at Infinity

Definition

Formal Definitions

Definition

A Finite Limit at Infinity Example

Hint

Definition

An Infinite Limit at Infinity

Hint

Key Equations

Licenses & Attributions

CC licensed content, Shared previously

Evaluating a Limit of the Form $K/0, \, K\ne 0$ Using the Limit Laws