math homework 3

S E C T I O N 1.1 Real Numbers, Functions, and Graphs 3

EXAMPLE 1 Describe [7, 13] using inequalities. Solution The midpoint of the interval [7, 13] is c = 12 (7 + 13) = 10 and its radius is r = 12 (13 − 7) = 3 (Figure 8). Therefore,137

3 3

10

FIGURE 8 The interval [7, 13] is described by |x − 10| ≤ 3.

[7, 13] = { x ∈ R : |x − 10| ≤ 3

}

EXAMPLE 2 Describe the set S = { x :

∣∣ 1 2x − 3

∣∣ > 4 }

in terms of intervals.

Solution It is easier to consider the opposite inequality ∣∣ 1

2x − 3 ∣∣ ≤ 4 first. By (2),

In Example 2 we use the notation ∪ to denote “union”: The union A ∪ B of sets A and B consists of all elements that belong to either A or B (or to both).

∣∣∣∣ 1 2 x − 3

∣∣∣∣ ≤ 4 ⇔ −4 ≤ 1 2 x − 3 ≤ 4

−1 ≤ 1 2 x ≤ 7 (add 3)

−2 ≤ x ≤ 14 (multiply by 2) Thus,

∣∣ 1 2x − 3

∣∣ ≤ 4 is satisfied when x belongs to [−2, 14]. The set S is the complement,−2 0 14 FIGURE 9 The set S =

{ x :

∣∣ 1 2x − 3

∣∣ > 4 } .

consisting of all numbers x not in [−2, 14]. We can describe S as the union of two intervals: S = (−∞, −2) ∪ (14, ∞) (Figure 9).

Graphing Graphing is a basic tool in calculus, as it is in algebra and trigonometry. Recall that rect- angular (or Cartesian) coordinates in the plane are defined by choosing two perpendicular axes, the x-axis and the y-axis.To a pair of numbers (a, b) we associate the point P located

The term “Cartesian” refers to the French philosopher and mathematician René Descartes (1596–1650), whose Latin name was Cartesius. He is credited (along with Pierre de Fermat) with the invention of analytic geometry. In his great work La Géométrie, Descartes used the letters x, y, z for unknowns and a, b, c for constants, a convention that has been followed ever since.

at the intersection of the line perpendicular to the x-axis at a and the line perpendicular to the y-axis at b [Figure 10(A)]. The numbers a and b are the x- and y-coordinates of P . The x-coordinate is sometimes called the “abscissa” and the y-coordinate the “ordinate.” The origin is the point with coordinates (0, 0).

xx

b

aa

yy

21−1−2 −1

−2

2

1

P = (a, b)

(A) (B)FIGURE 10 Rectangular coordinate system.

The axes divide the plane into four quadrants labeled I–IV, determined by the signs of the coordinates [Figure 10(B)]. For example, quadrant III consists of points (x, y) such that x < 0 and y < 0.

The distance d between two points P1 = (x1, y1) and P2 = (x2, y2) is computed

d

x1

P1 = (x1, y1)

P2 = (x2, y2)

x2

y1

y2

|y2 − y1|

|x2 − x1|

x

y

FIGURE 11 Distance d is given by the distance formula.

using the Pythagorean Theorem. In Figure 11, we see that P1P2 is the hypotenuse of a right triangle with sides a = |x2 − x1| and b = |y2 − y1|. Therefore,

d2 = a2 + b2 = (x2 − x1)2 + (y2 − y1)2

We obtain the distance formula by taking square roots.

Distance Formula The distance between P1 = (x1, y1) and P2 = (x2, y2) is equal to

d = √

(x2 − x1)2 + (y2 − y1)2

4 C H A P T E R 1 PRECALCULUS REVIEW

Once we have the distance formula, we can derive the equation of a circle of radius r and center (a, b) (Figure 12). A point (x, y) lies on this circle if the distance from (x, y)

a

(a, b)

(x, y)

r

b

x

y

FIGURE 12 Circle with equation (x − a)2 + (y − b)2 = r2.

to (a, b) is r: √

(x − a)2 + (y − b)2 = r Squaring both sides, we obtain the standard equation of the circle:

(x − a)2 + (y − b)2 = r2

We now review some definitions and notation concerning functions.

DEFINITION A function f from a set D to a set Y is a rule that assigns, to each element x in D, a unique element y = f (x) in Y . We write

f : D → Y

The set D, called the domain of f , is the set of “allowable inputs.” For x ∈ D, f (x) is called the value of f at x (Figure 13). The range R of f is the subset of Y consisting of all values f (x):

R = {y ∈ Y : f (x) = y for some x ∈ D} Informally, we think of f as a “machine” that produces an output y for every input xA function f : D → Y is also called a

“map.” The sets D and Y can be arbitrary. For example, we can define a map from the set of living people to the set of whole numbers by mapping each person to his or her year of birth. The range of this map is the set of years in which a living person was born. In multivariable calculus, the domain might be a set of points in the two-dimensional plane and the range a set of numbers, points, or vectors.

in the domain D (Figure 14).

f (x)x

Domain D Y

f

FIGURE 13 A function assigns an element f (x) in Y to each x ∈ D.

f (x) Output

x Input

Machine “f ”

FIGURE 14 Think of f as a “machine” that takes the input x and produces the output f (x).

The first part of this text deals with numerical functions f , where both the domain and the range are sets of real numbers. We refer to such a function as f and its value at x as f (x). The letter x is used often to denote the independent variablethat can take on any value in the domain D. We write y = f (x) and refer to y as the dependent variable (because its value depends on the choice of x).

When f is defined by a formula, its natural domain is the set of real numbers x for which the formula is meaningful. For example, the function f (x) =

√ 9 − x has domain

D = {x : x ≤ 9} because √

9 − x is defined if 9 − x ≥ 0. Here are some other examples of domains and ranges:

f (x) Domain D Range R

x2 R {y : y ≥ 0} cos x R {y : −1 ≤ y ≤ 1}

1 x + 1 {x : x ̸= −1} {y : y ̸= 0}

The graph of a function y = f (x) is obtained by plotting the points (a, f (a)) for a in the domain D (Figure 15). If you start at x = a on the x-axis, move up to the graph

x

y = f (x)

Zero of f

f (a) (a, f (a))

a c

y

FIGURE 15

and then over to the y-axis, you arrive at the value f (a). The absolute value |f (a)| is the distance from the graph to the x-axis.

A zero or root of a function f is a number c such that f (c) = 0. The zeros are the values of x where the graph intersects the x-axis.

In Chapter 4, we will use calculus to sketch and analyze graphs. At this stage, to sketch a graph by hand, we can make a table of function values, plot the corresponding points (including any zeros), and connect them by a smooth curve.

S E C T I O N 1.1 Real Numbers, Functions, and Graphs 5

EXAMPLE 3 Find the roots and sketch the graph of f (x) = x3 − 2x. Solution First, we solve

x3 − 2x = x(x2 − 2) = 0

The roots of f are x = 0 and x = ± √

2. To sketch the graph, we plot the roots and a few values listed in Table 1 and join them by a curve (Figure 16).

TABLE 1

x x3 − 2x

−2 −4 −1 1

0 0 1 −1 2 4

−1−2

−4

−1

4

1

2

1

2

2− x

y

FIGURE 16 Graph of f (x) = x3 − 2x.

Functions arising in applications are not always given by formulas. For example, data collected from observation or experiment define functions for which there may be no exact formula. Such functions can be displayed either graphically or by a table of values. Figure 17 and Table 2 display data collected by biologist Julian Huxley (1887–1975) in a study of the antler weight W of male red deer as a function of age t . We will see that many of the tools from calculus can be applied to functions constructed from data in this way.

Antler weight W (kg)

0 20 4 6 8 10 12

Age t (years)

1

2

3

4

5

6

7

8

FIGURE 17 Male red deer shed their antlers every winter and regrow them in the spring. This graph shows average antler weight as a function of age.

TABLE 2

t (years) W (kg) t (years) W (kg)

1 0.48 7 5.34 2 1.59 8 5.62 3 2.66 9 6.18 4 3.68 10 6.81 5 4.35 11 6.21 6 4.92 12 6.1

We can graph not just functions but, more generally, any equation relating y and x.

−1 1

(1, 1)

(1, −1)

x

y

1

−1

FIGURE 18 Graph of 4y2 − x3 = 3. This graph fails the Vertical Line Test, so it is not the graph of a function.

Figure 18 shows the graph of the equation 4y2 − x3 = 3; it consists of all pairs (x, y) satisfying the equation. This curve is not the graph of a function because some x-values are associated with two y-values. For example, x = 1 is associated with y = ±1. A curve is the graph of a function if and only if it passes the Vertical Line Test; that is, every vertical line x = a intersects the curve in at most one point.

We are often interested in whether a function is increasing or decreasing. Roughly speaking, a function f is increasing if its graph goes up as we move to the right and is decreasing if its graph goes down [Figures 19(A) and (B)]. More precisely, we define the notion of increase/decrease on an open interval.

A function f is:

• increasing on (a, b) if f (x1) < f (x2) for all x1, x2 ∈ (a, b) such that x1 < x2. • decreasing on (a, b) if f (x1) > f (x2) for all x1, x2 ∈ (a, b) such that x1 < x2.

6 C H A P T E R 1 PRECALCULUS REVIEW

We say that f is monotonic if it is either increasing or decreasing. In Figure 19(C), the function is not monotonic because it is neither increasing nor decreasing for all x.

A function f is called nondecreasing if f (x1) ≤ f (x2) for x1 < x2 (defined by ≤ rather than a strict inequality <). Nonincreasing functions are defined similarly. Function (D) in Figure 19 is nondecreasing, but it is not increasing on the intervals where the graph is horizontal. Function (E) is increasing everywhere even though it levels off momentarily.

(A) Increasing (C)(B) Decreasing Decreasing on (a, b) but not decreasing everywhere

(D) Nondecreasing but not increasing

(E) Increasing

x

y

x

y

x

y

a b x

y

x

y

FIGURE 19 Another important property is parity, which refers to whether a function is even or

odd:

• f is even if f (−x) = f (x) • f is odd if f (−x) = −f (x)

The graphs of functions with even or odd parity have a special symmetry:

• Even function: Graph is symmetric about the y-axis. This means that if P = (a, b) lies on the graph, then so does Q = (−a, b) [Figure 20(A)].

• Odd function: Graph is symmetric with respect to the origin. This means that if P = (a, b) lies on the graph, then so does Q = (−a, −b) [Figure 20(B)].

Many functions are neither even nor odd [Figure 20(C)].

(A) Even function: f (−x) = f (x) Graph is symmetric about the y-axis.

(B) Odd function: f (−x) = − f (x) Graph is symmetric about the origin.

(C) Neither even nor odd

(a, b)

(a, b)

(−a, b) b

a−a (−a, −b)

b

a

−a

−b

x x x

y

y

y

FIGURE 20

EXAMPLE 4 Determine whether the function is even, odd, or neither.

(a) f (x) = x4 (b) g(x) = x−1 (c) h(x) = x2 + x Solution

(a) f (−x) = (−x)4 = x4. Thus, f (x) = f (−x), and f is even. (b) g(−x) = (−x)−1 = −x−1. Thus, g(−x) = −g(x), and g is odd. (c) h(−x) = (−x)2 + (−x) = x2 − x. We see that h(−x) is not equal to h(x) or to −h(x) = −x2 − x. Therefore, h is neither even nor odd.

EXAMPLE 5 Using Symmetry Sketch the graph of f (x) = 1 x2 + 1 .

Solution The function f is positive [f (x) > 0] and even [f (−x) = f (x)]. Therefore, the graph lies above the x-axis and is symmetric with respect to the y-axis. Furthermore,

S E C T I O N 1.1 Real Numbers, Functions, and Graphs 7

f is decreasing for x ≥ 0 (because a larger value of x makes the denominator larger). We use this information and a short table of values (Table 3) to sketch the graph (Figure 21). Note that the graph approaches the x-axis as we move to the right or left because f (x) gets closer to 0 as |x| increases.

TABLE 3

x 1

x2 + 1

0 1

±1 12 ±2 15

1

−1−2 21 x

y

f (x) = 1 x2 + 1

FIGURE 21

Two important ways of modifying a graph are translation (or shifting) and scaling. Translation consists of moving the graph horizontally or vertically:

DEFINITION Translation (Shifting)

• Vertical translation y = f (x) + c: Shifts the graph by |c| units vertically, upward if c > 0 and downward if c < 0.

• Horizontal translation y = f (x + c): Shifts the graph by |c| units horizontally, to the right if c < 0 and c units to the left if c > 0.

Figure 22 shows the effect of translating the graph of f (x) = 1/(x2 + 1) vertically and

Remember that f (x) + c and f (x + c) are different. The graph of y = f (x) + c is a vertical translation and y = f (x + c) a horizontal translation of the graph of y = f (x).

horizontally.

−1−2 21 x

y

1

2 Shift 1 unit

upward Shift 1 unit to the left

−1−2−3 1 x

y

1

2

−1−2 21 x

y

1

2

(A) y = f (x) = 1 + 1 x2 + 1

(B) y = f (x) + 1 = 1 x2 + 1

(C) y = f (x + 1) = 1 (x + 1)2 + 1

FIGURE 22

EXAMPLE 6 Figure 23(A) is the graph of f (x) = x2, and Figure 23(B) is a horizontal and vertical shift of (A). What is the equation of graph (B)?

−1−2 2 31

2

1

4

3

−1

(A) f (x) = x2 (B)

x

y

−1−2 2 31

2

1

4

3

−1

x

y

FIGURE 23

Solution Graph (B) is obtained by shifting graph (A) 1 unit to the right and 1 unit down. We can see this by observing that the point (0, 0) on the graph of f is shifted to (1, −1). Therefore, (B) is the graph of g(x) = (x − 1)2 − 1.

8 C H A P T E R 1 PRECALCULUS REVIEW

Scaling (also called dilation) consists of compressing or expanding the graph in the

y = −2 f (x)

y = f (x) 2

1

−2

−4

x

y

FIGURE 24 Negative vertical scale factor k = −2.

vertical or horizontal directions:

DEFINITION Scaling

• Vertical scaling y = kf (x): If k > 1, the graph is expanded vertically by the factor k. If 0 < k < 1, the graph is compressed vertically. When the scale factor k is negative (k < 0), the graph is also reflected across the x-axis (Figure 24).

• Horizontal scaling y = f (kx): If k > 1, the graph is compressed in the horizontal direction. If 0 < k < 1, the graph is expanded. If k < 0, then the graph is also reflected across the y-axis.

The amplitude of a function is half the difference between its greatest value and its least value, if it has both a greatest value and least value. Thus, vertical scaling changes the amplitude by the factor |k|.

EXAMPLE 7 Sketch the graphs of f (x) = sin(πx) and its dilates f (3x) and 3f (x). Solution The graph of f (x) = sin(πx) is a sine curve with period 2. It completes one cycle over every interval of length 2—see Figure 25(A). It has amplitude 1.

• The graph of f (3x) = sin(3πx) is a compressed version of y = f (x), completing three cycles instead of one over intervals of length 2 [Figure 25(B)]. It also has amplitude 1.

• The graph of y = 3f (x) = 3 sin(πx) differs from y = f (x) only in amplitude: It is expanded in the vertical direction by a factor of 3 [Figure 25(C)], so its amplitude is 3.

(C) Vertical expansion: y = 3 f (x) = 3sin(πx)

(B) Horizontal compression: y = f (3x) = sin(3πx)

1

2

3

−3

−2

−1

1

−1 2 41 3 2 41 3

(A) y = f (x) = sin(πx)

1

−1 2 41 3 xx x

y

yy

One cycle Three cycles

FIGURE 25 Horizontal and vertical scaling of f (x) = sin(πx).

1.1 SUMMARY

• Absolute value: |a| = {

a if a ≥ 0 −a if a < 0

• Triangle inequality: |a + b| ≤ |a| + |b| • Four intervals with endpoints a and b:

(a, b), [a, b], [a, b), (a, b] • Writing open and closed intervals using inequalities:

(a, b) = {x : |x − c| < r}, [a, b] = {x : |x − c| ≤ r}

where c = 12 (a + b) is the midpoint and r = 12 (b − a) is the radius.

S E C T I O N 1.1 Real Numbers, Functions, and Graphs 9

• Distance d between (x1, y1) and (x2, y2):

d = √

(x2 − x1)2 + (y2 − y1)2

• Equation of circle of radius r with center (a, b):

(x − a)2 + (y − b)2 = r2

• A zero or root of a function f is a number c such that f (c) = 0. • Vertical Line Test: A curve in the plane is the graph of a function if and only if each

vertical line x = a intersects the curve in at most one point.

•

Increasing: f (x1) < f (x2) if x1 < x2 Nondecreasing: f (x1) ≤ f (x2) if x1 < x2 Decreasing: f (x1) > f (x2) if x1 < x2 Nonincreasing: f (x1) ≥ f (x2) if x1 < x2

• Even function: f (−x) = f (x) (graph is symmetric about the y-axis). • Odd function: f (−x) = −f (x) (graph is symmetric about the origin). • Four ways to transform the graph of f :

f (x) + c Shifts graph vertically |c| units (upward if c > 0, downward if c < 0) f (x + c) Shifts graph horizontally |c| units (to the right if c < 0, to the left if c > 0) kf (x) Scales graph vertically by factor k;

if k < 0, graph is reflected across x-axis

f (kx) Scales graph horizontally by factor k (compresses if k > 1); if k < 0, graph is reflected across y-axis

1.1 EXERCISES

Preliminary Questions 1. Give an example of numbers a and b such that a < b and |a| > |b|.

2. Which numbers satisfy |a| = a? Which satisfy |a| = −a? What about |−a| = a?

3. Give an example of numbers a and b such that |a + b| < |a| + |b|.

4. Are there numbers a and b such that |a + b| > |a| + |b|?

5. What are the coordinates of the point lying at the intersection of the lines x = 9 and y = −4?

6. In which quadrant do the following points lie? (a) (1, 4) (b) (−3, 2) (c) (4, −3) (d) (−4, −1) 7. What is the radius of the circle with equation

(x − 7)2 + (y − 8)2 = 9? 8. The equation f (x) = 5 has a solution if (choose one):

(a) 5 belongs to the domain of f . (b) 5 belongs to the range of f .

9. What kind of symmetry does the graph have if f (−x) = −f (x)? 10. Is there a function that is both even and odd?

Exercises 1. Use a calculator to find a rational number r such that

|r − π2| < 10−4.

2. Which of (a)–(f) are true for a = −3 and b = 2? (a) a < b (b) |a| < |b| (c) ab > 0

(d) 3a < 3b (e) −4a < −4b (f) 1 a

< 1 b

In Exercises 3–8, express the interval in terms of an inequality involving absolute value.

3. [−2, 2] 4. (−4, 4) 5. (0, 4)

6. [−4, 0] 7. [1, 5] 8. (−2, 8)

In Exercises 9–12, write the inequality in the form a < x < b.

9. |x| < 8 10. |x − 12| < 8

11. |2x + 1| < 5 12. |3x − 4| < 2

In Exercises 13–18, express the set of numbers x satisfying the given condition as an interval.

13. |x| < 4 14. |x| ≤ 9

15. |x − 4| < 2 16. |x + 7| < 2

17. |4x − 1| ≤ 8 18. |3x + 5| < 1

10 C H A P T E R 1 PRECALCULUS REVIEW

In Exercises 19–22, describe the set as a union of finite or infinite in- tervals.

19. {x : |x − 4| > 2} 20. {x : |2x + 4| > 3}

21. {x : |x2 − 1| > 2} 22. {x : |x2 + 2x| > 2}

23. Match (a)–(f) with (i)–(vi).

(a) a > 3 (b) |a − 5| < 1 3

(c) ∣∣∣∣a −

1 3

∣∣∣∣ < 5 (d) |a| > 5

(e) |a − 4| < 3 (f) 1 ≤ a ≤ 5

(i) a lies to the right of 3.

(ii) a lies between 1 and 7.

(iii) The distance from a to 5 is less than 13 .

(iv) The distance from a to 3 is at most 2.

(v) a is less than 5 units from 13 .

(vi) a lies either to the left of −5 or to the right of 5.

24. Describe { x : x

x + 1 < 0 }

as an interval. Hint: Consider the sign

of x and x + 1 individually.

25. Describe {x : x2 + 2x < 3} as an interval. Hint: Plot y = x2 + 2x − 3.

26. Describe the set of real numbers satisfying |x − 3| = |x − 2| + 1 as a half-infinite interval.

27. Show that if a > b, and a, b ̸= 0, then b−1 > a−1, provided that a and b have the same sign. What happens if a > 0 and b < 0?

28. Which x satisfies both |x − 3| < 2 and |x − 5| < 1?

29. Show that if |a − 5| < 12 and |b − 8| < 12 , then |(a + b) − 13| < 1. Hint: Use the triangle inequality (|a + b| ≤ |a| + |b|).

30. Suppose that |x − 4| ≤ 1. (a) What is the maximum possible value of |x + 4|? (b) Show that |x2 − 16| ≤ 9.

31. Suppose that |a − 6| ≤ 2 and |b| ≤ 3. (a) What is the largest possible value of |a + b|? (b) What is the smallest possible value of |a + b|?

32. Prove that |x| − |y| ≤ |x − y|. Hint: Apply the triangle inequality to y and x − y.

33. Express r1 = 0.27 as a fraction. Hint: 100r1 − r1 is an integer. Then express r2 = 0.2666 . . . as a fraction.

34. Represent 1/7 and 4/27 as repeating decimals.

35. The text states: If the decimal expansions of numbers a and b agree to k places, then |a − b| ≤ 10−k . Show that the converse is false: For all k there are numbers a and b whose decimal expansions do not agree at all but |a − b| ≤ 10−k .

36. Plot each pair of points and compute the distance between them: (a) (1, 4) and (3, 2) (b) (2, 1) and (2, 4)

(c) (0, 0) and (−2, 3) (d) (−3, −3) and (−2, 3)

37. Find the equation of the circle with center (2, 4): (a) with radius r = 3. (b) that passes through (1, −1). 38. Find all points in the xy-plane with integer coordinates located at a distance 5 from the origin. Then find all points with integer coordinates located at a distance 5 from (2, 3).

39. Determine the domain and range of the function

f : {r, s, t, u} → {A,B,C,D,E}

defined by f (r) = A, f (s) = B, f (t) = B, f (u) = E. 40. Give an example of a function whose domain D has three elements and whose range R has two elements. Does a function exist whose do- main D has two elements and whose range R has three elements?

In Exercises 41–48, find the domain and range of the function.

41. f (x) = −x 42. g(t) = t4

43. f (x) = x3 44. g(t) = √

2 − t

45. f (x) = |x| 46. h(s) = 1 s

47. f (x) = 1 x2

48. g(t) = cos 1 t

In Exercises 49–52, determine where f is increasing.

49. f (x) = |x + 1| 50. f (x) = x3

51. f (x) = x4 52. f (x) = 1 x4 + x2 + 1

In Exercises 53–58, find the zeros of f and sketch its graph by plot- ting points. Use symmetry and increase/decrease information where appropriate.

53. f (x) = x2 − 4 54. f (x) = 2x2 − 4

55. f (x) = x3 − 4x 56. f (x) = x3

57. f (x) = 2 − x3 58. f (x) = 1 (x − 1)2 + 1

59. Which of the curves in Figure 26 is the graph of a function?

(B)

(D)(C)

(A)

x

x

x

x

y

y

y

y

FIGURE 26

60. Determine whether the function is even, odd, or neither. (a) f (x) = x5 (b) g(t) = t3 − t2

(c) F(t) = 1 t4 + t2

S E C T I O N 1.1 Real Numbers, Functions, and Graphs 11

61. Determine whether the function is even, odd, or neither.

(a) f (t) = 1 t4 + t + 1 −

1

t4 − t + 1 (b) g(t) = 2 t − 2−t

(c) G(θ) = sin θ + cos θ (d) H(θ) = sin(θ2)

62. Write f (x) = 2x4 − 5x3 + 12x2 − 3x + 4 as the sum of an even and an odd function.

63. Show that f (x) = ln (

1 − x 1 + x

) is an odd function.

64. State whether the function is increasing, decreasing, or neither. (a) Surface area of a sphere as a function of its radius

(b) Temperature at a point on the equator as a function of time

(c) Price of an airline ticket as a function of the price of oil

(d) Pressure of the gas in a piston as a function of volume

In Exercises 65–70, let f be the function shown in Figure 27.

65. Find the domain and range of f .

66. Sketch the graphs of y = f (x + 2) and y = f (x) + 2.

67. Sketch the graphs of y = f (2x), y = f ( 1

2x ) , and y = 2f (x).

68. Sketch the graphs of y = f (−x) and y = −f (−x).

69. Extend the graph of f to [−4, 4] so that it is an even function.

70. Extend the graph of f to [−4, 4] so that it is an odd function.

1 2 3 4 0

1

2

3

4

x

y

FIGURE 27

71. Suppose that f has domain [4, 8] and range [2, 6]. Find the domain and range of:

(a) y = f (x) + 3 (b) y = f (x + 3) (c) y = f (3x) (d) y = 3f (x)

72. Let f (x) = x2. Sketch the graph over [−2, 2] of: (a) y = f (x + 1) (b) y = f (x) + 1 (c) y = f (5x) (d) y = 5f (x)

73. Suppose that the graph of f (x) = sin x is compressed horizontally by a factor of 2 and then shifted 5 units to the right.

(a) What is the equation for the new graph?

(b) What is the equation if you first shift by 5 and then compress by 2?

(c) Verify your answers by plotting your equations.

74. Figure 28 shows the graph of f (x) = |x| + 1. Match the functions (a)–(e) with their graphs (i)–(v).

(a) y = f (x − 1) (b) y = −f (x) (c) y = −f (x) + 2 (d) y = f (x − 1) − 2 (e) y = f (x + 1)

y = f (x) = |x| + 1 (i) (ii)

1 2 3

−1−2−3 −1 2 31

y

x 1 2 3

−1−2−3 −1 2 31

y

x 1 2 3

−1−2−3 −1 2 31

y

x

(iv) (v)(iii)

1 2 3

−1 −2 −3

−2−3 −1 2 31

y

x 1 2 3

−1 −2 −3

−2−3 −1 2 31

y

x 1 2 3

−1 −2 −3

−2−3 −1 2 31

y

x

FIGURE 28

75. Sketch the graph of y = f (2x) and y = f ( 1

2x ) , where f (x) =

|x| + 1 (Figure 28).

76. Find the function f whose graph is obtained by shifting the parabola y = x2 by 3 units to the right and 4 units down, as in Fig- ure 29.

y = f (x)

y = x2

−4

3

y

x

FIGURE 29

77. Define f (x) to be the larger of x and 2 − x. Sketch the graph of f . What are its domain and range? Express f (x) in terms of the absolute value function.

78. For each curve in Figure 30, state whether it is symmetric with respect to the y-axis, the origin, both, or neither.

(D)

(B)

(C)

(A)

yy

yy

xx

x x

FIGURE 30

79. Show that the sum of two even functions is even and the sum of two odd functions is odd.

12 C H A P T E R 1 PRECALCULUS REVIEW

80. Suppose that f and g are both odd. Which of the following func- tions are even? Which are odd? (a) y = f (x)g(x) (b) y = f (x)3

(c) y = f (x) − g(x) (d) y = f (x) g(x)

81. Prove that the only function whose graph is symmetric with respect to both the y-axis and the origin is the function f (x) = 0.

Further Insights and Challenges 82. Prove the triangle inequality (|a + b| ≤ |a| + |b|) by adding the two inequalities

−|a| ≤ a ≤ |a|, −|b| ≤ b ≤ |b|

83. Show that a fraction r = a/b in lowest terms has a finite decimal expansion if and only if

b = 2n5m for some n, m ≥ 0.

Hint: Observe that r has a finite decimal expansion when 10Nr is an integer for some N ≥ 0 (and hence b divides 10N ).

84. Let p = p1 . . . ps be an integer with digits p1, . . . , ps . Show that p

10s − 1 = 0.p1 . . . ps

Use this to find the decimal expansion of r = 211 . Note that

r = 2 11

= 18 102 − 1

85. A function f is symmetric with respect to the vertical line x = a if f (a − x) = f (a + x). (a) Draw the graph of a function that is symmetric with respect to x = 2. (b) Show that if f is symmetric with respect to x = a, then g(x) = f (x + a) is even.

86. Formulate a condition for f to be symmetric with respect to the point (a, 0) on the x-axis.

1.2 Linear and Quadratic Functions Linear functions are the simplest of all functions, and their graphs (lines) are the simplest of all curves. However, linear functions and lines play an enormously important role in calculus. For this reason, you should be thoroughly familiar with the basic properties of linear functions and the different ways of writing an equation of a line.

Let’s recall that a linear functionis a function of the form

f (x) = mx + b (m and b constants)

The graph of f is a line of slope m, and since f (0) = b, the graph intersects the y-axis at the point (0, b) (Figure 1). The number b is called the y-intercept.

The slope-intercept form of the line with slope m and y-intercept b is given by

y = mx + b

x1 x2

y -intercept

y = mx + b

m =

!y

!y

!x

!x

y2

y1

b

y

x

FIGURE 1 The slope m is the ratio “rise over run.”

We use the symbols #x and #y to denote the change (or increment) in x and y = f (x) over an interval [x1, x2] (Figure 1):

#x = x2 − x1, #y = y2 − y1 = f (x2) − f (x1)

The slope m of a line is equal to the ratio

m = #y #x

= vertical change horizontal change

= rise run

S E C T I O N 1.2 Linear and Quadratic Functions 13

This follows from the formula y = mx + b: #y

#x = y2 − y1

x2 − x1 = (mx2 + b) − (mx1 + b)

x2 − x1 = m(x2 − x1)

x2 − x1 = m

The slope m measures the rate of change of y with respect to x. In fact, by writing

#y = m#x

we see that a 1-unit increase in x (i.e., #x = 1) produces an m-unit change #y in y. For example, if m = 5, then y increases by 5 units per unit increase in x. The rate-of-change interpretation of the slope is fundamental in calculus. We discuss it in greater detail in Section 2.1.

Graphically, the slope m measures the steepness of the line y = mx + b. Figure 2(A) shows lines through a point of varying slope m. Note the following properties:

• Steepness: The larger the absolute value |m|, the steeper the line. • Positive slope: If m > 0, the line slants upward from left to right. • Negative slope: If m < 0, the line slants downward from left to right. • f (x) = mx + b is increasing if m > 0 and decreasing if m < 0. • The horizontal line y = b has slope m = 0 [Figure 2(B)]. • A vertical line has equation x = c, where c is a constant. The slope of a vertical line

is undefined. It is not possible to write the equation of a vertical line in slope-intercept form y = mx + b. A vertical line is not the graph of a function [Figure 2(B)].

125

0.5−0.5

0

(A) Lines of varying slopes through P

−1 −2 −5

y = b (slope 0)

x = c (slope undefined)

c

b P P

(B) Horizontal and vertical lines through P

yy

xx

FIGURE 2

Scale is especially important in applications because the steepness of a graph depends

CAUTION Graphs are often plotted using different scales for the x- and y-axes. This is necessary to keep the sizes of graphs within reasonable bounds. However, when the scales are different, lines do not appear with their true slopes.

on the choice of units for the x- and y-axes. We can create very different subjective impressions by changing the scale. Figure 3 shows the growth of company profits over a 4-year period. The two plots convey the same information, but the left-hand plot makes the growth look more dramatic.

100

125

150

100 20112010 2012 2013 201420112010 2012 2013 2014

125 150 175 200 225 250 275 300

FIGURE 3 Growth of company profits.

14 C H A P T E R 1 PRECALCULUS REVIEW

Slope = m

Slope = m Slope = m

(B) Perpendicular lines (A) Parallel lines

yy

xx

Slope = − 1m

FIGURE 4 Parallel and perpendicular lines.

Next, we recall the relation between the slopes of parallel and perpendicular lines (Figure 4):

• Lines of slopes m1 and m2 are parallel if and only if m1 = m2. • Lines of slopes m1 and m2 are perpendicular if and only if

m1 = − 1

m2 (or m1m2 = −1)

CONCEPTUAL INSIGHT The increments over an interval [x1, x2]:

#x = x2 − x1, #y = f (x2) − f (x1)

are defined for any function f (linear or not), but the ratio #y/#x may depend on the interval (Figure 5). The characteristic property of a linear function f (x) = mx + b is that #y/#x has the same value m for every interval. In other words, y has a constant rate of change with respect to x. We can use this property to test if two quantities are related by a linear equation.

x

y

x

Nonlinear function: the ratio y / x changes, depending

on the interval.

Linear function: the ratio y / x is the same over

all intervals.

x

y

x

y

y

FIGURE 5

EXAMPLE 1 Testing for a Linear Relationship Do the data in Table 1 suggest a linear relation between the pressure P and temperature T of a gas?

TABLE 1

Temperature (◦C) Pressure (kPa)

40 1365.80 45 1385.40 55 1424.60 70 1483.40 80 1522.60

S E C T I O N 1.2 Linear and Quadratic Functions 15

Solution We calculate #P/#T at successive data points and check whether this ratio is constant:

(T1, P1) (T2, P2) #P

#T

(40, 1365.80) (45, 1385.40) 1385.40 − 1365.80

45 − 40 = 3.92

(45, 1385.40) (55, 1424.60) 1424.60 − 1385.40

55 − 45 = 3.92

(55, 1424.60) (70, 1483.40) 1483.40 − 1424.60

70 − 55 = 3.92

(70, 1483.40) (80, 1522.60) 1522.60 − 1483.40

80 − 70 = 3.92

Because #P/#T has the constant value 3.92, the data points lie on a line with slope m = 3.92. This is confirmed in the plot in Figure 6.

Real experimental data are unlikely to reveal perfect linearity, even if the data points do essentially lie on a line. The method of “linear regression” is used to find the linear function that best fits the data.

40 60 80

1350

1400

1450

1500

1550 Pressure (kPa)

T (°C)

FIGURE 6 Line through pressure- temperature data points.

As mentioned above, it is important to be familiar with the standard ways of writing the equation of a line. The general linear equation is

ax + by = c 1

where a and b are not both zero. For b = 0, we obtain the vertical line ax = c. When b ̸= 0, we can rewrite Eq. (1) in slope-intercept form. For example, −6x + 2y = 3 can be rewritten as y = 3x + 32 .

Two other forms we will use frequently are the point-slope and point-point forms.

(a1, b1)

(a2, b2)

b2 − b1

a2 − a1

x

y

FIGURE 7 Slope of the line between P = (a1, b1) and Q = (a2, b2) is m = b2 − b1

a2 − a1 .

Given a point P = (a, b) and a slope m, the equation of the line through P with slope m is y − b = m(x − a). Similarly, the line through two distinct points P = (a1, b1) and Q = (a2, b2) has slope (Figure 7)

m = b2 − b1 a2 − a1

Therefore, we can write its equation as y − b1 = m(x − a1).

If a = 0, point-slope form becomes slope-intercept form y = mx + b.

Additional Equations for Lines

1. Point-slope form of the line through P = (a, b) with slope m:

y − b = m(x − a)

2. Point-point form of the line through P = (a1, b1) and Q = (a2, b2):

y − b1 = m(x − a1), where m = b2 − b1 a2 − a1

16 C H A P T E R 1 PRECALCULUS REVIEW

EXAMPLE 2 Line of Given Slope Through a Given Point Find the equation of the line through (9, 2) with slope − 23 .

P = (9, 2)

x + 8y = −

2

8

x

y

9 12

2 3

FIGURE 8 Line through P = (9, 2) with slope m = − 23 .

Solution In point-slope form:

y − 2 = −2 3 (x − 9)

In slope-intercept form: y = − 23 (x − 9) + 2 or y = − 23x + 8. See Figure 8.

EXAMPLE 3 Line Through Two Points Find the equation of the line through (2, 1) and (9, 5).

Solution The line has slope

m = 5 − 1 9 − 2 =

4 7

Because (2, 1) lies on the line, its equation in point-slope form is y − 1 = 47 (x − 2).

A quadratic function is a function defined by a quadratic polynomial

f (x) = ax2 + bx + c (a, b, c, constants with a ̸= 0) The graph of f is a parabola (Figure 9). The parabola opens upward if the leading coefficient a is positive and downward if a is negative. The discriminant of f (x) is the quantity

D = b2 − 4ac The roots of f are given by the quadratic formula (see Exercise 58):

Roots of f = −b ± √

b2 − 4ac 2a

= −b ± √

D

2a

The sign of D determines whether or not f has real roots (Figure 9). If D > 0, then f has two real roots, and if D = 0, it has one real root (a “double root”). If D < 0, then

√ D is

imaginary and f has no real roots.

Two real roots a > 0 and D > 0

Double root a > 0 and D = 0

No real roots a > 0 and D < 0

Two real roots a < 0 and D > 0

xxxx

yyyy

FIGURE 9 Graphs of quadratic functions f (x) = ax2 + bx + c.

When f has two real roots r1 and r2, then f (x) factors as

f (x) = a(x − r1)(x − r2) For example, f (x) = 2x2 − 3x + 1 has discriminant D = b2 − 4ac = 9 − 8 = 1 > 0, and by the quadratic formula, its roots are (3 ± 1)/4 or 1 and 12 . Therefore,

f (x) = 2x2 − 3x + 1 = 2(x − 1) (

x − 1 2

)

The technique of completing the square consists of writing a quadratic polynomial as a multiple of a square plus a constant. Then

x2 + bx + c = x2 + bx + (

b

2

)2 −

( b

2

)2 + c =

( x + b

2

)2 −

( b

2

)2 + c

If there is a constant a multiplying the x2 term, we factor that out first, as demonstrated in the following example.

S E C T I O N 1.2 Linear and Quadratic Functions 17

EXAMPLE 4 Completing the Square Complete the square for the quadratic polyno- mial f (x) = 4x2 − 12x + 3.

Cuneiform texts written on clay tablets show that the method of completing the square was known to ancient Babylonian mathematicians who lived some 4000 years ago.

Solution First factor out the leading coefficient:

4x2 − 12x + 3 = 4 (

x2 − 3x + 3 4

)

Then complete the square for the term x2 − 3x: Ignoring air resistance, a basketball follows a parabolic path (Figure 10).

FIGURE 10

x2 − 3x = x2 − 3x + (

3 2

)2 −

( 3 2

)2 =

( x − 3

2

)2 − 9

4

Therefore,

4x2 − 12x + 3 = 4 ((

x − 3 2

)2 − 9

4 + 3

4

)

= 4 (

x − 3 2

)2 − 6

The method of completing the square can be used to find the minimum or maximum value of a quadratic function.

EXAMPLE 5 Finding the Maximum of a Quadratic Function Complete the square and find the maximum value of f (x) = −x2 + 4x + 1.

5

1

2 x

y

FIGURE 11 Graph of f (x) = −x2 + 4x + 1.

Solution We have

f (x) = −(x2 − 4x − 1) = −(x2 − 4x + 4 − 4 − 1) = −((x − 2)2 − 5) = This term is ≤ 0︷ ︸︸ ︷ −(x − 2)2 + 5

Thus, f (x) ≤ 5 for all x, and the maximum value of f is f (2) = 5 (Figure 11).

1.2 SUMMARY

• A linear function is a function of the form f (x) = mx + b. • The general equation of a line is ax + by = c. The line y = c is horizontal and x = c

is vertical. • Three convenient ways of writing the equation of a nonvertical line:

– Slope-intercept form: y = mx + b (slope m and y-intercept b) – Point-slope form: y − b = m(x − a) [slope m, passes through (a, b)] – Point-point form: The line through two points P = (a1, b1) and Q = (a2, b2) has

slope m = b2 − b1 a2 − a1

and equation y − b1 = m(x − a1).

• Two lines of slopes m1 and m2 are parallel if and only if m1 = m2, and they are perpen- dicular if and only if m1 = −1/m2.

• Quadratic function: f (x) = ax2 + bx + c. The roots are x = (−b ± √

D)/2a, where D = b2 − 4ac is the discriminant. The roots are real and distinct if D > 0, there is a double root if D = 0, and there are no real roots if D < 0.

• Completing the square consists of writing a quadratic function as a multiple of a square plus a constant.

1.2 EXERCISES

Preliminary Questions 1. What is the slope of the line y = −4x − 9? 2. Are the lines y = 2x + 1 and y = −2x − 4 perpendicular? 3. When is the line ax + by = c parallel to the y-axis? To the x-axis?

4. Suppose y = 3x + 2. What is #y if x increases by 3? 5. What is the minimum of f (x) = (x + 3)2 − 4? 6. What is the result of completing the square for f (x) = x2 + 1?

18 C H A P T E R 1 PRECALCULUS REVIEW

Exercises In Exercises 1–4, find the slope, the y-intercept, and the x-intercept of the line with the given equation.

1. y = 3x + 12 2. y = 4 − x

3. 4x + 9y = 3 4. y − 3 = 12 (x − 6) In Exercises 5–8, find the slope of the line.

5. y = 3x + 2 6. y = 3(x − 9) + 2 7. 3x + 4y = 12 8. 3x + 4y = −8

In Exercises 9–20, find the equation of the line with the given descrip- tion.

9. Slope 3, y-intercept 8

10. Slope −2, y-intercept 3 11. Slope 3, passes through (7, 9)

12. Slope −5, passes through (0, 0) 13. Horizontal, passes through (0, −2) 14. Passes through (−1, 4) and (2, 7) 15. Parallel to y = 3x − 4, passes through (1, 1) 16. Passes through (1, 4) and (12, −3) 17. Perpendicular to 3x + 5y = 9, passes through (2, 3) 18. Vertical, passes through (−4, 9) 19. Horizontal, passes through (8, 4)

20. Slope 3, x-intercept 6

21. Find the equation of the perpendicular bisector of the segment join- ing (1, 2) and (5, 4) (Figure 12). Hint: The midpoint Q of the segment

joining (a, b) and (c, d) is (

a + c 2

, b + d

2

) .

Q

(1, 2)

(5, 4)

Perpendicular bisector

x

y

FIGURE 12

22. Intercept-Intercept Form Show that if a, b ̸= 0, then the line with x-intercept x = a and y-intercept y = b has equation (Figure 13)

x

a + y

b = 1

b

a x

y

FIGURE 13

23. Find an equation of the line with x-intercept x = 4 and y-intercept y = 3.

24. Find y such that (3, y) lies on the line of slope m = 2 through (1, 4).

25. Determine whether there exists a constant c such that the line x + cy = 1: (a) has slope 4. (b) passes through (3, 1). (c) is horizontal. (d) is vertical.

26. Assume that the number N of concert tickets that can be sold at a price of P dollars per ticket is a linear function N(P ) for 10 ≤ P ≤ 40. Determine N(P ) (called the demand function) if N(10) = 500 and N(40) = 0. What is the decrease #N in the number of tickets sold if the price is increased by #P = 5 dollars?

27. Suppose that the number of a certain type of computer that can be sold when its price is P (in dollars) is given by a linear function N(P ). Determine N(P ) if N(1000) = 10,000 and N(1500) = 7,500. What is the change #N in the number of computers sold if the price is increased by #P = 100 dollars?

28. Suppose that the demand for Colin’s kidney pies is linear in the price P . Determine the demand function N as a function of P giving the number of pies sold when the price is P if he can sell 100 pies when the price is $5.00 and he can sell 40 pies when the price is $10.00. De- termine the revenue (N × P) for prices P = 5, 6, 7, 8, 9, 10 and then choose a price to maximize the revenue.

29. Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount #T , then the rod’s length approximately changes by #L = αL0#T , where α is the thermal expansion coefficient and #T is not an extreme temperature change. For steel, α = 1.24 × 10−5 ◦C−1. (a) A steel rod has length L0 = 40 cm at T0 = 40◦C. Find its length at T = 90◦C. (b) Find its length at T = 50◦C if its length at T0 = 100◦C is 65 cm. (c) Express length L as a function of T if L0 = 65 cm at T0 = 100◦C.

30. Do the points (0.5, 1), (1, 1.2), (2, 2) lie on a line?

31. Find b such that (2, −1), (3, 2), and (b, 5) lie on a line.

32. Find an expression for the velocity v as a linear function of t that matches the following data:

t (s) 0 2 4 6

v (m/s) 39.2 58.6 78 97.4

33. The period T of a pendulum is measured for pendulums of several different lengths L. Based on the following data, does T appear to be a linear function of L?

L (cm) 20 30 40 50

T (s) 0.9 1.1 1.27 1.42

34. Show that f is linear of slope m if and only if

f (x + h) − f (x) = mh (for all x and h)

That is to say, prove the following two statements: (a) f is linear of slope m implies that f (x + h) − f (x) = mh (for all x and h).

S E C T I O N 1.3 The Basic Classes of Functions 19

(b) f (x + h) − f (x) = mh (for all x and h) implies that f is linear of slope m.

35. Find the roots of the quadratic polynomials: (a) f (x) = 4x2 − 3x − 1 (b) f (x) = x2 − 2x − 1 In Exercises 36–43, complete the square and find the minimum or max- imum value of the quadratic function.

36. y = x2 + 2x + 5 37. y = x2 − 6x + 9

38. y = −9x2 + x 39. y = x2 + 6x + 2

40. y = 2x2 − 4x − 7 41. y = −4x2 + 3x + 8

42. y = 3x2 + 12x − 5 43. y = 4x − 12x2

44. Sketch the graph of y = x2 − 6x + 8 by plotting the roots and the minimum point.

45. Sketch the graph of y = x2 + 4x + 6 by plotting the minimum point, the y-intercept, and one other point.

46. If the alleles A and B of the cystic fibrosis gene occur in a popu- lation with frequencies p and 1 − p (where p is a fraction between 0 and 1), then the frequency of heterozygous carriers (carriers with both alleles) is 2p(1 − p). Which value of p gives the largest frequency of heterozygous carriers?

47. For which values of c does f (x) = x2 + cx + 1 have a double root? No real roots?

48. Let f be a quadratic function and c a constant. Which of the following statements is correct? Explain graphically. (a) There is a unique value of c such that y = f (x) − c has a double root. (b) There is a unique value of c such that y = f (x − c) has a double root.

49. Prove that x + 1x ≥ 2 for all x > 0. Hint: Consider (x1/2 − x−1/2)2. 50. Let a, b > 0. Show that the geometric mean

√ ab is not larger than

the arithmetic mean (a + b)/2. Hint: Use a variation of the hint given in Exercise 49.

51. If objects of weights x and w1 are suspended from the balance in Figure 14(A), the cross-beam is horizontal if bx = aw1. If the lengths a and b are known, we may use this equation to determine an unknown weight x by selecting w1 such that the cross-beam is horizontal. If a and b are not known precisely, we might proceed as follows. First balance x by w1 on the left as in (A). Then switch places and balance x by w2 on the right as in (B). The average x̄ = 12 (w1 + w2) gives an estimate for x. Show that x̄ is greater than or equal to the true weight x.

w1

(A) (B)

a

w2x x

b a b

FIGURE 14

52. Find numbers x and y with sum 10 and product 24. Hint: Find a quadratic polynomial satisfied by x.

53. Find a pair of numbers whose sum and product are both equal to 8.

54. Show that the parabola y = x2 consists of all points P such that d1 = d2, where d1 is the distance from P to

( 0, 14

) and d2 is the dis-

tance from P to the line y = − 14 (Figure 15).

d1

d2

P = (x, x2)

y = x2

1 4

1 4

−

x

y

FIGURE 15

Further Insights and Challenges 55. Show that if f and g are linear, then so is f + g. Is the same true of fg?

56. Show that if f and g are linear functions such that f (0) = g(0) and f (1) = g(1), then f = g.

57. Show that #y/#x for the function f (x) = x2 over the interval [x1, x2] is not a constant, but depends on the interval. Determine the exact dependence of #y/#x on x1 and x2.

58. Complete the square and use the result to derive the quadratic for- mula for the roots of ax2 + bx + c = 0.

59. Let a, c ̸= 0. Show that the roots of

ax2 + bx + c = 0 and cx2 + bx + a = 0

are reciprocals of each other.

60. Show, by completing the square, that the parabola

y = ax2 + bx + c

is congruent to y = ax2 by a vertical and horizontal translation. 61. Prove Viète’s Formulas: The quadratic polynomial with α and β as roots is x2 + bx + c, where b = −α − β and c = αβ.

1.3 The Basic Classes of Functions It would be impossible (and useless) to describe all possible functions f . Since the values of a function can be assigned arbitrarily, a function chosen at random would likely be so complicated that we could neither graph it nor describe it in any reasonable way. However, calculus makes no attempt to deal with all functions. The techniques of calculus, powerful

20 C H A P T E R 1 PRECALCULUS REVIEW

and general as they are, apply only to functions that are sufficiently “well-behaved” (we will see what well-behaved means when we study the derivative in Chapter 3). Fortunately, such functions are adequate for a vast range of applications.

Most of the functions considered in this text are constructed from the following familiar classes of well-behaved functions:

polynomials rational functions algebraic functions

exponential functions trigonometric functions

logarithmic functions inverse trigonometric functions

We shall refer to these as the basic functions.

• Polynomials: For any real number m, f (x) = xm is called the power function with exponent m. Power functions include f (x) = x3, f (x) = x−7 and f (x) = xπ . The base is the variable and the exponent is a constant. For now, we are interested in power functions with exponents that are positive integers. A polynomial is a sum of multiples of power functions with exponents that are positive integers or zero (making the term a constant in that case) (Figure 1):

5

2−2 −1 1 x

y

FIGURE 1 The polynomial y = x5 − 5x3 + 4x.

f (x) = x5 − 5x3 + 4x, g(t) = 7t6 + t3 − 3t − 1, h(x) = x9

Thus, the function f (x) = x + x−1 is not a polynomial because it includes a power x−1 with a negative exponent. The general polynomial P in the variable x may be written

P(x) = anxn + an−1xn−1 + · · · + a1x + a0 – The numbers a0, a1, . . . , an are called coefficients. – The degree of P is n (assuming that an ̸= 0). – The coefficient an is called the leading coefficient. – The domain of P is R.

• A rational function is a quotient of two polynomials (Figure 2):

5

−3

−2 1 x

y

FIGURE 2 The rational function

f (x) = x + 1 x3 − 3x + 2 .

f (x) = P(x) Q(x)

[P(x) and Q(x) polynomials]

The domain of f is the set of numbers x such that Q(x) ̸= 0. For example,

f (x) = 1 x2

domain {x : x ̸= 0}

h(t) = 7t 6 + t3 − 3t − 1

t2 − 1 domain {t : t ̸= ±1}

Every polynomial is also a rational function [with Q(x) = 1]. • An algebraic function is produced by taking sums, products, and quotients of roots

of polynomials and rational functions (Figure 3):

2−2 x

y

FIGURE 3 The algebraic function f (x) =

√ 1 + 3x2 − x4.

f (x) = √

1 + 3x2 − x4, g(t) = ( √

t − 2)−2, h(z) = z + z −5/3

5z3 − √z A number x belongs to the domain of f if each term in the formula is defined and

Any function that is not algebraic is called transcendental. Exponential and trigonometric functions are examples, as are the Bessel and gamma functions that appear in engineering and statistics. The term “transcendental” goes back to the 1670s, when it was used by Gottfried Wilhelm Leibniz (1646–1716) to describe functions of this type.

the result does not involve division by zero. For example, g(t) is defined if t ≥ 0 and

√ t ̸= 2, so the domain of g is D = {t : t ≥ 0 and t ̸= 4}. More generally,

algebraic functions are defined by polynomial equations between x and y. In this case, we say that y is implicitly defined as a function of x. For example, the equation y4 + 2x2y + x4 = 1 defines y implicitly as a function of x.

• Exponential functions: The function f (x) = bx , where b > 0, is called the expo- nential function with base b. Some examples are

f (x) = 2x, g(t) = 10t , h(x) = (

1 3

)x , p(t) = (

√ 5)t

S E C T I O N 1.3 The Basic Classes of Functions 21

Exponential functions and their inverses, the logarithmic functions, are treated in greater detail in Section 1.6.

• Trigonometric functions are functions built from sin x and cos x. These functions and their inverses are discussed in the next two sections.

Constructing New Functions Given functions f and g, we can construct new functions by forming the sum, difference, product, and quotient functions:

(f + g)(x) = f (x) + g(x), (f − g)(x) = f (x) − g(x)

(fg)(x) = f (x) g(x), (

f

g

) (x) = f (x)

g(x) (where g(x) ̸= 0)

For example, if f (x) = x2 and g(x) = sin x, then (f + g)(x) = x2 + sin x, (f − g)(x) = x2 − sin x

(fg)(x) = x2 sin x, (

f

g

) (x) = x

2

sin x

We can also multiply functions by constants. A function of the form

h(x) = c1f (x) + c2g(x) (c1, c2 constants) is called a linear combination of f and g.

Composition is another important way of constructing new functions. The compo- sition of f and g is the function f ◦ g defined by (f ◦ g)(x) = f (g(x)). The domain of f ◦ g is the set of values of x in the domain of g such that g(x) lies in the domain of f .

EXAMPLE 1 Compute the composite functions f ◦ g and g ◦ f and discuss their domains, where

f (x) = √x, g(x) = 1 − x Solution We haveExample 1 shows that the composition of

functions is not commutative: The functions f ◦ g and g ◦ f may be (and usually are) different.

(f ◦ g)(x) = f (g(x)) = f (1 − x) = √

1 − x The square root

√ 1 − x is defined if 1 − x ≥ 0 or x ≤ 1, so the domain of f ◦ g is

{x : x ≤ 1}. On the other hand, (g ◦ f )(x) = g(f (x)) = g(√x) = 1 − √x

The domain of g ◦ f is {x : x ≥ 0}.

Elementary Functions As noted above, we can produce new functions by applying the operations of addition,Inverse functions are discussed in Section

1.5. subtraction, multiplication, division, and composition. It is convenient to refer to a function constructed in this way from the basic functions listed above as an elementary function. The following functions are elementary:

f (x) = √

2x + sin x, f (x) = 10 √

x, f (x) = 1 + x −1

1 + cos x

Piecewise-Defined Functions We can also create new functions by piecing together functions defined over limited domains, obtaining piecewise-defined functions. One example we have already seen is the absolute value function defined by

|x| = {−x when x < 0 x when x ≥ 0

22 C H A P T E R 1 PRECALCULUS REVIEW

EXAMPLE 2 Given the function f , determine its domain, range, and whether or not it is increasing or decreasing for different values of x.

f (x) = {

1 when x < 0 x + 1 when x ≥ 0

Solution The function f appears in Figure 4. It is defined for all values of x so the domain

y = x + 1

y = 1

x

y

x < 0 x ≥ 0

FIGURE 4 A function defined piecewise. is all real numbers. However, for x < 0 the range is just the single value of 1, and for x ≥ 0 the range is all x ≥ 1. Hence, the range of the function is {x : x ≥ 1}. The function is neither increasing nor decreasing for x < 0; however, the function is increasing for x ≥ 0.

1.3 SUMMARY

• For m a real number, f (x) = xm is called the power function with exponent m. A poly- nomial P is a sum of multiples of xm, where m is a whole number:

P(x) = anxn + an−1xn−1 + · · · + a1x + a0 This polynomial has degree n (assuming that an ̸= 0) and an is called the leading coef- ficient.

• A rational function is a quotient P/Q of two polynomials (defined when Q(x) ̸= 0). • An algebraic function is produced by taking sums, products, and nth roots of polynomials

and rational functions. • Exponential function: f (x) = bx , where b > 0 (b is called the base). • The composite function f ◦ g is defined by (f ◦ g)(x) = f (g(x)). The domain of f ◦ g

is the set of x in the domain of g such that g(x) belongs to the domain of f . • The elementary functions are obtained by taking products, sums, differences, quotients,

and compositions of the basic functions, which include polynomials, rational functions, algebraic functions, exponential functions, trigonometric functions, logarithmic func- tions, and inverse trigonometric functions.

• Apiecewise-defined function is obtained by defining a function over two or more distinct domains.

1.3 EXERCISES

Preliminary Questions 1. Give an example of a rational function.

2. Is y = |x| a polynomial function? What about y = |x2 + 1|? 3. What is unusual about the domain of the composite function f ◦ g

for the functions f (x) = x1/2 and g(x) = −1 − |x|? 4. Is f (x) =

( 1 2 )x increasing or decreasing?

5. Give an example of a transcendental function.

Exercises In Exercises 1–12, determine the domain of the function.

1. f (x) = x1/4 2. g(t) = t2/3

3. f (x) = x3 + 3x − 4 4. h(z) = z3 + z−3

5. g(t) = 1 t + 2 6. f (x) =

1

x2 + 4

7. G(u) = 1 u2 − 4 8. f (x) =

√ x

x2 − 9

9. f (x) = x−4 + (x − 1)−3 10. F(s) = sin (

s

s + 1

)

11. g(y) = 10 √

y+y−1 12. f (x) = x + x −1

(x − 3)(x + 4) In Exercises 13–24, identify each of the following functions as polyno- mial, rational, algebraic, or transcendental.

13. f (x) = 4x3 + 9x2 − 8 14. f (x) = x−4

15. f (x) = √x 16. f (x) = √

1 − x2

S E C T I O N 1.4 Trigonometric Functions 23

17. f (x) = x 2

x + sin x 18. f (x) = 2 x

19. f (x) = 2x 3 + 3x

9 − 7x2 20. f (x) = 3x − 9x−1/2

9 − 7x2

21. f (x) = sin(x2) 22. f (x) = x√ x + 1

23. f (x) = x2 + 3x−1 24. f (x) = sin(3x)

25. Is f (x) = 2x2 a transcendental function?

26. Show that f (x) = x2 + 3x−1 and g(x) = 3x3 − 9x + x−2 are ra- tional functions—that is, quotients of polynomials.

In Exercises 27–34, calculate the composite functions f ◦ g and g ◦ f , and determine their domains.

27. f (x) = √x, g(x) = x + 1

28. f (x) = 1 x

, g(x) = x−4

29. f (x) = 2x , g(x) = x2

30. f (x) = |x|, g(θ) = sin θ

31. f (θ) = cos θ , g(x) = x3 + x2

32. f (x) = 1 x2 + 1 , g(x) = x

−2

33. f (t) = 1√ t

, g(t) = −t2

34. f (t) = √t , g(t) = 1 − t3

In Exercises 35–38, draw the graphs of each of the piecewise-defined functions.

35.

f (x) = {

3 when x < 0 x2 + 3 when x ≥ 0

36.

f (x) = { x + 1 when x < 0 1 − x when x ≥ 0

37.

f (x) = { x2 when x < 0 −x2 when x ≥ 0

38.

f (x) = {

2x − 2 when x < 0 x when x ≥ 0

39. The population (in millions) of a country as a function of time t (years) is P(t) = 30 · 20.1t . Show that the population doubles every 10 years. Show more generally that for any positive constants a and k, the function g(t) = a2kt doubles after 1/k years.

40. Find all values of c such that f (x) = x + 1 x2 + 2cx + 4 has domain R.

Further Insights and Challenges In Exercises 41–47, we define the first difference δf of a function f by δf (x) = f (x + 1) − f (x).

41. Show that if f (x) = x2, then δf (x) = 2x + 1. Calculate δf for f (x) = x and f (x) = x3.

42. Show that δ(10x) = 9 · 10x and, more generally, that δ(bx) = (b − 1)bx .

43. Show that for any two functions f and g, δ(f + g) = δf + δg and δ(cf ) = cδ(f ), where c is any constant.

44. Suppose we can find a function P such that δP(x) = (x + 1)k and P(0) = 0. Prove that P(1) = 1k , P(2) = 1k + 2k , and, more gener- ally, for every whole number n,

P(n) = 1k + 2k + · · · + nk 1

45. First show that

P(x) = x(x + 1) 2

satisfies δP = (x + 1). Then apply Exercise 44 to conclude that

1 + 2 + 3 + · · · + n = n(n + 1) 2

46. Calculate δ(x3), δ(x2), and δ(x). Then find a polynomial P of degree 3 such that δP = (x + 1)2 and P(0) = 0. Conclude that P(n) = 12 + 22 + · · · + n2.

47. This exercise combined with Exercise 44 shows that for all whole numbers k, there exists a polynomial P satisfying Eq. (1). The solu- tion requires the Binomial Theorem and proof by induction (see Ap- pendix C). (a) Show that δ(xk+1) = (k + 1) xk + · · · , where the dots indicate terms involving smaller powers of x. (b) Show by induction that there exists a polynomial of degree k + 1 with leading coefficient 1/(k + 1):

P(x) = 1 k + 1x

k+1 + · · ·

such that δP = (x + 1)k and P(0) = 0.

1.4 Trigonometric Functions We begin our trigonometric review by recalling the two systems of angle measurement: radians and degrees. They are best described using the relationship between angles and rotation. As is customary, we often use the lowercase Greek letter θ (“theta”) to denote angles and rotations.

24 C H A P T E R 1 PRECALCULUS REVIEW

1

(A) (B) (C) (D)

O P

Q

P

Q

θ

1O P = Q

θ = 2π θ =

1O

θ = −

P

Q

1 O

π 4

π 2

FIGURE 1 The radian measure θ of a counterclockwise rotation is the length along the unit circle of the arc traversed by P as it rotates into Q.

Figure 1(A) shows a unit circle with radius OP rotating counterclockwise into radius

rO θ

θr

FIGURE 2 On a circle of radius r , the arc traversed by a counterclockwise rotation of θ radians has length θr .

OQ. The radian measure of this rotation is the length θ of the circular arc traversed by P as it rotates into Q. On a circle of radius r , the arc traversed by a counterclockwise rotation of θ radians has length θr (Figure 2).

The unit circle has circumference 2π . Therefore, a rotation through a full circle has radian measure θ = 2π [Figure 1(B)]. The radian measure of a rotation through one- quarter of a circle is θ = 2π/4 = π/2 [Figure 1(C)] and, in general, the rotation through one-nth of a circle has radian measure 2π/n (Table 1). A negative rotation (with θ < 0) is a rotation in the clockwise direction [Figure 1(D)].

The radian measure of an angle such as ̸ POQ in Figure 1(A) is defined as the radian

TABLE 1

Rotation through Radian measure

Two full circles 4π Full circle 2π Half circle π Quarter circle 2π/4 = π/2 One-sixth circle 2π/6 = π/3

measure of a rotation that carries OP to OQ. Notice, however, that the radian measure of an angle is not unique. The rotations through θ and θ + 2π both carry OP to OQ. Therefore, θ and θ + 2π represent the same angle even though the rotation through θ + 2π takes an extra trip around the circle. In general, two radian measures represent the same an- gle if the corresponding rotations differ by an integer multiple of 2π . For example, π/4, 9π/4, and −15π/4 all represent the same angle because they differ by multiples of 2π :

π

4 = 9π

4 − 2π = −15π

4 + 4π

Every angle has a unique radian measure satisfying 0 ≤ θ < 2π . With this choice, the angle θ subtends an arc of length θr on a circle of radius r (Figure 2).

Degrees are defined by dividing the circle (not necessarily the unit circle) into 360 equal parts. A degree is 1360 of a circle. A rotation through θ degrees (denoted θ

◦) is a rotation through the fraction θ/360 of the complete circle. For example, a rotation through 90◦ is a rotation through the fraction 90360 , or

1 4 , of a circle.

As with radians, the degree measure of an angle is not unique. Two degree measures represent that same angle if they differ by an integer multiple of 360. For example, the angles −45◦ and 675◦ coincide because 675 = −45 + 2(360). Every angle has a unique degree measure θ with 0 ≤ θ < 360.

To convert between radians and degrees, remember that 2π radians is equal to 360◦.

Radians Degrees

0 0◦ π

6 30◦

π

4 45◦

π

3 60◦

π

2 90◦ Therefore, 1 radian equals 360/2π or 180/π degrees.

• To convert from radians to degrees, multiply by 180/π . • To convert from degrees to radians, multiply by π/180.

EXAMPLE 1 Convert (a) 55◦ to radians and (b) 0.5 radians to degrees.

Solution

(a) 55◦ × π 180◦

≈ 0.9599 radians (b) 0.5 radians × 180 ◦

π ≈ 28.648◦

Convention Unless otherwise stated, we always measure angles in radians.

Radian measurement is usually the better choice for mathematical purposes, but there are good practical reasons for using degrees. The number 360 has many divisors (360 = 8 · 9 · 5), and consequently, many fractional parts of the circle can be expressed as an integer number of degrees. For example, one-fifth of the circle is 72◦, two-ninths is 80◦, three-eighths is 135◦, etc.

S E C T I O N 1.4 Trigonometric Functions 25

The trigonometric functions sine and cosine can be defined in terms of right triangles. Let θ be an acute angle in a right triangle, and let us label the sides as in Figure 3. Then

a

b c

Hypotenuse

Adjacent

Opposite

θ

FIGURE 3

sin θ = b c

= opposite hypotenuse

, cos θ = a c

= adjacent hypotenuse

A disadvantage of this definition is that it makes sense only if θ lies between 0 and π/2 (because an angle in a right triangle cannot exceed π/2). However, sine and cosine can be defined for all angles in terms of the unit circle. Let P = (x, y) be the point on the unit circle corresponding to the angle θ as in Figures 4(A) and (B), and define

cos θ = x-coordinate of P , sin θ = y-coordinate of P This agrees with the right-triangle definition when 0 < θ < π2 . On the circle of radius r (centered at the origin), the point corresponding to the angle θ has coordinates

(r cos θ, r sin θ)

Furthermore, we see from Figure 4(C) that f (x) = sin θ is an odd function and f (x) = cos θ is an even function:

sin(−θ) = − sin θ, cos(−θ) = cos θ

P = (cos θ, sin θ)

x

1 y

θ

(A)

P = (cos θ, sin θ)

x y

θ

(B) (C)

(x, y)

(x, −y)

θ −θ

FIGURE 4 The unit circle definition of sine and cosine is valid for all angles θ .

Although we use a calculator to evaluate sine and cosine for general angles, the standard values listed in Figure 5 and Table 2 appear often and should be memorized.

(0, 1)

π/6 π/2π/3π/4

( ) , .32 12 ( ) , .22 .22 ( ) ,

.3 2

1 2

FIGURE 5 Four standard angles: The x- and y-coordinates of the points are cos θ and sin θ .

TABLE 2

θ 0 π

6 π

4 π

3 π

2 2π 3

3π 4

5π 6

π

sin θ 0 1 2

√ 2

2

√ 3

2 1

√ 3

2

√ 2

2 1 2

0

cos θ 1

√ 3

2

√ 2

2 1 2

0 −1 2

− √

2 2

− √

3 2

−1

The graph of y = sin θ is the familiar “sine wave” shown in Figure 6. Observe how the graph is generated by the y-coordinate of the point P = (cos θ, sin θ) moving around the unit circle.

26 C H A P T E R 1 PRECALCULUS REVIEW

1

pq qq

2p

1 y y

x

FIGURE 6 The graph of y = sin θ is generated as the point P = (cos θ, sin θ) moves around the unit circle.

1

y = sin θ

I II III IV

−1

Quadrant of unit circle

π 2π π 2π θ θ

y = cos θ

I II III IV y y

π 4

π 2

π 4

π 2

3π 4

5π 4

7π 4

3π 4

5π 4

3π 2

3π 2

7π 4

FIGURE 7 Graphs of y = sin θ and y = cos θ over one period of length 2π .

The graph of y = cos θ has the same shape but is shifted to the left π/2 units (Figure 7). The signs of sin θ and cos θ vary as P = (cos θ, sin θ) changes quadrant.

A function f is called periodic with period T if f (x + T ) = f (x) (for all x) and T is the smallest positive number with this property. The sine and cosine functions are periodicWe often write sin x and cos x, using x

instead of θ . Depending on the application, we may think of x as an angle or simply as a real number.

with period T = 2π (Figure 8) because the radian measures x and x + 2πk correspond to the same point on the unit circle for any integer k:

sin x = sin(x + 2πk), cos x = cos(x + 2πk)

y = sin x

2π−2π 4π

y = cos x

2π−2π 4π xx

y 1

y

1

FIGURE 8 Sine and cosine have period 2π .

There are four other standard trigonometric functions, each defined in terms of sin x and cos x or as ratios of sides in a right triangle (Figure 9):

a x

b c

Hypotenuse

Adjacent

Opposite

FIGURE 9

Tangent: tan x = sin x cos x

= b a , Cotangent: cot x = cos x

sin x = a

b

Secant: sec x = 1 cos x

= c a , Cosecant: csc x = 1

sin x = c

b

These functions are periodic (Figure 10):y = tan x andy = cot x have periodπ ;y = sec x and y = csc x have period 2π (see Exercise 57).

x

−1

y = csc xy = sec x

1 1

−1 xx

yy

−1

y = tan x

1

π 2π 2π 2π 2ππ−−π −π π π−π −π x

y

y = cot x

1

−1 x

y

π 2

−π 2

−π 2

π 2

−π 2

π 2

π 2

π 2

3π 2

3π 2

5π 2

3π 2

3π 2

5π 2

FIGURE 10 Graphs of the standard trigonometric functions.

S E C T I O N 1.4 Trigonometric Functions 27

EXAMPLE 2 Computing Values of Trigonometric Functions Find the values of the six trigonometric functions at x = 4π/3. Solution The point P on the unit circle corresponding to the angle x = 4π/3 lies opposite the point with angle π/3 (Figure 11). Thus, we see that (refer to Table 2)

P =

1

(− − ) , .3212

( ), .3212 4π 3

π 3

FIGURE 11

sin 4π 3

= − sin π 3

= − √

3 2

, cos 4π 3

= − cos π 3

= −1 2

The remaining values are

tan 4π 3

= sin 4π/3 cos 4π/3

= − √

3/2 −1/2 =

√ 3, cot

4π 3

= cos 4π/3 sin 4π/3

= √

3 3

sec 4π 3

= 1 cos 4π/3

= 1−1/2 = −2, csc 4π 3

= 1 sin 4π/3

= −2 √

3 3

EXAMPLE 3 Find the angles x such that sec x = 2. Solution Because sec x = 1/ cos x, we must solve cos x = 12 . From Figure 12 we see

1 2

1−

π 3 π 3

FIGURE 12 cos x = 12 for x = ±π3 that x = π/3 and x = −π/3 are solutions. We may add any integer multiple of 2π , so the general solution is x = ±π/3 + 2πk for any integer k.

EXAMPLE 4 Trigonometric Equation Solve sin 4x + sin 2x = 0 for x ∈ [0, 2π). Solution We must find the angles x such that sin 4x = − sin 2x. First, let’s determine when angles θ1 and θ2 satisfy sin θ2 = − sin θ1. Figure 13 shows that this occurs if θ2 = −θ1 or θ2 = θ1 + π . Because the sine function is periodic with period 2π ,

sin θ2 = − sin θ1 ⇔ θ2 = −θ1 + 2πk or θ2 = θ1 + π + 2πk

where k is an integer. Taking θ2 = 4x and θ1 = 2x, we see that

sin 4x = − sin 2x ⇔ 4x = −2x + 2πk or 4x = 2x + π + 2πk

The first equation gives 6x = 2πk or x = (π/3)k and the second equation gives 2x = π + 2πk or x = π/2 + πk. We obtain eight solutions in [0, 2π) (Figure 14):

x = 0, π 3

, 2π 3

, π, 4π 3

, 5π 3

and x = π 2

, 3π 2

−sin θ1

sin θ1

θ1

θ1

θ2 = −θ1θ2 = θ1 + π

FIGURE 13 sin θ2 = − sin θ1 when θ2 = −θ1 or θ2 = θ1 + π .

−1

1

y = sin 4x + sin 2x

0 2ππ x

y

5π 3

3π 2

4π 3

2π 3

π 2

π 3

FIGURE 14 Solutions of sin 4x + sin 2x = 0.

EXAMPLE 5 Sketch the graph of f (x) = 3 cos ( 2

( x + π2

)) over [0, 2π ].

Solution The graph is obtained by scaling and shifting the graph of y = cos x in three steps (Figure 15):

• Compress horizontally by a factor of 2: y = cos 2x • Shift to the left π/2 units: y = cos

( 2

( x + π

2

))

Shift left π/2 units

Compress horizontally by

a factor of 2

Expand vertically by a factor of 3

(B) y = cos 2x (periodic with period π)

(C) y = cos 2(x +

2ππ 2ππ

1

3

−1

−3

2ππ xxx

y

1

3

−1

−3

y

1

3

−1

−3

y

(A) y = cos x

2ππ x

1

3

−1

−3

y

) π2 π 2 (D) y

= 3 cos 2(x + )

π 2

π 2

π 2

FIGURE 15

28 C H A P T E R 1 PRECALCULUS REVIEW

• Expand vertically by a factor of 3: y = 3 cos (

2 ( x + π

2

))

The vertical height taken on by such a function is the amplitude. So in this last

CAUTION To shift the graph of y = cos 2x to the left π/2 units, we must replace x by x + π2 to obtain cos

( 2

( x + π2

)) . It is

incorrect to take cos ( 2x + π2

) . Note that

to shift left (in the −x direction), we add π/2. To shift right (in the +x direction), we subtract π/2, counter to what you might expect.

example, the amplitude was 3.

Trigonometric Identities A key feature of trigonometric functions is that they satisfy a large number of identities. First and foremost, sine and cosine satisfy a fundamental identity, which is equivalent to the Pythagorean Theorem:The expression (sin x)k is usually denoted

sink x. For example, sin2 x is the square of sin x. We use similar notation for the other trigonometric functions. However, we reserve sin−1 x for the inverse sine function discussed in the next section, rather than for 1sin x .

sin2 x + cos2 x = 1 1

Equivalent versions are obtained by dividing Eq. (1) by cos2 x or sin2 x:

tan2 x + 1 = sec2 x, 1 + cot2 x = csc2 x 2

Here is a list of some other commonly used identities. The identities for complementary angles are justified by Figure 16.

Basic Trigonometric Identities

Complementary angles: sin (π

2 − x

) = cos x, cos

(π 2

− x )

= sin x

Addition formulas: sin(x + y) = sin x cos y + cos x sin y cos(x + y) = cos x cos y − sin x sin y

Double-angle formulas: sin2 x = 1 2 (1 − cos 2x), cos2 x = 1

2 (1 + cos 2x)

cos 2x = cos2 x − sin2 x, sin 2x = 2 sin x cos x

Shift formulas: sin ( x + π

2

) = cos x, cos

( x + π

2

) = − sin x

EXAMPLE 6 Suppose that cos θ = 25 . Calculate tan θ in the following two cases: (a) 0 < θ < π2 and (b) π < θ < 2π .

a

b c

x

π 2

− x

FIGURE 16 For complementary angles, the sine of one angle is equal to the cosine of the complementary angle.

Solution First, using the identity cos2 θ + sin2 θ = 1, we obtain

sin θ = ± √

1 − cos2 θ = ± √

1 − 4 25

= ± √

21 5

(a) If 0 < θ < π2 , then sin θ is positive and we take the positive square root:

Opposite Hypotenuse

5

Adjacent 2 θ

21

FIGURE 17

tan θ = sin θ cos θ

= √

21/5 2/5

= √

21 2

To visualize this computation, draw a right triangle with angle θ such that cos θ = 25 as in Figure 17. The opposite side then has length

√ 21 =

√ 52 − 22 by the Pythagorean

Theorem.

(b) If π < θ < 2π , then sin θ is negative and tan θ = − √

21 2 .

We conclude this section by quoting the Law of Cosines (Figure 18), which is a generalization of the Pythagorean Theorem (see Exercise 60).

S E C T I O N 1.4 Trigonometric Functions 29

THEOREM 1 Law of Cosines If a triangle has sides a, b, and c, and θ is the angle opposite side c, then

c2 = a2 + b2 − 2ab cos θ a

b

c

θ

FIGURE 18 If θ = π/2, then cos θ = 0 and the Law of Cosines reduces to the Pythagorean Theorem.

1.4 SUMMARY

• An angle of θ radians subtends an arc of length θr on a circle of radius r . • To convert from radians to degrees, multiply by 180/π . • To convert from degrees to radians, multiply by π/180. • Unless otherwise stated, all angles in this text are given in radians. • The functions f (x) = cos θ and f (x) = sin θ are defined in terms of right triangles for

acute angles and as coordinates of a point on the unit circle for general angles (Figure 19):

1

(cos θ, sin θ)

a

c b

θ

θ

FIGURE 19

sin θ = b c

= opposite hypotenuse

, cos θ = a c

= adjacent hypotenuse

• Basic properties of sine and cosine:

– Periodicity: sin(θ + 2π) = sin θ , cos(θ + 2π) = cos θ – Parity: sin(−θ) = − sin θ , cos(−θ) = cos θ – Basic identity: sin2 θ + cos2 θ = 1

• The four additional trigonometric functions:

tan θ = sin θ cos θ

, cot θ = cos θ sin θ

, sec θ = 1 cos θ

, csc θ = 1 sin θ

1.4 EXERCISES

Preliminary Questions 1. How is it possible for two different rotations to define the same

angle?

2. Give two different positive rotations that define the angle π/4.

3. Give a negative rotation that defines the angle π/3.

4. The definition of cos θ using right triangles applies when (choose the correct answer):

(a) 0 < θ < π

2 (b) 0 < θ < π (c) 0 < θ < 2π

5. What is the unit circle definition of sin θ?

6. How does the periodicity of f (x) = sin θ and f (x) = cos θ follow from the unit circle definition?

Exercises 1. Find the angle between 0 and 2π equivalent to 13π/4.

2. Describe θ = π/6 by an angle of negative radian measure.

3. Convert from radians to degrees:

(a) 1 (b) π

3 (c)

5 12

(d) −3π 4

4. Convert from degrees to radians:

(a) 1◦ (b) 30◦ (c) 25◦ (d) 120◦

5. Find the lengths of the arcs subtended by the angles θ and φ radians in Figure 20.

4 θ = 0.9

φ = 2

FIGURE 20 Circle of radius 4.

30 C H A P T E R 1 PRECALCULUS REVIEW

6. Calculate the values of the six standard trigonometric functions for the angle θ in Figure 21.

15

8 17

θ

FIGURE 21

7. Fill in the remaining values of (cos θ, sin θ) for the points in Figure 22.

π

11π 67π

45π 33π

2

4π 3

7π 6

5π 4

5π 6

3π 4

2π 3

π 4

π 3

π 2

( ),22 22 π 6 ( ),32 12

( ), 3212

0 (1, 0)

FIGURE 22

8. Find the values of the six standard trigonometric functions at θ = 11π/6. In Exercises 9–14, use Figure 22 to find all angles between 0 and 2π satisfying the given condition.

9. cos θ = 1 2

10. tan θ = 1

11. tan θ = −1 12. csc θ = 2

13. sin x = √

3 2

14. sec t = 2

15. Fill in the following table of values:

θ π

6 π

4 π

3 π

2 2π 3

3π 4

5π 6

tan θ

sec θ

16. Complete the following table of signs:

θ sin θ cos θ tan θ cot θ sec θ csc θ

0 < θ < π

2 + +

π

2 < θ < π

π < θ < 3π 2

3π 2

< θ < 2π

17. Show that if tan θ = c and 0 ≤ θ < π/2, then cos θ = 1/ √

1 + c2. Hint: Draw a right triangle whose opposite and adjacent sides have lengths c and 1.

18. Suppose that cos θ = 13 . (a) Show that if 0 ≤ θ < π/2, then sin θ = 2

√ 2/3 and tan θ = 2

√ 2.

(b) Find sin θ and tan θ if 3π/2 ≤ θ < 2π . In Exercises 19–24, assume that 0 ≤ θ < π/2. 19. Find sin θ and tan θ if cos θ = 513 . 20. Find cos θ and tan θ if sin θ = 35 . 21. Find sin θ , sec θ , and cot θ if tan θ = 27 . 22. Find sin θ , cos θ , and sec θ if cot θ = 4. 23. Find cos 2θ if sin θ = 15 . 24. Find sin 2θ and cos 2θ if tan θ =

√ 2.

25. Find cos θ and tan θ if sin θ = 0.4 and π/2 ≤ θ < π . 26. Find cos θ and sin θ if tan θ = 4 and π ≤ θ < 3π/2. 27. Find cos θ if cot θ = 43 and sin θ < 0. 28. Find tan θ if sec θ =

√ 5 and sin θ < 0.

29. Find the values of sin θ , cos θ , and tan θ for the angles correspond- ing to the eight points on the unit circles in Figure 23(A) and (B).

(0.3965, 0.918)

(A) (B)

(0.3965, 0.918)

FIGURE 23

30. Refer to Figure 24(A). Express the functions sin θ , tan θ , and csc θ in terms of c.

31. Refer to Figure 24(B). Compute cos ψ , sin ψ , cot ψ , and csc ψ .

c 1 1

0.3

(B)(A)

θ ψ

FIGURE 24

32. Express cos ( θ + π2

) and sin

( θ + π2

) in terms of cos θ and sin θ .

Hint: Find the relation between the coordinates (a, b) and (c, d) in Figure 25.

(c, d)

(a, b)

1

θ

FIGURE 25

S E C T I O N 1.4 Trigonometric Functions 31

33. Use the addition formula to compute cos (π

3 + π4 )

exactly.

34. Use the addition formula to compute sin (π

3 − π4 )

exactly.

In Exercises 35–38, sketch the graph over [0, 2π ].

35. f (θ) = 2 sin 4θ 36. f (θ) = cos (

2 ( θ − π

2

))

37. f (θ) = cos (

2θ − π 2

)

38. f (θ) = sin (

2 ( θ − π

2

) + π

) + 2

39. Determine a function that would have a graph as in Figure 26(A), stating the period and amplitude.

(A)

−3

3

π 2π 3π−π x

y

−2

2

π x

y

(B)

−π 3

π 3 3

2π

FIGURE 26

40. Determine a function that would have a graph as in Figure 26(B), stating the period and amplitude.

41. How many points lie on the intersection of the horizontal line y = c and the graph of y = sin x for 0 ≤ x < 2π? Hint: The answer depends on c.

42. How many points lie on the intersection of the horizontal line y = c and the graph of y = tan x for 0 ≤ x < 2π? In Exercises 43–46, solve for 0 ≤ θ < 2π (see Example 4). 43. sin 2θ + sin 3θ = 0 44. sin θ = sin 2θ

45. cos 4θ + cos 2θ = 0 46. sin θ = cos 2θ In Exercises 47–56, derive the identity using the identities listed in this section.

47. cos 2θ = 2 cos2 θ − 1 48. cos2 θ 2

= 1 + cos θ 2

49. sin θ

2 =

√ 1 − cos θ

2 50. sin(θ + π) = − sin θ

51. cos(θ + π) = − cos θ 52. tan x = cot (π

2 − x

)

53. tan(π − θ) = − tan θ 54. tan 2x = 2 tan x 1 − tan2 x

55. tan x = sin 2x 1 + cos 2x

56. sin2 x cos2 x = 1 − cos 4x 8

57. Use Exercises 50 and 51 to show that tan θ and cot θ are periodic with period π .

58. Use the identity of Exercise 48 to show that cos π8 is equal to√ 1 2

+ √

2 4

.

59. Use the Law of Cosines to find the distance from P to Q in Fig- ure 27.

8

10

P

Q

7π/9

FIGURE 27

Further Insights and Challenges 60. Use Figure 28 to derive the Law of Cosines from the Pythagorean Theorem.

a

θ

b c

a − b cos θ FIGURE 28

61. Use the addition formula to prove

cos 3θ = 4 cos3 θ − 3 cos θ

62. Use the addition formulas for sine and cosine to prove

tan(a + b) = tan a + tan b 1 − tan a tan b

cot(a − b) = cot a cot b + 1 cot b − cot a

63. Let θ be the angle between the line y = mx + b and the x-axis [Figure 29(A)]. Prove that m = tan θ .

y = mx + b

θ x

r

s

(A)

y

θ x

(B)

y L2

L1

FIGURE 29

64. Let L1 and L2 be the lines of slopem1 and m2 [Figure 29(B)]. Show

that the angle θ between L1 and L2 satisfies cot θ = m2m1 + 1 m2 − m1

.

65. Perpendicular Lines Use Exercise 64 to prove that two lines with nonzero slopes m1 and m2 are perpendicular if and only if m2 = −1/m1. 66. Apply the double-angle formula to prove:

(a) cos π

8 = 1

2

√ 2 +

√ 2

(b) cos π

16 = 1

2

√ 2 +

√ 2 +

√ 2

Guess the values of cos π

32 and of cos

π

2n for all n.

32 C H A P T E R 1 PRECALCULUS REVIEW

1.5 Inverse Functions Many important functions, such as logarithms, roots, and the arcsine, are defined as inverse

D

f

f −1

R

FIGURE 1 A function and its inverse.

functions. In this section, we review inverse functions and their graphs, and we discuss the inverse trigonometric functions.

The inverse of f , denoted f −1, is the function that reverses the effect of f (Figure 1). For example, the inverse of f (x) = x3 is the cube root function f −1(x) = x1/3. Given a table of function values for f , we obtain a table for f −1 by interchanging the x and yIn general, f −1(x) ̸= 1

f (x) . The expression

f −1(x) is simply a notation for the inverse function, and the ‘−1’ does not represent an exponent.

columns, assuming the resulting f −1 is a function:

Function

x f (x) = x3

−2 −8 −1 −1

0 0 1 1 2 8 3 27

(Interchange columns) /⇒

Inverse

x f −1(x) = x1/3

−8 −2 −1 −1

0 0 1 1 8 2

27 3

If we apply both f and f −1 to a number x in either order, we get back x. For instance,

Apply f and then f −1: 2 (Apply x3)−→ 8 (Apply x

1/3)−→ 2

Apply f −1 and then f : 8 (Apply x1/3)−→ 2 (Apply x

3)−→ 8

This property is used in the formal definition of the inverse function.

REMINDER The “domain” is the set of numbers x such that f (x) is defined (the set of allowable inputs), and the “range” is the set of all values f (x) (the set of outputs).

DEFINITION Inverse Let f have domain D and range R. If there is a function g with domain R such that

g ( f (x)

) = x for x ∈ D and f

( g(x)

) = x for x ∈ R

then f is said to be invertible. The function g is called the inverse function and is denoted f −1.

EXAMPLE 1 Show that f (x) = 2x − 18 is invertible. What are the domain and range of f −1?

Solution We show that f is invertible by computing the inverse function in two steps.

Step 1. Solve the equation y = f (x) for x in terms of y.

y = 2x − 18 y + 18 = 2x

x = 1 2 y + 9

This gives us the inverse as a function of the variable y: f −1(y) = 12y + 9. Step 2. Interchange variables.

We usually prefer to write the inverse as a function of x, so we interchange the roles of x and y (Figure 2):

x

y

y = f (x) = 2x − 18

−18

−18

y = f −1(x) = x + 91 2

FIGURE 2 f −1(x) = 1

2 x + 9

S E C T I O N 1.5 Inverse Functions 33

To check our calculation, let’s verify that f −1(f (x)) = x and f (f −1(x)) = x:

f −1 ( f (x)

) = f −1(2x − 18) = 1

2 (2x − 18) + 9 = (x − 9) + 9 = x

f ( f −1(x)

) = f

( 1 2 x + 9

) = 2

( 1 2 x + 9

) − 18 = (x + 18) − 18 = x

Because f −1 is a linear function, its domain and range are R.

The inverse function, if it exists, is unique. However, some functions do not have an inverse. Consider f (x) = x2. When we interchange the columns in a table of values (which should give us a table of values for f −1), the resulting table does not define a function:

Function

x f (x) = x2

−2 4 −1 1

0 0 1 1 2 4

(Interchange columns) /⇒

Inverse (?)

x f −1(x)

4 −2 1 −1 0 0 1 1 4 2

⎫ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

f −1(1) has two values: 1 and −1.

The problem is that every positive number occurs twice as an output of f (x) = x2. For example, 1 occurs twice as an output in the first table and therefore occurs twice as an input in the second table. So the second table gives us two possible values for f −1(1), namely f −1(1) = 1 and f −1(1) = −1. Neither value satisfies the inverse property. For instance, if we set f −1(1) = 1, then f −1(f (−1)) = f −1(1) = 1, but an inverse would have to satisfy f −1(f (−1)) = −1.

So when does a function f have an inverse? The answer is: If f is one-to-one, which means that f takes on each value at most once (Figure 3). Here is the formal definition:

Another standard term for one-to-one is injective.

DEFINITION One-to-One Function A function f is one-to-one on a domain D if, for every value c, the equation f (x) = c has at most one solution for x ∈ D. Or, equivalently, if for all a, b ∈ D,

f (a) ̸= f (b) unless a = b

f (x) = c has at most one solution for all c

aca

One-to-one

f (x) = c has two solutions: x = a and x = a

a´

´

c

Not one-to-one

FIGURE 3 A one-to-one function takes on each value at most once.

When f is one-to-one on its domain D, the inverse function f −1 exists and its do-Think of a function as a device for “labeling” members of the range by members of the domain. When f is one-to-one, this labeling is unique and f −1 maps each number in the range back to its label.

main is equal to the range R of f (Figure 4). Indeed, for every c ∈ R, there is precisely one element a ∈ D such that f (a) = c and we may define f −1(c) = a. With this defini- tion, f (f −1(c)) = f (a) = c and f −1(f (a)) = f −1(c) = a. This proves the following theorem.

THEOREM 1 Existence of Inverses The inverse function f −1 exists if and only if f is one-to-one on its domain D. Furthermore,

• Domain of f = range of f −1. • Range of f = domain of f −1.

34 C H A P T E R 1 PRECALCULUS REVIEW

a cDomain of f = range of f −1 f −1

f

Range of f = domain of f −1

FIGURE 4 In passing from f to f −1, the domain and range are interchanged.

EXAMPLE 2 Show that f (x) = 3x + 2 5x − 1 is invertible. Determine the domain and

range of f and f −1.

Solution The domain of f is D = { x : x ̸= 15

} (Figure 5). Assume that x ∈ D, and let’s

solve y = f (x) for x in terms of y: 3 5

1 5

x

y

FIGURE 5 Graph of f (x) = 3x + 2 5x − 1 .

y = 3x + 2 5x − 1

y(5x − 1) = 3x + 2

5xy − y = 3x + 2

5xy − 3x = y + 2 (gather terms involving x)

x(5y − 3) = y + 2 (factor out x in order to solve for x) 1

x = y + 2 5y − 3 (divide by 5y − 3) 2

The last step is valid if 5y − 3 ̸= 0—that is, if y ̸= 35 . But note that y = 35 is not in the Often, it is impossible to find a formula for the inverse because we cannot solve for x explicitly in the equation y = f (x). For example, the function f (x) = x + ex has an inverse, but we must make do without an explicit formula for it.

range of f . For if it were, Eq. (1) would yield the false equation 0 = 35 + 2. Now Eq. (2) shows that for all y ̸= 35 there is a unique value x such that f (x) = y. Therefore, f is one- to-one on its domain. By Theorem 1, f is invertible. The range of f is R =

{ x : x ̸= 35

}

and

f −1(x) = x + 2 5x − 3

The inverse function has domain R and range D.

We can tell whether f is one-to-one from its graph. The horizontal line y = c intersects the graph of f at points (a, f (a)), where f (a) = c (Figure 6). There is at most one such

x

y

(a, f (a))

y = f (x)

y = c

a

FIGURE 6 The line y = c intersects the graph at points where f (a) = c.

point if f (x) = c has at most one solution. This gives us the

Horizontal Line Test A function f is one-to-one if and only if every horizontal line intersects the graph of f in at most one point.

In Figure 7, we see that f (x) = x3 passes the Horizontal Line Test and therefore is one-to-one, whereas f (x) = x2 fails the test and is not one-to-one.

EXAMPLE 3 Increasing Functions Are One-to-One Show that increasing functions are one-to-one. Then show that f (x) = x5 + 4x + 3 is one-to-one. Solution An increasing function satisfies f (a) < f (b) if a < b. Therefore, f cannot take on any value more than once, and thus f is one-to-one. Note similarly that decreasing functions are one-to-one.

Now observe that

• If n is odd and c > 0, then cxn is increasing. • A sum of increasing functions is increasing.

S E C T I O N 1.5 Inverse Functions 35

y y

x

x

b

b1/3 b

(A) f (x) = x3 is one-to-one. (B) f (x) = x2 is not one-to-one.

− b b

FIGURE 7

Thus, x5, 4x, and hence the sum x5 + 4x are increasing. It follows that the function f (x) = x5 + 4x + 3 is increasing and therefore one-to-one (Figure 8). However, determining an explicit formula for its inverse would be difficult.

We can make a function one-to-one by restricting its domain suitably.

f (x) = x5 + 4x + 3 x

y

2

30

−2

FIGURE 8 The increasing function f (x) = x5 + 4x + 3 satisfies the Horizontal Line Test.

EXAMPLE 4 Restricting the Domain Find a domain on which f (x) = x2 is one-to- one and determine its inverse on this domain.

Solution The function f (x) = x2 is one-to-one on the domain D = {x : x ≥ 0}, for if

One-to-one for x ≥ 0

x

y

2

2

4

1−2 −1

FIGURE 9 f (x) = x2 satisfies the Horizontal Line Test on the domain {x : x ≥ 0}.

a2 = b2 where a and b are both nonnegative, then a = b (Figure 9). The inverse of f on D is the positive square root f −1(x) = √x. Alternatively, we may restrict f to the domain {x : x ≤ 0}, on which the inverse function is f −1(x) = −√x.

Next we describe the graph of the inverse function. The reflection of a point (a, b) through the line y = x is defined to be the point (b, a) (Figure 10). Note that if the x- and y-axes are drawn to the same scale, then (a, b) and (b, a) are equidistant from the line y = x and the segment joining them is perpendicular to y = x.

The graph of f −1 is the reflection of the graph of f through y = x (Figure 11). To check this, note that (a, b) lies on the graph of f if f (a) = b. But f (a) = b if and only if f −1(b) = a, and in this case, (b, a) lies on the graph of f −1.

y = x

x

y

(a, b)

(b, a)

FIGURE 10 The reflection (a, b) through the line y = x is the point (b, a).

x

y

y =

y =

y =

x(b, a)

(a, b)

f −1(x)

f (x)

FIGURE 11 The graph of f −1 is the reflection of the graph of f through the line y = x.

EXAMPLE 5 Sketching the Graph of the Inverse Sketch the graph of the inverse of f (x) =

√ 4 − x.

Solution Let g(x) = f −1(x). Observe that f has domain {x : x ≤ 4} and range {x : x ≥ 0}. We do not need a formula for g(x) to draw its graph. We simply reflect the graph of f through the line y = x as in Figure 12. If desired, however, we can easily solve y =

√ 4 − x to obtain x = 4 − y2 and thus g(x) = 4 − x2 with domain {x : x ≥ 0}.

x

y g(x) = f −1(x) = 4 − x2

2 4 −2

4

2 f (x) = 4 − x

y = x

FIGURE 12 Graph of the inverse g of g(x) =

√ 4 − x.

36 C H A P T E R 1 PRECALCULUS REVIEW

Inverse Trigonometric Functions

We have seen that the inverse function f −1 exists if and only if f is one-to-one on its domain. Because the trigonometric functions are not one-to-one, we must restrict their domains to define their inverses.

First consider the sine function. Figure 13 shows that f (θ) = sin θ is one-to-one onDo not confuse the inverse sin−1 x with the reciprocal

(sin x)−1 = 1 sin x

= csc x

The inverse functions sin−1 x, cos−1 x, . . . are often denoted arcsin x, arccos x, etc.

[ −π2 , π2

] . With this interval as domain, the inverse is called the arcsine function and is

denoted θ = sin−1 x or θ = arcsin x. By definition,

θ = sin−1 x is the unique angle in [ −π

2 , π

2

] such that sin θ = x

− θ

f (θ) = sin θ 1

−1

1

−1

sin θ with restricted domain

θ

θ

θ = sin−1 x

1 x

−1π 2

−π 2

π 2

π 2

−π 2

π 2

y y

FIGURE 13

The range of f (x) = sin x is [−1, 1], so f −1(x) = sin−1 x has domain [−1, 1]. A table

Summary of inverse relation between the sine and arcsine functions:

sin(sin−1 x) = x for −1 ≤ x ≤ 1

sin−1(sin θ) = θ for −π 2

≤ θ ≤ π 2

of values for the arcsine (Table 1) is obtained by reversing the columns in a table of values for sin x.

TABLE 1

x −1 − √

3 2 −

√ 2

2 − 12 0 12 √

2 2

√ 3

2 1

θ = sin−1 x −π2 −π3 −π4 −π6 0 π6 π4 π3 π2

EXAMPLE 6 (a) Show that sin−1 ( sin

( π 4

)) = π4 .

(b) Explain why sin−1 ( sin

( 5π 4

)) ̸= 5π4 .

Solution The equation sin−1(sin θ) = θ is valid only if θ lies in [ −π2 , π2

] .

(a) Because π4 lies in the required interval, sin −1 (sin

( π 4

)) = π4 .

(b) Let θ = sin−1 ( sin

( 5π 4

)) . By definition, θ is the angle in

[ −π2 , π2

] such that sin θ =

sin ( 5π

4

) . By the identity sin θ = sin(π − θ) (Figure 14),

−

y

x 5π 4

π 4

FIGURE 14 sin ( 5π

4 )

= sin ( −π4

) .

sin (

5π 4

) = sin

( π − 5π

4

) = sin

( −π

4

)

The angle −π4 lies in the required interval, so θ = sin−1 ( sin

( 5π 4

)) = −π4 .

The cosine function is one-to-one on [0, π ] rather than [ −π2 , π2

] (Figure 15). With

this domain, the inverse is called the arccosine function and is denoted θ = cos−1 x or θ = arccos x. It has domain [−1, 1]. By definition,

Summary of inverse relation between the cosine and arccosine:

cos(cos−1 x) = x for −1 ≤ x ≤ 1

cos−1(cos θ) = θ for 0 ≤ θ ≤ π θ = cos −1 x is the unique angle in [0, π ] such that cos θ = x

When we study the calculus of inverse trigonometric functions in Section 3.8, we will need to simplify composite expressions such as cos(sin−1 x) and tan(sin−1 x). This can be done in two ways: by referring to the appropriate right triangle or by using trigonometric identities.

S E C T I O N 1.5 Inverse Functions 37

cos θ with restricted domain

θ 2ππ

1

−1

θ

θ π

1

−1

1

π

θ = cos−1 x

x −1

f (θ) = cos θ

FIGURE 15

EXAMPLE 7 Find an alternative form in terms of x for each of cos(sin−1 x) and tan(sin−1 x).

Solution This problem asks for the values of cos θ and tan θ at the angle θ = sin−1 x. Consider a right triangle with hypotenuse of length 1 and angle θ such that sin θ = x, as in Figure 16. By the Pythagorean Theorem, the adjacent side has length

√ 1 − x2. Now

θ

1 x

.1 − x2 FIGURE 16 Right triangle constructed such that sin θ = x.

we can read off the values from Figure 16:

cos(sin−1 x) = cos θ = adjacent hypotenuse

= √

1 − x2

tan(sin−1 x) = tan θ = opposite adjacent

= x√ 1 − x2

Alternatively, we may argue using trigonometric identities. Because sin θ = x,

cos(sin−1 x) = cos θ = √

1 − sin2 θ = √

1 − x2

We are justified in taking the positive square root in either approach because θ = sin−1 x lies in

[ −π2 , π2

] and cos θ is positive in this interval.

We now address the remaining trigonometric functions. The function f (θ) = tan θ is one-to-one on

( −π2 , π2

) , and f (θ) = cot θ is one-to-one on (0, π) (see Figure 10 in

Section 1.4). We define their inverses by restricting them to these domains:

θ = tan−1 x is the unique angle in ( −π

2 , π

2

) such that tan θ = x

θ = cot−1 x is the unique angle in (0, π) such that cot θ = x

The range of both f (θ) = tan θ and f (θ) = cot θ is the set of all real numbers R. There- fore, θ = tan−1 x and θ = cot−1 x have domain R (Figure 17).

x

y

π

y = tan−1 x

y = cot−1 x

y

x

−

π 2

π 2

π 2

FIGURE 17

The function f (θ) = sec θ is not defined at θ = π2 , but we see in Figure 18 that it is one-to-one on both

[ 0, π2

) and

( π 2 , π

] . Similarly, f (θ) = csc θ is not defined at θ = 0,

but it is one-to-one on [ −π2 , 0

) and

( 0, π2

] . We define the inverse functions as follows:

θ = sec−1 x is the unique angle in [ 0,

π

2

) ∪

(π 2

, π ]

such that sec θ = x

θ = csc−1 x is the unique angle in [ −π

2 , 0

) ∪

( 0,

π

2

] such that csc θ = x

38 C H A P T E R 1 PRECALCULUS REVIEW

Figure 18 shows that the range of f (θ) = sec θ is the set of real numbers x such that |x| ≥ 1. The same is true of f (θ) = csc θ . It follows that both θ = sec−1 x and θ = csc−1 x have domain {x : |x| ≥ 1}.

π

π

x

θ = sec−1 x

f (θ) = sec θ

θ −

1

1−1

−1

θ

π 2

−π 2

π 2π

2

FIGURE 18 f (θ) = sec θ is one-to-one on the interval [0, π ] with π2 removed.

1.5 SUMMARY

• A function f is one-to-one on a domain D if for every value c, the equation f (x) = c has at most one solution for x ∈ D, or, equivalently, if for all a, b ∈ D, f (a) ̸= f (b) unless a = b.

• Let f have domain D and range R. The inverse f −1 (if it exists) is the unique function with domain R and range D satisfying f (f −1(x)) = x and f −1(f (x)) = x.

• The inverse of f exists if and only if f is one-to-one on its domain. • To find the inverse function, solve y = f (x) for x in terms of y to obtain x = g(y). The

inverse is the function g. • Horizontal Line Test: f is one-to-one if and only if every horizontal line intersects the

graph of f in at most one point. • The graph of f −1 is obtained by reflecting the graph of f through the line y = x. • The arcsine and arccosine are defined for −1 ≤ x ≤ 1:

θ = sin−1 x is the unique angle in [ −π

2 , π

2

] such that sin θ = x

θ = cos−1 x is the unique angle in [0, π ] such that cos θ = x • tan−1 x and cot−1 x are defined for all x:

θ = tan−1 x is the unique angle in ( −π

2 , π

2

) such that tan θ = x

θ = cot−1 x is the unique angle in (0, π) such that cot θ = x • sec−1 x and csc−1 x are defined for |x| ≥ 1:

θ = sec−1 x is the unique angle in [ 0,

π

2

) ∪

(π 2

, π ]

such that sec θ = x

θ = csc−1 x is the unique angle in [ −π

2 , 0

) ∪

( 0,

π

2

] such that csc θ = x

1.5 EXERCISES

Preliminary Questions 1. Which of the following satisfy f −1(x) = f (x)?

(a) f (x) = x (b) f (x) = 1 − x (c) f (x) = 1 (d) f (x) = √x (e) f (x) = |x| (f) f (x) = x−1

2. The function f maps teenagers in the United States to their last names. Explain why the inverse function f −1 does not exist.

3. The following fragment of a train schedule for the New Jersey Tran- sit System defines a function f from towns to times. Is f one-to-one?

S E C T I O N 1.5 Inverse Functions 39

What is f −1(6:27)?

Trenton 6:21

Hamilton Township 6:27

Princeton Junction 6:34

New Brunswick 6:38

4. A homework problem asks for a sketch of the graph of the inverse of f (x) = x + cos x. Frank, after trying but failing to find a formula

for f −1(x), says it’s impossible to graph the inverse. Bianca hands in an accurate sketch without solving for f −1. How did Bianca complete the problem?

5. Which of the following quantities is undefined?

(a) sin−1 ( − 12

) (b) cos−1(2)

(c) csc−1 ( 1

2 )

(d) csc−1(2)

6. Give an example of an angle θ such that cos−1(cos θ) ̸= θ . Does this contradict the definition of inverse function?

Exercises 1. Show that f (x) = 7x − 4 is invertible and find its inverse. 2. Is f (x) = x2 + 2 one-to-one? If not, describe a domain on which

it is one-to-one.

3. What is the largest interval containing zero on which f (x) = sin x is one-to-one?

4. Show that f (x) = x − 2 x + 3 is invertible and find its inverse.

(a) What is the domain of f ? The range of f −1? (b) What is the domain of f −1? The range of f ?

5. Verify that f (x) = x3 + 3 and g(x) = (x − 3)1/3 are inverses by showing that f (g(x)) = x and g(f (x)) = x.

6. Repeat Exercise 5 for f (t) = t + 1 t − 1 and g(t) =

t + 1 t − 1 .

7. The escape velocity from a planet of radius R is v(R) = √

2GM R

,

where G is the universal gravitational constant and M is the mass. Find the inverse of v(R) expressing R in terms of v.

In Exercises 8–15, find a domain on which f is one-to-one and a for- mula for the inverse of f restricted to this domain. Sketch the graphs of f and f −1.

8. f (x) = 3x − 2 9. f (x) = 4 − x

10. f (x) = 1 x + 1 11. f (x) =

1 7x − 3

12. f (s) = 1 s2

13. f (x) = 1√ x2 + 1

14. f (z) = z3 15. f (x) = √

x3 + 9 16. For each function shown in Figure 19, sketch the graph of the inverse (restrict the function’s domain if necessary).

x

(A)

y

x

(F)

y

x (B)

y

x

(D)

y

x (C)

y

x

(E)

y

FIGURE 19

17. Which of the graphs in Figure 20 is the graph of a function satis- fying f −1 = f ?

(A) x

(B)

y

x (D)

y

(C)

x

y

x

y

FIGURE 20

18. Let n be a nonzero integer. Find a domain on which f (x) = (1 − xn)1/n coincides with its inverse. Hint: The answer depends on whether n is even or odd.

19. Let f (x) = x7 + x + 1. (a) Show that f −1 exists (but do not attempt to find it). Hint: Show that f is increasing.

(b) What is the domain of f −1? (c) Find f −1(3).

20. Show that f (x) = (x2 + 1)−1 is one-to-one on (−∞, 0], and find a formula for f −1 for this domain of f .

21. Let f (x) = x2 − 2x. Determine a domain on which f −1 exists, and find a formula for f −1 for this domain of f .

22. Show that f (x) = x + x−1 is one-to-one on [1, ∞), and find the corresponding inverse f −1. What is the domain of f −1?

For each of the piecewise-defined functions in Exercises 23–26, deter- mine whether or not the function is one-to-one, and if it is, determine its inverse function.

23.

f (x) = { x when x < 0 2x when x ≥ 0

40 C H A P T E R 1 PRECALCULUS REVIEW

24.

g(x) = {−x when x < −1 x when x ≥ −1

25.

f (x) = { x2 when x < 0 x when x ≥ 0

26.

g(x) = { x when x < 0 x2 when x ≥ 0

In Exercises 27–32, evaluate without using a calculator.

27. cos−1 1 28. sin−1 1 2

29. cot−1 1 30. sec−1 2√ 3

31. tan−1 √

3 32. sin−1(−1) In Exercises 33–42, compute without using a calculator.

33. sin−1 (

sin π

3

) 34. sin−1

( sin

4π 3

)

35. cos−1 (

cos 3π 2

) 36. sin−1

( sin

( −5π

6

))

37. tan−1 (

tan 3π 4

) 38. tan−1(tan π)

39. sec−1(sec 3π) 40. sec−1 (

sec 3π 2

)

41. csc−1 ( csc(−π)

) 42. cot−1

( cot

( −π

4

))

In Exercises 43–46, simplify by referring to the appropriate triangle or trigonometric identity.

43. tan(cos−1 x) 44. cos(tan−1 x)

45. cot(sec−1 x) 46. cot(sin−1 x)

In Exercises 47–54, refer to the appropriate triangle or trigonometric identity to compute the given value.

47. cos ( sin−1 23

) 48. tan

( cos−1 23

)

49. tan ( sin−1 0.8

) 50. cos

( cot−1 1

)

51. cot ( csc−1 2

) 52. tan

( sec−1(−2)

)

53. cot ( tan−1 20

) 54. sin

( csc−1 20

)

Further Insights and Challenges 55. Show that if f is odd and f −1 exists, then f −1 is odd. Show, on the other hand, that an even function does not have an inverse.

56. A cylindrical tank of radius R and length L lying horizontally as in Figure 21 is filled with oil to height h. Show that the volume V (h) of oil in the tank as a function of height h is

V (h) = L (

R2 cos−1 (

1 − h R

) − (R − h)

√ 2hR − h2

)

h

L

R

FIGURE 21 Oil in the tank has level h.

1.6 Exponential and Logarithmic Functions An exponential function is a function of the form f (x) = bx , where b > 0 and b ̸= 1. The number b is called the base. Some examples are f (x) = 2x , g(x) = (1.4)x , and h(x) = 10x . The case b = 1 is excluded because f (x) = 1x is a constant function. Calculators give good decimal approximations to values of exponential functions:

24 = 16, 2−3 = 0.125, (1.4)0.8 ≈ 1.309, 104.6 ≈ 39,810.717

Three properties of exponential functions should be singled out from the start (see Figure 1 for the case b = 2):

• Exponential functions are positive: bx > 0 for all x. • The range of f (x) = bx is the set of all positive real numbers. • f (x) = bx is increasing if b > 1 and decreasing if 0 < b < 1. If b > 1, the exponential function f (x) = bx is not merely increasing but is, in a

certain sense, rapidly increasing. Although the term “rapid increase” is perhaps subjective, the following precise statement is true: f (x) = bx increases more rapidly than the power function g(x) = xn for all n (we will prove this in Section 4.5). For example, Figure 2

S E C T I O N 1.6 Exponential and Logarithmic Functions 41

x

y

21−2 −1

4

1

x

y

21−2 −1

4

1

y = 2x is increasing y = ( )x is decreasing

y = 2x

1 2

y = ( )x12

FIGURE 1

xxx

yyy 3x

3x 3x

x3

x4 x5

1 000,051000,051000,05

010101

f (x) =

g(x) =

f (x) = g(x) = g(x) =

f (x) =

FIGURE 2 Comparison of f (x) = 3x and power functions.

shows that f (x) = 3x eventually overtakes and increases faster than the power functions g(x) = x3, g(x) = x4, and g(x) = x5. Table 1 compares f (x) = 3x and g(x) = x5.

We now review the Laws of Exponents. The most important law is

TABLE 1

x x5 3x

1 1 3 5 3125 243

10 100,000 59,049 15 759,375 14,348,907 25 9,765,625 847,288,609,443

bxby = bx+y

In other words, under multiplication, the exponents add, provided that the bases are the

Gordon Moore (1929– ). Moore, who later became chairman of Intel Corporation, predicted that in the decades following 1965, the number of transistors per integrated circuit would grow “exponentially.” This prediction has held up for nearly five decades and may well continue for several more years. Moore has said, “Moore’s Law is a term that got applied to a curve I plotted in the mid-sixties showing the increase in complexity of integrated circuits versus time. It’s been expanded to include a lot more than that, and I’m happy to take credit for all of it.” (AP Photo/Paul Sakuma)

same. This law does not apply to a product such as 32 · 54.

Laws of Exponents (b > 0)

Rule Example

Exponent zero b0 = 1 Products bxby = bx+y 25 · 23 = 25+3 = 28

Quotients bx

by = bx−y 4

7

42 = 47−2 = 45

Negative exponents b−x = 1 bx

3−4 = 1 34

= 1 81

Power to a power ( bx

)y = bxy

( 32

)4 = 32(4) = 38

Roots b1/n = n √

b 51/2 = √

5

EXAMPLE 1 Rewrite as a whole number or fraction: Be sure you are familiar with the Laws of Exponents. They are used throughout calculus.

(a) 16−1/2 (b) 272/3 (c) 416 · 4−18 (d) 9 3

37

Solution

(a) 16−1/2 = 1 161/2

= 1√ 16

= 1 4

(b) 272/3 = ( 271/3

)2 = 32 = 9

(c) 416 · 4−18 = 4−2 = 1 42

= 1 16

(d) 93

37 =

( 32

)3

37 = 3

6

37 = 3−1 = 1

3

42 C H A P T E R 1 PRECALCULUS REVIEW

In the next example, we use the fact that f (x) = bx is one-to-one. In other words, if bx = by , then x = y.

EXAMPLE 2 Solve for the unknown:

(a) 23x+1 = 25 (b) b3 = 56 (c) 7t+1 = (

1 7

)2t

Solution

(a) If 23x+1 = 25, then 3x + 1 = 5 and thus x = 43 . (b) Raise both sides of b3 = 56 to the 13 power. By the “power to a power” rule,

b = ( b3

)1/3 = ( 56

)1/3 = 56/3 = 52 = 25

(c) Since 17 = 7−1, the right-hand side of the equation is ( 1

7

)2t = (7−1)2t = 7−2t . The equation becomes 7t+1 = 7−2t . Therefore, t + 1 = −2t , or t = − 13 .

The Number e In Chapter 3, we will use calculus to study exponential functions. One of the surprisingAlthough written references to the number

π go back more than 4000 years, mathematicians first became aware of the special role played by e in the seventeenth century. The notation e was introduced by Leonhard Euler, who discovered many fundamental properties of this important number.

insights of calculus is that the most convenient or “natural” base for an exponential function is not b = 10 or b = 2, as one might think at first, but rather a certain irrational number, denoted by e, whose value is approximately e ≈ 2.718. A calculator is used to evaluate specific values of f (x) = ex . For example,

e3 ≈ 20.0855, e−1/4 ≈ 0.7788

In calculus, when we speak of the exponential function, it is understood that the base is e. Another common notation for ex is exp(x).

How is e defined? There are many different definitions, but they all rely on the calculus concept of a limit. We shall discuss one way of defining e in Section 3.2.Another definition is described in Example 4 of Section 1.7. For now, we mention the following two graphical descriptions:

• Using Figure 3(A): Among all exponential functions y = bx , b = e is the unique base for which the slope of the tangent line to the graph at (0, 1) is equal to 1.

• Using Figure 3(B): The number e is the unique number such that the area of the region under the hyperbola y = 1/x for 1 ≤ x ≤ e is equal to 1.

(0, 1)

(A) (B)

y = x + 1

y = ex

Region has area 1

Tangent line has slope 1

1 2 x

2 41 e 3 x

4

3

2

y

4

3

2

1

y

y = 1x

FIGURE 3

From these descriptions it is not clear why e is important. As we will learn, however, the exponential function f (x) = ex plays a fundamental role because it behaves in a particularly simple way with respect to the basic operations of calculus: differentiation and integration.

S E C T I O N 1.6 Exponential and Logarithmic Functions 43

EXAMPLE 3 Draw the graph of y = −e(x−2). Solution Figure 3(A) shows the graph of y = ex . The graph of y = −ex is simply the reflection of this graph over the y-axis, making all values negative instead of positive, as in Figure 4. The graph of y = −e(x−2) is obtained by translating this graph 2 units to the right.

x

y

(2, −1)(0, −1)

(0, 1)y = ex

y = −ex

y = −e (x−2)

FIGURE 4 Graphing y = −e(x−2).

Logarithms Logarithmic functions are inverses of exponential functions. More precisely, if b > 0 and

Seismograph of the 2010 Haiti earthquake, which registered 7.0 on the Richter scale. The Richter scale is based on the logarithm (to base 10) of the amplitude of seismic waves. Each whole-number increase in Richter magnitude corresponds to a tenfold increase in amplitude and approximately 31.6 times more energy. (University of South Carolina and the IRIS Consortium)

b ̸= 1, then the logarithm to the base b, denoted logb x, is the inverse of f (x) = bx . By definition, y = logb x if by = x, so we have

blogb x = x and logb(bx) = x

In other words, logb x is the number to which b must be raised in order to get x. For example,

log2(8) = 3 because 23 = 8 log10(1) = 0 because 100 = 1

log3

( 1 9

) = −2 because 3−2 = 1

32 = 1

9

The logarithm to the base e, denoted ln x, plays a special role and is called the natural logarithm.

ln x = loge x

We use a calculator to evaluate logarithms numerically. For example,

In this text, the natural logarithm is denoted ln x. Other common notations are log x and Log x.

ln 17 ≈ 2.83321 because e2.83321 ≈ 17

As in Figure 5, f (x) = ln x and g(x) = ex are inverse functions, so we have

eln x = x and ln(ex) = x

Recall that the domain of f (x) = bx is R and its range is the set of positive real numbers {x : x > 0}. Since the domain and range are reversed in the inverse function,

• The domain of f (x) = logb x is {x : x > 0}. • The range of f (x) = logb x is the set of all real numbers R.

If b > 1, then logb x is positive for x > 1 and negative for 0 < x < 1. Figure 5 illustrates these facts for the base b = e. Keep in mind that the logarithm of a negative number does not exist. For example, log10(−2) does not exist because 10y = −2 has no solution.

y = ex

y = ln x

y = x

1

1 x

y

FIGURE 5 y = ln x is the inverse of y = ex .

For each law of exponents, there is a corresponding law for logarithms. The rule bx+y = bxby corresponds to the rule

logb(xy) = logb x + logb y

In words: The log of a product is the sum of the logs. To verify this rule, observe that

blogb(xy) = xy = blogb x · blogb y

= blogb x+logb y

The exponents logb(xy) and logb x + logb y are equal as claimed because f (x) = bx is one-to-one. The logarithm laws are collected in the following table.

44 C H A P T E R 1 PRECALCULUS REVIEW

Laws of Logarithms

Law Example

Log of 1 logb(1) = 0 Log of b logb(b) = 1 Products logb(xy) = logb x + logb y log5(2 · 3) = log5 2 + log5 3

Quotients logb

( x

y

) = logb x − logb y log2

( 3 7

) = log2 3 − log2 7

Reciprocals logb

( 1 x

) = − logb x log2

( 1 7

) = − log2 7

Powers (any n) logb(x n) = n logb x log10(82) = 2 · log10 8

We note also that all logarithmic functions are proportional. More precisely, the fol- lowing change-of-base formula holds (see Exercise 51):

logb x = loga x loga b

, logb x = ln x ln b

1

EXAMPLE 4 Using the Logarithm Laws Evaluate:

(a) log6 9 + log6 4 (b) ln (

1√ e

) (c) 10 logb(b

3) − 4 logb( √

b)

Solution

(a) log6 9 + log6 4 = log6(9 · 4) = log6(36) = log6(62) = 2

(b) ln (

1√ e

) = ln(e−1/2) = −1

2 ln(e) = −1

2

(c) 10 logb(b 3) − 4 logb(

√ b) = 10(3) − 4 logb(b1/2) = 30 − 4

( 1 2

) = 28

EXAMPLE 5 Solving an Exponential Equation The bacteria population in a bottle at time t (in hours) has size P(t) = 1000e0.35t . After how many hours will there be 5000 bacteria?

Solution We must solve P(t) = 1000e0.35t = 5000 for t (Figure 6):

1 4.62 3 4

Bacteria population P

t (h) 1000 2000 3000 4000 5000 6000

FIGURE 6 Bacteria population as a function of time.

e0.35t = 5000 1000

= 5

ln(e0.35t ) = ln 5 (take logarithm of both sides) 0.35t = ln 5 ≈ 1.609 [because ln(ea) = a]

t ≈ 1.609 0.35

≈ 4.6 hours

FIGURE 7 The St. Louis Arch has the shape of an inverted hyperbolic cosine. (© Corbis)

Hyperbolic Functions The hyperbolic functions are certain special combinations of ex and e−x that play a role in engineering and physics (see Figure 7 for a real-life example). The hyperbolic sine and cosine, often called “cinch” and “cosh,” are defined as follows:

sinh x = e x − e−x

2 , cosh x = e

x + e−x 2

As the terminology suggests, there are similarities between the hyperbolic and trigono- metric functions. Here are some examples:

S E C T I O N 1.6 Exponential and Logarithmic Functions 45

• Parity: The trigonometric functions and their hyperbolic analogs have the same parity. Thus, f (x) = sin x and f (x) = sinh x are both odd, and f (x) = cos x and f (x) = cosh x are both even (Figure 8):

sinh(−x) = − sinh x, cosh(−x) = cosh x • Identities: The basic trigonometric identity sin2 x + cos2 x = 1 has a hyperbolic

analog:

cosh2 x − sinh2 x = 1 2

The addition formulas satisfied by sin x and cos x also have hyperbolic analogs:

sinh(x + y) = sinh x cosh y + cosh x sinh y cosh(x + y) = cosh x cosh y + sinh x sinh y

• Hyperbola instead of the circle: Because of the identity cosh2 t − sinh2 t = 1, the

x

y

y = sinh x

321−3 −2 −1

1

2

−1

−2

y

x

y = cosh x 321−3 −2 −1

2

4

3

FIGURE 8 y = sinh x is an odd function; y = cosh x is an even function.

point (cosh t, sinh t) lies on the hyperbola x2 − y2 = 1, just as (cos t, sin t) lies on the unit circle x2 + y2 = 1 (Figure 9).

x

y

(cosh t, sinh t)

32−3 −2

1

2

3

−1

−2

−3

(cos t, sin t)

x

y

1−1

x2 − y2 = 1 x2 + y2 = 1

FIGURE 9

• Other hyperbolic functions: The hyperbolic tangent, cotangent, secant, and co- secant functions (see Figures 10 and 11) are defined like their trigonometric coun- terparts:

x

y

2

−2

2

−2

y = csch x

x

y

2 4−4 −2

1

y = sech x

FIGURE 10 The hyperbolic secant and cosecant.

tanh x = sinh x cosh x

= e x − e−x

ex + e−x , sech x = 1

cosh x = 2

ex + e−x

coth x = cosh x sinh x

= e x + e−x

ex − e−x , csch x = 1

sinh x = 2

ex − e−x y

y = tanh x

−2

y = coth x

x

y

2−2 x

2

1

−1

1

−1

FIGURE 11 The hyperbolic tangent and cotangent.

EXAMPLE 6 Verifying the Basic Identity Verify Eq. (2): cosh2 x − sinh2 x = 1.

Solution Because cosh x = 1 2 (ex + e−x) and sinh x = 1

2 (ex − e−x), we have

cosh x + sinh x = ex, cosh x − sinh x = e−x

46 C H A P T E R 1 PRECALCULUS REVIEW

We obtain Eq. (2) by multiplying these two equations together:

cosh2 x − sinh2 x = (cosh x + sinh x)(cosh x − sinh x) = ex · e−x = 1

Inverse Hyperbolic Functions Each of the hyperbolic functions, except y = cosh x and y = sech x, is one-to-one on its domain and therefore has a well-defined inverse. The functions y = cosh x and y = sech x are one-to-one on the restricted domain {x : x ≥ 0}. We let y = cosh−1 x and y = sech−1 x denote the corresponding inverses.

Inverse hyperbolic functions

Function Domain

y = sinh−1 x all x y = cosh−1 x x ≥ 1 y = tanh−1 x |x| < 1 y = coth−1 x |x| > 1 y = sech−1 x 0 < x ≤ 1 y = csch−1 x x ̸= 0

Einstein’s Law of Velocity Addition The inverse hyperbolic tangent plays a role in the Special Theory of Relativity, developed

u = 200,000,000 m/s

FIGURE 12 What is the missile’s velocity relative to the earth?

byAlbert Einstein in 1905. One consequence of this theory is that no object can travel faster than the speed of light, c ≈ 3 × 108 m/s. Einstein realized that this contradicts a law stated by Galileo more than 250 years earlier, namely that velocities add. Imagine a train traveling at u = 50 m/s and a man walking down the aisle in the train at v = 2 m/s. According to Galileo, the man’s velocity relative to the ground is u + v = 52 m/s. This agrees with our everyday experience. But now imagine an (unrealistic) rocket traveling away from the earth at u = 2 × 108 m/s, and suppose that the rocket fires a missile with velocity v = 1.5 × 108 m/s (relative to the rocket). If Galileo’s Law were correct, the velocity of the missile relative to the earth would be u + v = 3.5 × 108 m/s, which exceeds Einstein’s maximum speed limit of c ≈ 3 × 108 m/s.

However, Einstein’s theory replaces Galileo’s Law with a new law stating that the

Einstein’s Law of Velocity Addition [Eq. (3)] reduces to Galileo’s Law, w = u + v, when u and v are small relative to the velocity of light c. See Exercise 52 for another way of expressing Eq. (3).

inverse hyperbolic tangents of velocities add. More precisely, if u is the rocket’s velocity relative to the earth and v is the missile’s velocity relative to the rocket, then the velocity of the missile relative to the earth (Figure 12) is w, where

tanh−1 (w

c

) = tanh−1

(u c

) + tanh−1

(v c

) 3

EXAMPLE 7 A rocket travels away from the earth at a velocity of 2 × 108 m/s. A missile is fired from the rocket at a velocity of 1.5 × 108 m/s (relative to the rocket) away from the earth. Use Einstein’s Law to find the velocity w of the missile relative to the earth.

Solution According to Eq. (3),

tanh−1 (w

c

) = tanh−1

( 2 × 108 3 × 108

) + tanh−1

( 1.5 × 108 3 × 108

) ≈ 0.805 + 0.549 ≈ 1.354

Therefore, w/c ≈ tanh(1.354) ≈ 0.875, and w ≈ 0.875c ≈ 2.6 × 108 m/s. This value obeys the Einstein speed limit of 3 × 108 m/s.

EXAMPLE 8 Low Velocities A plane traveling at 300 m/s fires a missile forward at a velocity of 200 m/s. Calculate the missile’s velocity w relative to the earth using both Einstein’s Law and Galileo’s Law.

Solution According to Einstein’s Law,

tanh−1 (w

c

) = tanh−1

( 300 c

) + tanh−1

( 200 c

)

w = c · tanh (

tanh−1 (

300 c

) + tanh−1

( 200 c

)) ≈ 499.99999999967 m/s

This is practically indistinguishable from the value w = 300 + 200 = 500 m/s obtained using Galileo’s Law.

S E C T I O N 1.6 Exponential and Logarithmic Functions 47

1.6 SUMMARY

• f (x) = bx is the exponential function with base b (where b > 0 and b ̸= 1). • f (x) = bx is increasing if b > 1 and decreasing if b < 1. • Important exponent laws:

(i) bxby = bx+y (ii) bxby = bx−y (iii) b−x = 1bx (iv) (bx)y = bxy

• The number e ≈ 2.718. • For b > 0 with b ̸= 1, the logarithmic function f (x) = logb x is the inverse of f (x) =

bx ; y = logb x ⇔ x = by

• The natural logarithm is the logarithm with base e and is denoted ln x. • eln x = x for x > 0 and ln(ex) = x for all x. • Important logarithm laws:

(i) logb(xy) = logb x + logb y (ii) logb (

x

y

) = logb x − logb y

(iii) logb(x n) = n logb x (iv) logb 1 = 0 and logb b = 1

• The hyperbolic sine and cosine:

sinh x = e x − e−x

2 (odd function), cosh x = e

x + e−x 2

(even function)

The remaining hyperbolic functions:

tanh x = sinh x cosh x

, coth x = cosh x sinh x

, sech x = 1 cosh x

, csch x = 1 sinh x

• Basic identity: cosh2 x − sinh2 x = 1. • The inverse hyperbolic functions and their domains:

f (x) = sinh−1 x, for all x f (x) = coth−1 x, for |x| > 1 f (x) = cosh−1 x, for x ≥ 1 f (x) = sech−1 x, for 0 < x ≤ 1 f (x) = tanh−1 x, for |x| < 1 f (x) = csch−1 x, for x ̸= 0

1.6 EXERCISES

Preliminary Questions 1. Which of the following equations is incorrect?

(a) 32 · 35 = 37 (b) ( √

5)4/3 = 52/3

(c) 32 · 23 = 1 (d) (2−2)−2 = 16

2. Compute logb2(b 4). 3. When is ln x negative?

4. What is ln(−3)? Explain.

5. Explain the phrase, “The logarithm converts multiplication into addition.”

6. What are the domain and range of f (x) = ln x? 7. Which hyperbolic functions take on only positive values?

8. Which hyperbolic functions are increasing on their domains?

9. Describe three properties of hyperbolic functions that have trigo- nometric analogs.

Exercises 1. Rewrite as a whole number (without using a calculator):

(a) 70 (b) 102(2−2 + 5−2)

(c)

( 43

)5 ( 45

) 3 (d) 27

4/3

(e) 8−1/3 · 85/3 (f) 3 · 41/4 − 12 · 2−3/2

In Exercises 2–10, solve for the unknown variable.

2. 92x = 98 3. e2x = ex+1

4. et 2 = e4t−3 5. 3x =

( 1 3 )x+1

6. ( √

5)x = 125 7. 4−x = 2x+1

8. b4 = 1012 9. k3/2 = 27 10.

( b2

)x+1 = b−6

In Exercises 11–26, calculate without using a calculator.

11. log3 27 12. log5 1

25

48 C H A P T E R 1 PRECALCULUS REVIEW

13. ln 1 14. log5(5 4)

15. log2(2 5/3) 16. log2(8

5/3)

17. log64 4 18. log7(49 2)

19. log8 2 + log4 2 20. log25 30 + log25 56 21. log4 48 − log4 12 22. ln(

√ e · e7/5)

23. ln(e3) + ln(e4) 24. log2 43 + log2 24

25. 7log7(29) 26. 83 log8(2)

27. Write as the natural log of a single expression: (a) 2 ln 5 + 3 ln 4 (b) 5 ln(x1/2) + ln(9x)

28. Solve for x: ln(x2 + 1) − 3 ln x = ln(2). In Exercises 29–34, solve for the unknown.

29. 7e5t = 100 30. 6e−4t = 2 31. 2x

2−2x = 8 32. e2t+1 = 9e1−t

33. ln(x4) − ln(x2) = 2 34. log3 y + 3 log3(y2) = 14

35. Find the inverse of y = e2x−3. 36. Find the inverse of y = ln(x2 − 2) for x >

√ 2.

37. Use a calculator to compute sinh x and cosh x for x = −3, 0, 5. 38. Compute sinh(ln 5) and tanh(3 ln 5) without using a calculator.

39. Show, by producing a counterexample, that ln(ab) is not equal to (ln a)(ln b).

40. For which values of x are y = sinh x and y = cosh x increasing and decreasing?

41. Show that y = tanh x is an odd function. 42. The population of a city (in millions) at time t (years) is P(t) = 2.4e0.06t , where t = 0 is the year 2000.

(a) What is the population at time t = 0? (b) When will the population double from its size at t = 0?

43. The Gutenberg–Richter Law states that the number N of earth- quakes per year worldwide of Richter magnitude at least M satisfies an approximate relation log10 N = a − M for some constant a. Find a, assuming that there is one earthquake of magnitude M ≥ 8 per year. How many earthquakes of magnitude M ≥ 5 occur per year?

44. The energy E (in joules) radiated as seismic waves from an earth- quake of Richter magnitude M is given by the formula log10 E = 4.8 + 1.5M . (a) Express E as a function of M . (b) Show that when M increases by 1, the energy increases by a factor of approximately 31.6.

45. Refer to the graphs to explain why the equation sinh x = t has a unique solution for every t and why cosh x = t has two solutions for every t > 1.

46. Compute cosh x and tanh x, assuming that sinh x = 0.8.

47. Prove the addition formula for cosh x given by cosh(x + y) = cosh x cosh y + sinh x sinh y.

48. Use the addition formulas to prove

sinh(2x) = 2 cosh x sinh x

cosh(2x) = cosh2 x + sinh2 x

49. A train moves along a track at velocity v. Bionica walks down the aisle of the train with velocity u in the direction of the train’s motion. Compute the velocity w of Bionica relative to the ground using the laws of both Galileo and Einstein in the following cases: (a) v = 500 m/s and u = 10 m/s. Is your calculator accurate enough to detect the difference between the two laws? (b) v = 107 m/s and u = 106 m/s.

Further Insights and Challenges 50. Show that loga b logb a = 1 for all a, b > 0 such that a ̸= 1 and b ̸= 1.

51. Verify that for all x, the formula holds. logb x = loga x loga b

for a, b >

0 such that a ̸= 1, b ̸= 1.

52. (a) Use the addition formulas for sinh x and cosh x to prove

tanh(u + v) = tanh u + tanh v 1 + tanh u tanh v

(b) Use (a) to show that Einstein’s Law of Velocity Addition [Eq. (3)] is equivalent to

w = u + v 1 + uv

c2

53. Prove that every function f can be written as a sum f (x) = f+(x) + f−(x) of an even function f+(x) and an odd function f−(x). Express f (x) = 5ex + 8e−x in terms of cosh x and sinh x. Hint: y = f (x) + f (−x) is an even function, and y = f (x) − f (−x) is an odd function.

1.7 Technology: Calculators and Computers Computer technology has vastly extended our ability to calculate and visualize mathemat- ical relationships. In applied settings, computers are indispensable for solving complex systems of equations and analyzing data, as in weather prediction and medical imaging. Mathematicians use computers to study complex structures such as the Mandelbrot Set (Figures 1 and 2). We take advantage of this technology to explore the ideas of calculus visually and numerically.

S E C T I O N 1.7 Technology: Calculators and Computers 49

FIGURE 1 Computer-generated image of the Mandelbrot Set, which occurs in the mathematical theory of chaos and fractals. (Scott Camazine/Science Source)

FIGURE 2 Even greater complexity is revealed when we zoom in on a portion of the Mandelbrot Set. (Laguna Design/Science Photo Library)

When we plot a function with a graphing calculator or computer algebra system, the graph is contained within a viewing rectangle, the region determined by the range of x- and y-values in the plot. We write [a, b] × [c, d] to denote the rectangle where a ≤ x ≤ b and c ≤ y ≤ d .

The appearance of the graph depends heavily on the choice of viewing rectangle. Different choices may convey very different impressions that are sometimes misleading. Compare the three viewing rectangles for the graph of f (x) = 12 − x − x2 in Figure 3. Only (A) successfully displays the shape of the graph as a parabola. In (B), the graph is cut off, and no graph at all appears in (C). Keep in mind that the scales along the axes may change with the viewing rectangle. For example, the unit increment along the y-axis is larger in (B) than in (A), so the graph in (B) is steeper.

18

−18

4

−4

1

−3 (A) [−6, 5] × [−18, 18] (B) [−6, 5] × [−4, 4] (C) [−1, 2] × [−3, 1]

−6 −65 5

−1 2

FIGURE 3 Viewing rectangles for the graph of f (x) = 12 − x − x2.

There is no single “correct” viewing rectangle. The goal is to select the viewing rectangle that displays the properties you wish to investigate. This usually requires exper- imentation.

EXAMPLE 1 How Many Roots and Where? How many real roots does the functionTechnology is indispensable but also has its limitations. When shown the computer- generated results of a complex calculation, the Nobel prize–winning physicist Eugene Wigner (1902–1995) is reported to have said: “It is nice to know that the computer understands the problem, but I would like to understand it too.”

f (x) = x9 − 20x + 1 have? Find their approximate locations. Solution We experiment with several viewing rectangles (Figure 4). Our first attempt (A) displays a cut-off graph, so we try a viewing rectangle that includes a larger range of y-values. Plot (B) shows that the roots of f probably lie somewhere in the interval [−3, 3], but it does not reveal how many real roots there are. Therefore, we try the viewing

10

−10 (A) [−12, 12] × [−10, 10]

−12 12

10,000

−10,000 (B) [−12, 12] × [−10,000, 10,000]

−12 12

100

−100 (C) [−2, 2] × [−100, 100]

−2 2

25

−20

−1.5−1−.5 .1 .5 1 1.5

(D) [−2, 2] × [−20, 25]

−2 2

FIGURE 4 Graphs of f (x) = x9 − 20x + 1.

50 C H A P T E R 1 PRECALCULUS REVIEW

rectangle in (C). Now we can see clearly that f has three roots. A further zoom in (D) shows that these roots are located near −1.5, 0.1, and 1.5. Further zooming would provide their locations with greater accuracy.

EXAMPLE 2 Does a Solution Exist? Does cos x = tan x have a solution? Describe the set of all solutions.

Solution The solutions of cos x = tan x are the x-coordinates of the points where the graphs of y = cos x and y = tan x intersect. Figure 5(A) shows that there are two solutions in the interval [0, 2π ]. By zooming in on the graph as in (B), we see that the first positive root lies between 0.6 and 0.7 and the second positive root lies between 2.4 and 2.5. Further zooming shows that the first root is approximately 0.67 [Figure 5(C)]. Continuing this process, we find that the first two roots are x ≈ 0.666 and x ≈ 2.475.

Since f (x) = cos x and f (x) = tan x are periodic, the picture repeats itself with period 2π . All solutions are obtained by adding multiples of 2π to the two solutions in [0, 2π ]:

x ≈ 0.666 + 2πk and x ≈ 2.475 + 2πk (for any integer k)

2π 2π 4π

5

−5 (A) [−7, 13] × [−5, 5]

−7 13

5

−5 (B) [0, 3] × [−5, 5] (C) [0.5, 0.7] × [0.55, 0.85]

0 3

y = cos x

y = tan x

0.6 0.7 2.4 2.5

0.55 0.6 0.65 0.7

FIGURE 5 Graphs of y = cos x and y = tan x.

EXAMPLE 3 Functions with Asymptotes Plot the function f (x) = 1 − 3x x − 2 and de-scribe its asymptotic behavior.

Solution First, we plot f in the viewing rectangle [−10, 20] × [−5, 5] as in Figure 7(A). The vertical line x = 2 is called a vertical asymptote. Many graphing calcula- tors display this line, but it is not part of the graph (and it can usually be eliminated by choosing a smaller range of y-values). We see that f (x) tends to ∞ as x approaches 2 from the left, and to −∞ as x approaches 2 from the right. To display the horizon-

CAUTION When considering the graph of a function such as y = ln x (Figure 6), it may appear to approach a horizontal asymptote, but in fact, it does not. For any given horizontal line, the graph eventually rises above it.In this respect, graphing calculators and computer graphing systems must be used judiciously.

x

y

FIGURE 6 y = ln x.

tal asymptotic behavior of f , we use the viewing rectangle [−10, 20] × [−10, 5] [Fig- ure 7(B)]. Here, we see that the graph approaches the horizontal line y = −3, called a hori- zontal asymptote (which we have added as a dashed horizontal line in the figure).

(A) [−10, 20] × [−5, 5]

20−10

5

−5

−3

2

2

(B) [−10, 20] × [−10, 5]

20−10

5

−10

−3

FIGURE 7 Graphs of f (x) = 1 − 3x x − 2 .

Calculators and computer algebra systems give us the freedom to experiment numer- ically. For instance, we can explore the behavior of a function by constructing a table of values. In the next example, we investigate a function related to exponential functions and compound interest (see Section 5.8).

S E C T I O N 1.7 Technology: Calculators and Computers 51

EXAMPLE 4 Investigating the Behavior of a Function How does f (n) = (1 + 1/n)n behave for large whole-number values of n? Does f (n) tend to infinity as n gets larger?

Solution First, we make a table of values of f (n) for larger and larger values of n. Table 1 suggests that f (n) does not tend to infinity. Rather, as n grows larger, f (n) appears to get closer to some value near 2.718 (a number resembling e). This is an example of limiting behavior that we will discuss in Chapter 2. Next, replace n by the variable x and plot the function f (x) = (1 + 1/x)x . The graphs in Figure 8 confirm that f (x) approaches a limit of approximately 2.7. We will prove that f (n) approaches e as n tends to infinity in Section 5.9.

TABLE 1

n

( 1 + 1

n

)n

10 2.59374 102 2.70481 103 2.71692 104 2.71815 105 2.71827 106 2.71828

(A) [0, 10] × [0, 3]

100

3

0

(B) [0, 1000] × [0, 3]

5

2.7

1

10000

3

0 500

2.7

1

FIGURE 8 Graphs of f (x) = (

1 + 1 x

)x .

EXAMPLE 5 Bird Flight: Finding a Minimum Graphically According to one model of bird flight, the power consumed by a pigeon flying at velocity v (in meters per second) is P(v) = 17v−1 + 10−3v3 (in joules per second). Use a graph of P to find the velocity that minimizes power consumption.

Solution The velocity that minimizes power consumption corresponds to the lowest point on the graph of P . We plot P first in a large viewing rectangle (Figure 9). This figure reveals the general shape of the graph and shows that P takes on a minimum value for v somewhere between v = 8 and v = 9. In the viewing rectangle [8, 9.2] × [2.6, 2.65], we see that the minimum occurs at approximately v = 8.65 m/s.

(A) [0, 20] × [0, 12]

200

12

P (J/s) P (J/s)

v (m/s) v (m/s) 0

(B) [8, 9.2] × [2.6, 2.65]

10 155

10

9.28

2.65

2.6

2.64

2.63

2.62

2.61

98.4 8.88.2 8.6

FIGURE 9 Power consumption P(v) as a function of velocity v.

Local linearity is an important concept in calculus that is based on the idea that many functions are nearly linear over small intervals. Local linearity can be illustrated effectively with a graphing calculator.

EXAMPLE 6 Illustrating Local Linearity Illustrate local linearity for the function f (x) = xsin x at x = 1.

Solution First, we plot f (x) = xsin x in the viewing window of Figure 10(A). The graph moves up and down and appears very wavy. However, as we zoom in, the graph straight- ens out. Figures (B)–(D) show the result of zooming in on the point (1, f (1)). When viewed up close, the graph looks like a straight line. This illustrates the local linearity of f at x = 1.

52 C H A P T E R 1 PRECALCULUS REVIEW

8

1

1

0

(A) (B) (C)

120

2

1

0.8 0.95 1.051 1.2

0

20

1.2 1.05

0.8 0.95

1.20.8

(D)

1.050.95

FIGURE 10 Zooming in on the graph of f (x) = xsin x near x = 1.

1.7 SUMMARY

• The appearance of a graph on a graphing calculator depends on the choice of viewing rectangle. Experiment with different viewing rectangles until you find one that displays the information you want. Keep in mind that the scales along the axes may change as you vary the viewing rectangle.

• The following are some ways in which graphing calculators and computer algebra systems can be used in calculus:

– Visualizing the behavior of a function – Finding solutions graphically or numerically – Conducting numerical or graphical experiments – Illustrating theoretical ideas (such as local linearity)

1.7 EXERCISES

Preliminary Questions 1. Is there a definite way of choosing the optimal viewing rectangle,

or is it best to experiment until you find a viewing rectangle appropriate to the problem at hand?

2. Describe the calculator screen produced when the function y = 3 + x2 is plotted with a viewing rectangle: (a) [−1, 1] × [0, 2] (b) [0, 1] × [0, 4]

3. According to the evidence in Example 4, it appears that f (n) = (1 + 1/n)n never takes on a value greater than 3 for n > 0. Does this evidence prove that f (n) ≤ 3 for n > 0?

4. How can a graphing calculator be used to find the minimum value of a function?

Exercises The exercises in this section should be done using a graphing calculator or computer algebra system.

1. Plot f (x) = 2x4 + 3x3 − 14x2 − 9x + 18 in the appropriate viewing rectangles and determine its roots.

2. How many solutions does x3 − 4x + 8 = 0 have?

3. How many positive solutions does x3 − 12x + 8 = 0 have?

4. Does cos x + x = 0 have a solution? A positive solution?

5. Find all the solutions of sin x = √x for x > 0.

6. How many solutions does cos x = x2 have?

7. Let f (x) = (x − 100)2 + 1000. What will the display show if you graph f in the viewing rectangle [−10, 10] by [−10, 10]? Find an ap- propriate viewing rectangle.

8. Plot f (x) = 8x + 1 8x − 4 in an appropriate viewing rectangle. What are

the vertical and horizontal asymptotes?

9. Plot the graph of f (x) = x/(4 − x) in a viewing rectangle that clearly displays the vertical and horizontal asymptotes.

10. Illustrate local linearity for f (x) = x2 by zooming in on the graph at x = 0.5 (see Example 6).

11. Plot f (x) = cos(x2) sin x for 0 ≤ x ≤ 2π . Then illustrate local linearity at x = 3.8 by choosing appropriate viewing rectangles.

12. If P0 dollars are deposited in a bank account paying 5% interest

compounded monthly, then the account has value P0 (

1 + 0.0512 )N

af- ter N months. Find, to the nearest integer N , the number of months after which the account value doubles.

In Exercises 13–18, investigate the behavior of the function as n or x grows large by making a table of function values and plotting a graph (see Example 4). Describe the behavior in words.

13. f (n) = n1/n 14. f (n) = 4n + 1 6n − 5

15. f (n) = (

1 + 1 n

)n2 16. f (x) =

( x + 6 x − 4

)x

17. f (x) = (

x tan 1 x

)x 18. f (x) =

( x tan

1 x

)x2

Chapter Review Exercises 53

19. The graph of f (θ) = A cos θ + B sin θ is a sinusoidal wave for any constants A and B. Confirm this for (A, B) = (1, 1), (1, 2), and (3, 4) by plotting f .

20. Find the maximum value of f for the graphs produced in Exer- cise 19. Can you guess the formula for the maximum value in terms of A and B?

21. Find the intervals on which f (x) = x(x + 2)(x − 3) is positive by plotting a graph.

22. Find the set of solutions to the inequality (x2 − 4)(x2 − 1) < 0 by plotting a graph.

Further Insights and Challenges 23. Let f1(x) = x and define a sequence of functions by fn+1(x) = 12 (fn(x) + x/fn(x)). For example, f2(x) = 12 (x + 1). Use a computer algebra system to compute fn(x) for n = 3, 4, 5 and plot y = fn(x) together with y =

√ x for x ≥ 0. What do you notice?

24. Set P0(x) = 1 and P1(x) = x. The Chebyshev polynomials (use- ful in approximation theory) are defined inductively by the formula Pn+1(x) = 2xPn(x) − Pn−1(x).

(a) Show that P2(x) = 2x2 − 1.

(b) Compute Pn(x) for 3 ≤ n ≤ 6 using a computer algebra system or by hand, and plot y = Pn(x) over [−1, 1].

(c) Check that your plots confirm two interesting properties: (a) y = Pn(x) has n real roots in [−1, 1] and (b) for x ∈ [−1, 1], Pn(x) lies between −1 and 1.

CHAPTER REVIEW EXERCISES

1. Express (4, 10) as a set {x : |x − a| < c} for suitable a and c. 2. Express as an interval:

(a) {x : |x − 5| < 4} (b) {x : |5x + 3| ≤ 2} 3. Express {x : 2 ≤ |x − 1| ≤ 6} as a union of two intervals. 4. Give an example of numbers x, y such that |x| + |y| = x − y. 5. Describe the pairs of numbers x, y such that |x + y| = x − y. 6. Sketch the graph of y = f (x + 2) − 1, where f (x) = x2 for

−2 ≤ x ≤ 2. In Exercises 7–10, let f (x) be the function shown in Figure 1.

7. Sketch the graphs of y = f (x) + 2 and y = f (x + 2). 8. Sketch the graphs of y = 12f (x) and y = f

( 1 2x

) .

9. Continue the graph of f to the interval [−4, 4] as an even function. 10. Continue the graph of f to the interval [−4, 4] as an odd function.

1 2 3 4

1

2

0

3

x

y

FIGURE 1

In Exercises 11–14, find the domain and range of the function.

11. f (x) = √

x + 1 12. f (x) = 4 x4 + 1

13. f (x) = 2 3 − x 14. f (x) =

√ x2 − x + 5

15. Determine whether the function is increasing, decreasing, or nei- ther:

(a) f (x) = 3−x (b) f (x) = 1 x2 + 1

(c) g(t) = t2 + t (d) g(t) = t3 + t

16. Determine whether the function is even, odd, or neither: (a) f (x) = x4 − 3x2 (b) g(x) = sin(x + 1) (c) f (x) = 2−x2

In Exercises 17–24, find the equation of the line.

17. Line passing through (−1, 4) and (2, 6) 18. Line passing through (−1, 4) and (−1, 6) 19. Line of slope 6 through (9, 1)

20. Line of slope − 32 through (4, −12) 21. Line through (2, 1) perpendicular to the line given by y = 3x + 7 22. Line through (3, 4) perpendicular to the line given by y = 4x − 2 23. Line through (2, 3) parallel to y = 4 − x 24. Horizontal line through (−3, 5) 25. Does the following table of market data suggest a linear relation- ship between price and number of homes sold during a one-year period? Explain.

Price (thousands of $) 180 195 220 240

No. of homes sold 127 118 103 91

26. Does the following table of revenue data for a computer manufac- turer suggest a linear relation between revenue and time? Explain.

Year 2005 2009 2011 2014

Revenue (billions of $) 13 18 15 11

27. Suppose that a cell phone plan that is offered at a price of P dollars per month attracts C customers, where C(P ) is a linear demand function for $100 ≤ P ≤ $500. If C(100) = 1,000,000 and C(500) = 100,000, determine the demand function C. What is the decrease in the number of customers for each increase of $100 in the price?

28. Suppose that Internet domain names are sold at a price of $P per month for $2 ≤ P ≤ $100. The number of customers C who buy the domain names is a linear function of the price. If 10,000 customers buy

54 C H A P T E R 1 PRECALCULUS REVIEW

a domain name when the price is $2 per month and 1000 customers buy when the price is $100 per month, determine the demand function C. What is the decrease in the number of customers for every $1 increase in the cost of the domain names?

29. Find the roots of f (x) = x4 − 4x2 and sketch its graph. On which intervals is f decreasing?

30. Let h(z) = −2z2 + 12z + 3. Complete the square and find the maximum value of h.

31. Let f (x) be the square of the distance from the point (2, 1) to a point (x, 3x + 2) on the line y = 3x + 2. Show that f is a quadratic function, and find its minimum value by completing the square.

32. Prove that x2 + 3x + 3 ≥ 0 for all x. In Exercises 33–38, sketch the graph by hand.

33. y = t4 34. y = t5

35. y = sin θ 2

36. y = 10−x

37. y = x1/3 38. y = 1 x2

39. Show that the graph of y = f ( 1

3x − b )

is obtained by shifting the

graph of y = f ( 1

3x )

to the right 3b units. Use this observation to sketch

the graph of y = ∣∣ 1

3x − 4 ∣∣.

40. Let h(x) = cos x and g(x) = x−1. Compute the composite func- tions h(g) and g(h), and find their domains.

41. Find functions f and g such that the function

f (g(t)) = (12t + 9)4

42. Sketch the points on the unit circle corresponding to the follow- ing three angles, and find the values of the six standard trigonometric functions at each angle:

(a) 2π 3

(b) 7π 4

(c) 7π 6

43. What are the periods of these functions? (a) y = sin 2θ (b) y = sin θ2 (c) y = sin 2θ + sin θ2 44. Assume that sin θ = 45 , where π/2 < θ < π . Find:

(a) tan θ (b) sin 2θ (c) csc θ

2

45. Give an example of values a, b such that

(a) cos(a + b) ̸= cos a + cos b (b) cos a 2

̸= cos a 2

46. Let f (x) = cos x. Sketch the graph of y = 2f (

1 3x − π4

) for

0 ≤ x ≤ 6π . 47. Solve sin 2x + cos x = 0 for 0 ≤ x < 2π . 48. How does h(n) = n2/2n behave for large whole-number values of n? Does h(n) tend to infinity?

49. Use a graphing calculator to determine whether the equa- tion cos x = 5x2 − 8x4 has any solutions.

50. Using a graphing calculator, find the number of real roots and estimate the largest root to two decimal places: (a) f (x) = 1.8x4 − x5 − x (b) g(x) = 1.7x4 − x5 − x

51. Match each quantity (a)–(d) with (i), (ii), or (iii) if possible, or state that no match exists.

(a) 2a3b (b) 2a

3b

(c) (2a)b (d) 2a−b3b−a

(i) 2ab (ii) 6a+b (iii) ( 2

3 )a−b

52. Match each quantity (a)–(d) with (i), (ii), or (iii) if possible, or state that no match exists.

(a) ln (a

b

) (b)

ln a ln b

(c) eln a−ln b (d) (ln a)(ln b)

(i) ln a + ln b (ii) ln a − ln b (iii) a b

53. Find the inverse of f (x) = √

x3 − 8 and determine its domain and range.

54. Find the inverse of f (x) = x − 2 x − 1 and determine its domain and

range.

55. Find a domain on which h(t) = (t − 3)2 is one-to-one and deter- mine the inverse on this domain.

56. Show that g(x) = x x − 1 is equal to its inverse on the domain

{x : x ̸= 1}.

57. Let

f (x) = { −x2 when x < 0 x when x ≥ 0

(a) Is f increasing? (b) Does f have an inverse? If so, what is it?

58. Let

f (x) = { x − 1 when x < 1 ln x when x ≥ 1

(a) Is f increasing? (b) Does f have an inverse? If so, what is it?

59. Suppose that g is the inverse of f . Match the functions (a)–(d) with their inverses (i)–(iv).

(a) f (x) + 1 (b) f (x + 1) (c) 4f (x) (d) f (4x)

(i) g(x)/4 (ii) g(x/4) (iii) g(x − 1) (iv) g(x) − 1

60. Plot f (x) = xe−x and use the zoom feature to find two solutions of f (x) = 0.3.

This “strange attractor” represents limit

behavior that appeared first in weather models

studied by meteorologist E. Lorenz in 1963. (Scott Camazine/Science Source)

2 LIMITS

C alculus is usually divided into two branches, differential and integral, partly for his-torical reasons. The subject grew out of efforts in the seventeenth century to solve two important geometric problems: finding tangent lines to curves (differential calculus) and computing areas under curves (integral calculus). However, calculus is a broad subject with no clear boundaries. It includes other topics, such as the theory of infinite series, and it has an extraordinarily wide range of applications. What makes these methods and applications part of calculus is that they all rely on the concept of a limit. We will see throughout the text how limits allow us to make computations and solve problems that cannot be solved using algebra alone.

This chapter introduces the limit concept and sets the stage for our study of the derivative in Chapter 3. The first section, intended as motivation, discusses how limits arise in the study of rates of change and tangent lines.

2.1 Limits, Rates of Change, and Tangent Lines Limits are about how a function f behaves, as x approaches a number a. Does f (x) get closer and closer to a number L? If so, we say L is the limit of f (x) as x approaches a. If not, we say the limit does not exist. Limits play a key role throughout calculus. In this section, we discuss their role in understanding rates of change.

Rates of change become critical whenever we study the relationship between two changing quantities. Velocity is a familiar example (the rate of change of position with respect to time), but there are many others, such as

• Rate of change in the number of infected individuals with respect to time • Rate of change in consumer price index with respect to time • Rate of change of atmospheric temperature with respect to altitude

Roughly speaking, if y and x are related quantities, the rate of change should tell us how much y changes in response to a unit change in x. For example, if an automobile travels at a velocity of 80 km/h, then its position changes by 80 km for each unit change in time (the unit being 1 hour). If the trip lasts only half an hour, its position changes by 40 km, and in general, the change in position is 80t km, where t is the change in time (i.e., the time elapsed in hours). In other words,

Change in position = velocity × change in time

However, this simple formula is not valid or even meaningful if the velocity is not constant. After all, if the automobile is accelerating or decelerating, which velocity would we use in the formula?

The problem of extending this formula to account for changing velocity lies at the heart of calculus. As we will learn, differential calculus uses the limit concept to define instantaneous velocity, and integral calculus enables us to compute the change in position in terms of instantaneous velocity. But these ideas are very general. They apply to all rates of change, making calculus an indispensable tool for modeling an amazing range of real-world phenomena.

In this section, we discuss velocity and other rates of change, emphasizing their graphical interpretation in terms of tangent lines. Although at this stage, we do not define precisely what a tangent line is—this will have to wait until Chapter 3—you can think of a tangent line as a line that skims a curve at a point, as in Figures 1(A) and (B) but not (C).

55

56 C H A P T E R 2 LIMITS

(A) (B) (C) FIGURE 1 The line is tangent in (A) and

(B) but not in (C).

HISTORICAL

PERSPECTIVE

(N A

SA /J

P L-

C al

te ch

/S TS

cI )

Philosophy is written in this grand book—I mean the universe— which stands continually open to our

gaze, but it cannot be understood unless one first learns to comprehend the language … in which it is written. It is written in the language of mathematics.…

—Galileo Galilei, 1623

The scientific revolution of the sixteenth and seventeenth centuries reached its high point in the work of Isaac Newton (1643–1727), who was the first scientist to show that the physical world, despite its complexity and diversity, is governed by a small number of universal laws. One of Newton’s great insights was that the uni- versal laws are dynamical, describing how the world changes over time in response to forces, rather than how the world actually is at any given moment in time. These laws are expressed best in the language of calculus, which is the mathe- matics of change.

More than 50 years before the work of Newton, the astronomer Johannes Kepler (1571–1630) discovered his three laws of plan- etary motion, the most famous of which states that the path of a planet around the sun is an ellipse. Kepler arrived at these laws through a painstaking analysis of astronomical data, but he could not explain why they were true. Accord- ing to Newton, the motion of any object—planet or pebble—is determined by the forces acting on it. The planets, if left undisturbed, would travel in straight lines. Since their paths are el- liptical, some force—in this case, the gravita- tional force of the sun—must be acting to make them change direction continuously. In his mag- num opus Principia Mathematica, published in 1687, Newton proved that Kepler’s laws follow from Newton’s own universal laws of motion and gravity.

For these discoveries, Newton gained widespread fame in his lifetime. His fame con- tinued to increase after his death, assuming a nearly mythic dimension, and his ideas had a profound influence, not only in science but also in the arts and literature, as expressed in this epi- taph by British poet Alexander Pope: “Nature and Nature’s Laws lay hid in Night. God said, Let Newton be! and all was Light.”

(© Je

re m

y P

em br

ey /A

la m

y)

This statue of Isaac Newton in Cambridge University was described in The Prelude, a poem by William Wordsworth (1770–1850):

Newton with his prism and silent face, The marble index of a mind for ever Voyaging through strange seas of Thought,

alone. Velocity When we speak of velocity, we usually mean instantaneous velocity, which indicates theIn linear motion, velocity may be positive or

negative (indicating forward or backward direction of motion). Speed, by definition, is the absolute value of velocity and is always positive.

speed and direction of an object at a particular moment. The idea of instantaneous velocity makes intuitive sense, but care is required to define it precisely.

Consider an object traveling in a straight line (linear motion). The average velocity over a given time interval has a straightforward definition as the ratio

Average velocity = change in position length of time interval

For example, if an automobile travels 200 km in 4 hours, then its average velocity during this 4-hour period is 2004 = 50 km/h. At any given moment, the automobile may be going faster or slower than the average.

We cannot define instantaneous velocity as a ratio because we would have to divide by the length of the time interval (which is zero). However, we should be able to estimate instantaneous velocity by computing average velocity over successively smaller time intervals. The guiding principle is: Average velocity over a very small time interval is very close to instantaneous velocity. To explore this idea further, we introduce some notation.

The Greek letter ! (delta) is commonly used to denote the change in a function or variable. If s(t) is the position of an object (distance from the origin) at time t and [t0, t1] is a time interval, we set

!s = s(t1) − s(t0) = change in position !t = t1 − t0 = change in time (length of time interval)

S E C T I O N 2.1 Limits, Rates of Change, and Tangent Lines 57

The change in position !s is also called the displacement, or net change in position. For t1 ̸= t0,

Average velocity over [t0, t1] = !s

!t = s(t1) − s(t0)

t1 − t0

One type of motion we will study is the motion of an object falling to earth under

t = 0 t = 0.5

0

4.9

1.225

t = 1

FIGURE 2 Distance traveled by a falling object after t seconds is s(t) = 4.9t2 meters.

the influence of gravity (assuming no air resistance). Galileo discovered that if the object is released at time t = 0 from a state of rest (Figure 2), then the distance traveled after t seconds is given by the formula

s(t) = 4.9t2 m

EXAMPLE 1 A hammer falls off a scaffolding at time t = 0. Estimate the instanta- neous velocity at t = 0.8 s. Solution We use Galileo’s formula s(t) = 4.9t2 to compute the average velocity over theTABLE 1

Time intervals Average velocity

[0.8, 0.81] 7.889 [0.8, 0.805] 7.8645 [0.8, 0.8001] 7.8405 [0.8, 0.80005] 7.84024 [0.8, 0.800001] 7.840005

five short time intervals listed in Table 1. Consider the first interval [t0, t1] = [0.8, 0.81]:

!s = s(0.81) − s(0.8) = 4.9(0.81)2 − 4.9(0.8)2 ≈ 3.2149 − 3.1360 = 0.7889 m !t = 0.81 − 0.8 = 0.01 s

The average velocity over [0.8, 0.81] is the ratio !s

!t = s(0.81) − s(0.8)

0.81 − 0.8 = 0.07889

0.01 = 7.889 m/s

Table 1 shows the results of similar calculations for intervals of successively shorterThere is nothing special about the particular time intervals in Table 1. We are looking for a trend, and we could have chosen any intervals [0.8, t] for values of t approaching 0.8. We could also have chosen intervals [t, 0.8] for t < 0.8.

lengths. It looks like these average velocities are getting closer to 7.84 m/s as the length of the time interval shrinks:

7.889, 7.8645, 7.8405, 7.84024, 7.840005

This suggests that 7.84 m/s is a good candidate for the instantaneous velocity at t = 0.8.

We express our conclusion in the previous example by saying the following:

Average velocity converges to instantaneous velocity. Instantaneous velocity is the limit of average velocity as the length of the time interval shrinks to zero.

We might write this as instantaneous velocity = lim !t→0

(average velocity).

Graphical Interpretation of Velocity The idea that average velocity converges to instantaneous velocity as we shorten the time interval has a vivid interpretation in terms of secant lines. The term secant line refers to a line through two points on a curve.

Consider the graph of position s(t) for an object traveling in a straight line such as a bicycle (Figure 3). The ratio defining average velocity over [t0, t1] is nothing more than the slope of the secant line through the points (t0, s(t0)) and (t1, s(t1)). For t1 ̸= t0,

Average velocity = slope of secant line = !s !t

= s(t1) − s(t0) t1 − t0

By interpreting average velocity as a slope, we can visualize what happens as the time interval gets smaller. Figure 4 shows the graph of position for the falling stone of Example 1, where s(t) = 4.9t2. As the time interval shrinks, the secant lines get closer to—and seem to rotate into—the tangent line at t = 0.8.

58 C H A P T E R 2 LIMITS

Secant line

t (time)

s(t0)

s(t1)

!s = s(t1) − s(t0)

!t = t1 − t0

t1t0

s (position)

s(t0)

s(t1)

FIGURE 3 The average velocity over [t0, t1] is equal to the slope of the secant line.

Time Average interval velocity

[0.8, 0.805] 7.8645 [0.8, 0.8001] 7.8405 [0.8, 0.80005] 7.8402

0.800050.8 0.8001

0.805

Tangent line at t = 0.8

Slopes of secants 7.8645

7.8405

7.8402 7.84 Slope of tangent

t (s)

s (m)

FIGURE 4 The secant lines “rotate into” the tangent line as the time interval shrinks. Note: The graph is not drawn to scale.

And since the secant lines approach the tangent line, the slopes of the secant lines get closer and closer to the slope of the tangent line. In other words, the statement

As the time interval shrinks to zero, the average velocity approaches the instantaneous velocity.

has the graphical interpretation

As the time interval shrinks to zero, the slope of the secant line approaches the slope of the tangent line.

We conclude the following:

Instantaneous velocity is equal to the slope of the tangent line to the graph of position as a function of time.

This conclusion and its generalization to other rates of change are of fundamental importance in differential calculus.

EXAMPLE 2 Describe the motion and velocities of a shuttle train that runs on a straight track at the airport, ferrying passengers from one terminal to another according to the graph given in Figure 5.

800

0 4 95

Position x (meters)

Time t (minutes)FIGURE 5 The position of a shuttle train traveling along a straight track.

Solution At time t = 0, the train is at position x = 0 on the track at the first terminal. It gradually begins moving, speeding up until around 2 min. It then begins slowing down, and reaches its destination 800 m down the track at t = 4 min. At this point, it stops. Notice that for t in the interval (0, 4), the slopes of the tangent lines, which represent

S E C T I O N 2.1 Limits, Rates of Change, and Tangent Lines 59

the velocity, are positive. So the instantaneous velocities are positive. The slopes of the tangent lines begin very small, increase, and then decrease to 0. In the interval [4, 5], the graph has a horizontal tangent line with slope 0. In this interval, the train is not moving, having instantaneous velocity 0. The shuttle train then moves in the reverse direction. The slopes of the tangent lines in the interval (5, 9) are all negative. The speed increases and then decreases in this interval but the velocities are all negative. The train has returned to the original terminal at x = 0 on the track at time t = 9 min.

Other Rates of Change Velocity is only one of many examples of a rate of change. Our reasoning applies to any quantity y that depends on a variable x—say, y = f (x). For any interval [x0, x1], we set

!f = f (x1) − f (x0), !x = x1 − x0 For x1 ̸= x0, the average rate of change of y with respect to x over [x0, x1] is the ratioSometimes, we write !y and !y/!x

instead of !f and !f /!x.

Average rate of change = !f !x

= f (x1) − f (x0) x1 − x0︸ ︷︷ ︸

Slope of secant line

The instantaneous rate of change at x = x0 is the limit of the average rates of change.The word “instantaneous” is often dropped. When we use the term “rate of change,” it is understood that the instantaneous rate is the intended not average rate.

We estimate it by computing the average rate over smaller and smaller intervals. In Example 1 above, we considered only right-hand intervals [x0, x1]. In the next

example, we compute the average rate of change for intervals lying to both the left and the right of x0.

EXAMPLE 3 Bacteria in a Petri Dish Suppose the formula B = 5 √

T yields a good approximation to the amount of bacteria, as measured by its weight, that grows in a particular petri dish (in milligrams) over one day as a function of the temperature T (in degrees centigrade) for temperatures in the range 0 ≤ T ≤ 25. Estimate the instantaneous rate of change of B with respect to T when T = 10◦C. What are the units of this rate? Solution To estimate the instantaneous rate of change at T = 10, we compute the average rate for several intervals lying to the left and right of T = 10. For example, the average rate of change over [9.5, 10] is

B(10) − B(9.5) 10 − 9.5 =

5 √

10 − 5 √

9.5 0.5

≈ 0.8007

Tables 2 and 3 suggest that the instantaneous rate is approximately 0.790. This is the rate of increase in the weight of the bacteria with respect to temperature, so it has units of mg/◦C, or milligrams per degree centigrade. The secant lines corresponding to the values in the tables are shown in Figures 6 and 7.

TABLE 2 Left-Hand Intervals

Temperature Average rate interval of change

[9.5, 10] 0.8007 [9.8, 10] 0.7946 [9.9, 10] 0.7926 [9.99, 10] 0.7908

TABLE 3 Right-Hand Intervals

Temperature Average rate interval of change

[10, 10.5] 0.7809 [10, 10.2] 0.7867 [10, 10.1] 0.7886 [10, 10.01] 0.7904

TB = 5

109.5

Tangent line

0.8007 0.7946 0.7926

T (K)

B (mg)

FIGURE 6 Secant lines for intervals lying to the left of T = 10.

TB = 5

10 10.5 T (K)

Tangent line 0.78670.7809

0.7866 B (mg)

FIGURE 7 Secant lines for intervals lying to the right of T = 10.

60 C H A P T E R 2 LIMITS

To conclude this section, we recall an important point discussed in Section 1.2: For any linear function f (x) = mx + b, the average rate of change over every interval is equal to the slope m (Figure 8). We verify as follows:

!f

!x = f (x1) − f (x0)

x1 − x0 = (mx1 + b) − (mx0 + b)

x1 − x0 = m(x1 − x0)

x1 − x0 = m

The instantaneous rate of change at x = x0, which is the limit of these average rates, is

!x

!f

!x

!f

x

y

FIGURE 8 For a linear function f (x) = mx + b, the ratio !f/!x is equal to the slope m for every interval.

also equal to m. This makes sense graphically because all secant lines and all tangent lines to the graph of f coincide with the graph itself.

2.1 SUMMARY

• The limit as x approaches a of a function f is a number L that f (x) approaches, if such an L exists.

• The average rate of change of y = f (x) over an interval [x0, x1]:

Average rate of change = !f !x

= f (x1) − f (x0) x1 − x0

(x1 ̸= x0)

• The instantaneous rate of change is the limit of the average rates of change as the time interval shrinks.

• Graphical interpretation:

– Average rate of change is the slope of the secant line through the points (x0, f (x0)) and (x1, f (x1)) on the graph of f.

– Instantaneous rate of change is the slope of the tangent line at x0.

• To estimate the instantaneous rate of change at x = x0, compute the average rate of change over several intervals [x0, x1] (or [x1, x0]) for x1 close to x0.

• The velocity of an object in linear motion is the rate of change of position s(t). • Linear function f (x) = mx + b: The average rate of change over every interval and the

instantaneous rate of change at every point are equal to the slope m.

2.1 EXERCISES

Preliminary Questions 1. Average velocity is equal to the slope of a secant line through two

points on a graph. Which graph?

2. Can instantaneous velocity be defined as a ratio? If not, how is instantaneous velocity computed?

3. What is the graphical interpretation of instantaneous velocity at a specific time t = t0?

4. Find a graphical interpretation of the following statement: The av- erage rate of change approaches the instantaneous rate of change as the interval [x0, x1] shrinks to x0. 5. The rate of change of atmospheric temperature with respect to al-

titude is equal to the slope of the tangent line to a graph. Which graph? What are possible units for this rate?

Exercises 1. A ball dropped from a state of rest at time t = 0 travels a distance

s(t) = 4.9t2 m in t seconds. (a) How far does the ball travel during the time interval [2, 2.5]? (b) Compute the average velocity over [2, 2.5]. (c) Compute the average velocity for the time intervals in the table and estimate the ball’s instantaneous velocity at t = 2.

Interval [2, 2.01] [2, 2.005] [2, 2.001] [2, 2.00001] Average velocity

2. A wrench dropped from a state of rest at time t = 0 travels a dis- tance s(t) = 4.9t2 m in t seconds. Estimate the instantaneous velocity at t = 3.

3. Let B = 5 √

T as in Example 3. Estimate the instantaneous rate of change of B with respect to T when T = 20◦C.

4. Compute !y/!x for the interval [2, 5], where y = 4x − 9. What is the instantaneous rate of change of y with respect to x at x = 2? In Exercises 5–6, a stone is tossed vertically into the air from ground level with an initial velocity of 15 m/s. Its height at time t is h(t) = 15t − 4.9t2 m.

S E C T I O N 2.1 Limits, Rates of Change, and Tangent Lines 61

5. Compute the stone’s average velocity over the time interval [0.5, 2.5] and indicate the corresponding secant line on a sketch of the graph of h.

6. Compute the stone’s average velocity over the time intervals [1, 1.01], [1, 1.001], [1, 1.0001] and [0.99, 1], [0.999, 1], [0.9999, 1], and then estimate the instantaneous velocity at t = 1.

7. With an initial deposit of $100, and an interest rate of 8%, the balance in a bank account after t years is f (t) = 100(1.08)t dollars. (a) What are the units of the rate of change of f (t)? (b) Find the average rate of change over [0, 0.5] and [0, 1]. (c) Estimate the instantaneous rate of change at t = 0.5 by computing the average rate of change over intervals to the left and right of t = 0.5.

8. The position of a particle at time t is s(t) = t3 + t . Compute the average velocity over the time interval [1, 4] and estimate the instanta- neous velocity at t = 1.

9. Figure 9 shows the estimated percentage P of the Chilean population that uses the Internet, based on data from the United Nations Statistics Division. (a) Estimate the rate of change of P at t = 2010. (b) Does the rate of change increase or decrease as t increases? Explain graphically. (c) Let R be the average rate of change over [2008, 2012]. Com- pute R. (d) Is the rate of change at t = 2012 greater than or less than the aver- age rate R? What about the rate at t = 2008? Explain graphically.

2008 2009 2010 2011 2012

35

50

65

P (percent of Chilean population that used Internet)

t (years)

FIGURE 9

10. The atmospheric temperature T (in degrees Celsius) at alti- tude h meters above a certain point on Earth can be approximated by T = 15 − 0.0065h for h ≤ 12,000 m. What are the average and in- stantaneous rates of change of T with respect to h? Why are they the same? Sketch the graph of T for h ≤ 12,000. In Exercises 11–18, estimate the instantaneous rate of change at the point indicated.

11. P(x) = 3x2 − 5; x = 2 12. f (t) = 12t − 7; t = −4

13. y(x) = 1 x + 2 ; x = 2 14. y(t) =

√ 3t + 1; t = 1

15. f (x) = ex ; x = 0 16. f (x) = ex ; x = e

17. f (x) = ln x; x = 3 18. f (x) = tan−1 x; x = π 4

19. The height (in centimeters) at time t (in seconds) of a small mass oscillating at the end of a spring is h(t) = 8 cos(12π t). (a) Calculate the mass’s average velocity over the time intervals [0, 0.1] and [3, 3.5]. (b) Estimate its instantaneous velocity at t = 3.

20. Assume that the period T (in seconds) of a pendulum (the time required for a complete back-and-forth cycle) is T = 32

√ L, where

L is the pendulum’s length (in meters). (a) What are the units for the rate of change of T with respect to L? Explain what this rate measures. (b) Which quantities are represented by the slopes of lines A and B in Figure 10? (c) Estimate the instantaneous rate of change of T with respect to L when L = 3 m.

Period (s)

Length (m) 1 3

AB

2

FIGURE 10 The period T is the time required for a pendulum to swing back and forth.

21. The number P(t) of E. coli cells at time t (hours) in a petri dish is plotted in Figure 11. (a) Calculate the average rate of change of P(t) over the time interval [1, 3] and draw the corresponding secant line. (b) Estimate the slope m of the line in Figure 11. What does m repre- sent?

t (h) 321

10,000

8000

6000

4000

2000 1000

P (t)

FIGURE 11 Number of E. coli cells at time t .

22. The graphs in Figure 12 represent the positions of moving particles as functions of time. (a) Do the instantaneous velocities at times t1, t2, t3 in (A) form an increasing or a decreasing sequence? (b) Is the particle speeding up or slowing down in (A)? (c) Is the particle speeding up or slowing down in (B)?

Distance Distance

t2 t3t1

(A)

Time Time

(B)

FIGURE 12

62 C H A P T E R 2 LIMITS

23. An advertising campaign boosted sales of Crunchy Crust frozen pizza to a peak level of S0 dollars per month. A marketing study showed that after t months, monthly sales declined to

S(t) = S0g(t), where g(t) = 1√

1 + t Do sales decline more slowly or more rapidly as time increases? An- swer by referring to a sketch of the graph of g together with several tangent lines.

24. The fraction of a city’s population infected by a flu virus is plotted as a function of time (in weeks) in Figure 13. (a) Which quantities are represented by the slopes of lines A and B? Estimate these slopes. (b) Is the flu spreading more rapidly at t = 1, 2, or 3? (c) Is the flu spreading more rapidly at t = 4, 5, or 6?

Fraction infected

Weeks

B

A

1 2 3 4 5 6

0.3

0.2

0.1

FIGURE 13

25. The graphs in Figure 14 represent the positions s of moving par- ticles as functions of time t . Match each graph with a description: (a) Speeding up (b) Speeding up and then slowing down (c) Slowing down (d) Slowing down and then speeding up

(B)(A)

(D)(C)

t

s

t

s

t

s

t

s

FIGURE 14

26. An epidemiologist finds that the percentage N(t) of susceptible children who are sick on day t during the first 3 weeks of a measles outbreak is given, to a reasonable approximation, by the formula (Fig- ure 15)

N(t) = 100t 2

t3 + 5t2 − 100t + 380

Percent infected

Time (days) 2 6 10 14 184 8 12 16 20

20

15

10

5

FIGURE 15 Graph of N .

(a) Draw the secant line whose slope is the average rate of change in infected children over the intervals [4, 6] and [12, 14]. Then compute these average rates (in units of percent per day). (b) Is the rate of decline greater at t = 8 or t = 16? (c) Estimate the rate of change of N(t) on day 12.

27. The fungus Fusarium exosporium infects a field of flax plants through the roots and causes the plants to wilt. Eventually, the entire field is infected. The percentage f (t) of infected plants as a function of time t (in days) since planting is shown in Figure 16. (a) What are the units of the rate of change of f (t) with respect to t? What does this rate measure? (b) Use the graph to rank (from smallest to largest) the average infec- tion rates over the intervals [0, 12], [20, 32], and [40, 52]. (c) Use the following table to compute the average rates of infection over the intervals [30, 40], [40, 50], [30, 50]:

Days 0 10 20 30 40 50 60 Percent infected 0 18 56 82 91 96 98

(d) Draw the tangent line at t = 40 and estimate its slope.

Percent infected

Days after planting 10 20 30 40 50 60

100

80

60

40

20

FIGURE 16

28. Let v = 5 √

T as in Example 3. Is the rate of change of v with respect to T greater at low temperatures or high temperatures? Explain in terms of the graph.

29. If an object in linear motion (but with changing veloc- ity) covers !s meters in !t seconds, then its average velocity is v0 = !s/!t m/s. Show that it would cover the same distance if it trav- eled at constant velocity v0 over the same time interval. This justifies our calling !s/!t the average velocity.

30. Sketch the graph of f (x) = x(1 − x) over [0, 1]. Refer to the graph and, without making any computations, find: (a) the average rate of change over [0, 1]. (b) the (instantaneous) rate of change at x = 12 . (c) the values of x at which the rate of change is positive.

S E C T I O N 2.2 Limits: A Numerical and Graphical Approach 63

31. Which graph in Figure 17 has the following property: For all x, the average rate of change over [0, x] is greater than the instan- taneous rate of change at x. Explain.

(B)

x

y

(A)

x

y

FIGURE 17

Further Insights and Challenges 32. The height of a projectile fired in the air vertically with initial ve- locity 25 m/s is

h(t) = 25t − 4.9t2 m

(a) Compute h(1). Show that h(t) − h(1) can be factored with (t − 1) as a factor. (b) Using part (a), show that the average velocity over the interval [1, t] is 20.1 − 4.9t . (c) Use this formula to find the average velocity over several intervals [1, t] with t close to 1. Then estimate the instantaneous velocity at time t = 1. 33. Let Q(t) = t2. As in the previous exercise, find a formula for the average rate of change of Q over the interval [1, t] and use it to estimate the instantaneous rate of change at t = 1. Repeat for the interval [2, t] and estimate the rate of change at t = 2. 34. Show that the average rate of change of f (x) = x3 over [1, x] is equal to

x2 + x + 1 Use this to estimate the instantaneous rate of change of f (x) at x = 1.

35. Find a formula for the average rate of change of f (x) = x3 over [2, x] and use it to estimate the instantaneous rate of change at x = 2.

36. Let T = 32 √

L as in Exercise 20. The numbers in the sec- ond column of Table 4 are increasing, and those in the last column are decreasing. Explain why in terms of the graph of T as a function of L. Also, explain graphically why the instantaneous rate of change at L = 3 lies between 0.43299 and 0.43303.

TABLE 4 Average Rates of Change of T with Respect to L

Average rate Average rate Interval of change Interval of change

[3, 3.2] 0.42603 [2.8, 3] 0.44048 [3, 3.1] 0.42946 [2.9, 3] 0.43668 [3, 3.001] 0.43298 [2.999, 3] 0.43305 [3, 3.0005] 0.43299 [2.9995, 3] 0.43303

2.2 Limits: A Numerical and Graphical Approach The goal in this section is to define limits and study them using numerical and graphical techniques. We begin with the following question: How do the values of a function f behave when x approaches a number c, whether or not f (c) is defined?

To explore this question, we’ll experiment with the function

f (x) = sin x x

(x in radians)

Notice that f (0) is not defined. In fact, when we set x = 0 inThe undefined expression 0/0 is referred to as an “indeterminate form.”

f (x) = sin x x

we obtain the undefined expression 0/0 because sin 0 = 0. Nevertheless, we can compute f (x) for values of x close to 0. When we do this, a clear trend emerges.

To describe the trend, we use the phrase “x approaches 0” or “x tends to 0” to indicate that x takes on values (both positive and negative) that get closer and closer to 0. The notation for this is x → 0, and more specifically we write

• x → 0+ if x approaches 0 from the right (on the number line). • x → 0− if x approaches 0 from the left (on the number line).

Now consider the values listed in Table 1. The table gives the unmistakable impression that f (x) gets closer and closer to 1 as x → 0+ and as x → 0−.

This conclusion is supported by the graph of f in Figure 1. The point (0, 1) is missing from the graph because f (x) is not defined at x = 0, but the graph approaches

64 C H A P T E R 2 LIMITS

TABLE 1

x sin x

x x

sin x x

1 0.841470985 −1 0.841470985 0.5 0.958851077 −0.5 0.958851077 0.1 0.998334166 −0.1 0.998334166 0.05 0.999583385 −0.05 0.999583385 0.01 0.999983333 −0.01 0.999983333 0.005 0.999995833 −0.005 0.999995833 0.001 0.999999833 −0.001 0.999999833

x → 0+ f (x) → 1 x → 0− f (x) → 1

2π 3π

1

−π−2π−3π π x

y

FIGURE 1 Graph of f (x) = sin x x

.

this missing point as x approaches 0 from the left and right. We say that the limit of f (x) as x → 0 is equal to 1, and we write

lim x→0

f (x) = 1

We also say that f (x) approaches or converges to 1 as x → 0.

CONCEPTUAL INSIGHT The numerical and graphical evidence may convince us that

f (x) = sin x x

converges to 1 as x → 0, but since f (0) yields the undefined expres- sion 0/0, could we not arrive at this conclusion more simply by saying that 0/0 is equal to 1? The answer is no. Algebra does not allow us to divide by 0 under any circumstances, and it is not correct to say that 0/0 equals 1 or any other number.

What we have learned, however, is that a function f may approach a limit as x → c even if the formula for f (c) produces the undefined expression 0/0. The limit of f (x) = sin x

x turns out to be 1. We will encounter other examples where the formula

for f (c) produces 0/0 but the limit is a number other than 1 (or the limit does not exist).

The fact that sin x x

approaches 1 for x approaching 0 can be interpreted to mean that the two functions y = sin x and y = x behave similarly as x approaches 0. Their graphs in Figure 2 make it clear this is the case. The closer we get to x = 0, the closer the two graphs become.

y

x

y = sin x

y = x

FIGURE 2 Definition of a Limit To define limits, let us recall that the distance between two numbers a and b is the absolute value |a − b|. So we can express the idea that f (x) is close to L by saying that |f (x) − L| is small.

DEFINITION Limit Assume thatf (x) is defined for allx in an open interval containing c, but not necessarily at c itself. We say that

the limit of f (x) as x approaches c is equal to the number L

if |f (x) − L| can be made arbitrarily small by taking x sufficiently close (but not equal) to c. In this case, we write

lim x→cf (x) = L

We also say thatf (x)approaches or converges toL asx → c [and we writef (x) → L].

In other words, as x approaches c, f (x) approaches L. See Figure 3 for the graphical interpretation. If the values of f (x) do not converge to any number L as x → c, we say

The concept of a limit was not fully clarified until the nineteenth century. The French mathematician Augustin-Louis Cauchy (1789–1857, pronounced Koh-shee) gave the following verbal definition: “When the values successively attributed to the same variable approach a fixed value indefinitely, in such a way as to end up differing from it by as little as one could wish, this last value is called the limit of all the others. So, for example, an irrational number is the limit of the various fractions which provide values that approximate it more and more closely.” (Translated by J. Grabiner)

that lim x→c f (x) does not exist. It is important to note that the value f (c) itself, which may

or may not be defined, plays no role in the limit. All that matters are the values of f (x) for x close to c. Furthermore, if f (x) approaches a limit as x → c, then the limiting value L is unique.

S E C T I O N 2.2 Limits: A Numerical and Graphical Approach 65

EXAMPLE 1 Use the definition above to verify the following limits:

| f (x) − L |

c x

f (x)

y = f (x)

L

x

y

FIGURE 3 As x → c, f (x) → L. A similar picture holds for x < c as x → c.

(a) lim x→7

5 = 5 (b) lim x→4

(3x + 1) = 13

Solution

(a) Let f (x) = 5. To show that lim x→7

f (x) = 5, we must show that |f (x) − 5| becomes arbitrarily small when x is sufficiently close (but not equal) to 7. But observe that |f (x) − 5| = |5 − 5| = 0 for all x, so what we are required to show is automatic (and it is not necessary to take x close to 7).

(b) Let f (x) = 3x + 1. To show that lim x→4

(3x + 1) = 13, we must show that |f (x) − 13| becomes arbitrarily small when x is sufficiently close (but not equal) to 4. We have

|f (x) − 13| = |(3x + 1) − 13| = |3x − 12| = 3|x − 4|

Because |f (x) − 13| is a multiple of |x − 4|, we can make |f (x) − 13| arbitrarily small by taking x sufficiently close to 4.

Reasoning as in Example 1 but with arbitrary constants, we obtain the following simple but important results:

THEOREM 1 For any constants k and c, (a) lim x→c k = k, (b) limx→c x = c.

To deal with more complicated limits and, especially, to provide mathematicallyHere is one version of the rigorous definition of a limit: lim

x→c f (x) = L if, for

every number n, we can find a value of m so that for any x such that 0 < |x − c| < 10−m, then |f (x) − L| < 10−n.

rigorous proofs, a more precise version of the above limit definition is needed. This more precise version is discussed in Section 2.9, where inequalities are used to pin down the exact meaning of the phrases “arbitrarily small” and “sufficiently close.”

Graphical and Numerical Investigation Our goal in the rest of this section is to develop a better intuitive understanding of limits by investigating them graphically and numerically.

Graphical Investigation Use a graphing utility to produce a graph of f. The graph should give a visual impression of whether or not a limit exists. It can often be used to estimate the value of the limit.

Numerical Investigation We writex → c− to indicate thatx approaches c through values less than c (i.e., from the left), and we write x → c+ to indicate that x approaches c through values greater than c (i.e., from the right). To investigate lim

x→c f (x),

(i) Make a table of values of f (x) for x close to but less than c—that is, as x → c−. (ii) Make a second table of values of f (x) for x close to but greater than c—that is, as x → c+. (iii) If both tables indicate convergence to the same number L, we take L to be an estimate for the limit.

The tables should contain enough values to reveal a clear trend of convergence to a value L.

Keep in mind that graphical and numerical investigations provide evidence for a limit, but they do not prove that the limit exists or has a given value. This is done using the Limit Laws established in the following sections.

If f (x) approaches a limit, the successive values of f (x) will generally agree to more and more decimal places as x is taken closer to c. If no pattern emerges, then the limit may not exist.

66 C H A P T E R 2 LIMITS

EXAMPLE 2 Investigate lim x→9

x − 9√ x − 3 graphically and numerically.

Solution The function f (x) = x − 9√ x − 3 is undefined at x = 9 because the formula for

f (9) leads to the undefined expression 0/0. Therefore, the graph in Figure 4 has a gap at x = 9. However, the graph suggests that f (x) approaches 6 as x → 9.

For numerical evidence, we consider a table of values of f (x) for x approaching 9 from both the left and the right. Table 2 supports our impression that

lim x→9

x − 9√ x − 3 = 6

We will see shortly that, in fact, we can write lim x→9

x − 9√ x − 3 = limx→9

(√ x − 3

) (√ x + 3

) √

x − 3 = lim x→9

( √

x + 3) = 6. Thus, the algebraic solution will confirm our numerical and geometric conclusion.

3 6 9 12

3

6

9

x

y

FIGURE 4 Graph of f (x) = x − 9√ x − 3 .

TABLE 2

x → 9− x − 9√ x − 3 x → 9

+ x − 9√ x − 3

8.9 5.98329 9.1 6.01662 8.99 5.99833 9.01 6.001666 8.999 5.99983 9.001 6.000167 8.9999 5.9999833 9.0001 6.0000167

EXAMPLE 3 Limit Equals Value of the Function Investigate lim x→4

x2.

Solution Figure 5 and Table 3 both suggest that lim x→4

x2 = 16. But f (x) = x2 is defined at x = 4 and f (4) = 16, so in this case, the limit is equal to the function value. This pleasant conclusion is valid whenever f is a continuous function, a concept treated in Section 2.4.

16

2 4 6 x

y

FIGURE 5 Graph of f (x) = x2. The limit is equal to the value of the function f (4) = 16.

TABLE 3

x → 4− x2 x → 4+ x2

3.9 15.21 4.1 16.81 3.99 15.9201 4.01 16.0801 3.999 15.992001 4.001 16.008001 3.9999 15.99920001 4.0001 16.00080001

EXAMPLE 4 Defining Property of e Verify numerically that lim h→0

eh − 1 h

= 1.

Solution The function f (h) = (eh − 1)/h is undefined at h = 0, but both Figure 6 and Table 4 suggest that lim

h→0 (eh − 1)/h = 1. We will ultimately see that the fact this limit is

equal to 1 reflects the fact that the tangent line to the graph of y = ex at x = 0 has slope 1, as in Figure 3 from Section 1.6.

S E C T I O N 2.2 Limits: A Numerical and Graphical Approach 67

2

y

1−1 2−2 h

1

y = h

eh − 1

FIGURE 6

TABLE 4

h → 0− e h − 1 h

h → 0+ e h − 1 h

−0.02 0.990 0.02 1.0101 −0.005 0.99750 0.005 1.00250 −0.001 0.999500 0.001 1.000500 −0.0001 0.99995000 0.0001 1.00005000

EXAMPLE 5 A Limit That Does Not Exist Investigate lim x→0

sin π

x graphically and nu-CAUTION Numerical investigations are often

suggestive, but may be misleading in some cases. If, in Example 5, we had chosen to

evaluate f (x) = sin π x

at the values

x = 0.1, 0.01, 0.001, . . . , we might have concluded incorrectly that f (x) approaches the limit 0 as x → 0. The problem is that f (10−n) = sin(10nπ) = 0 for every whole number n, but f (x) itself does not approach any limit.

merically.

Solution The function f (x) = sin πx is not defined at x = 0, but Figure 7 suggests that it oscillates between +1 and −1 infinitely often as x → 0. It appears, therefore, that lim x→0

sin πx does not exist. This impression is confirmed by Table 5, which shows that the

values of f (x) bounce around and do not tend toward any limit L as x → 0.

−2

y

x −1

1 21 2

1 3

1 4

1 3

1 2

−

−

FIGURE 7 Graph of f (x) = sin πx .

TABLE 5 The Function f (x) = sin πx Does Not Approach a Limit as x → 0

x → 0− sin π x

x → 0+ sin π x

−0.1 0 0.1 0 −0.03 0.866 0.03 −0.866 −0.007 −0.434 0.007 0.434 −0.0009 0.342 0.0009 −0.342 −0.00065 −0.935 0.00065 0.935

One-Sided Limits The limits discussed so far are two-sided. To show that lim

x→c f (x) = L, it is necessary to check that f (x) converges to L as x approaches c through values both larger and smaller than c. In some instances, f (x) may approach L from one side of c without necessarily approaching it from the other side, or f (x) may be defined on only one side of c. For this reason, we define the one-sided limits

lim x→c−

f (x) (left-hand limit), lim x→c+

f (x) (right-hand limit)

The limit itself exists if both one-sided limits exist and are equal.

EXAMPLE 6 Left- and Right-Hand Limits Not Equal Investigate the one-sided limits of f (x) = x|x| as x → 0. Does limx→0 f (x) exist?1

−1 −3 −2 −1 321

x

y

FIGURE 8 Graph of f (x) = x|x| .

Solution Figure 8 shows what is going on. For x < 0,

f (x) = x|x| = x

−x = −1

Therefore, the left-hand limit is lim x→0−

f (x) = −1. But for x > 0,

f (x) = x|x| = x

x = 1

Therefore, lim x→0+

f (x) = 1. These one-sided limits are not equal, so lim x→0

f (x) does not exist.

68 C H A P T E R 2 LIMITS

EXAMPLE 7 The function f in Figure 9 is not defined at c = 0, 2, 4. Investigate the one- and two-sided limits at these points.3

1

−1

−1

4321 x

y

2

FIGURE 9

Solution

• c = 0: The left-hand limit lim x→0−

f (x) does not seem to exist because f (x) appears

to oscillate infinitely often to the left of x = 0. On the other hand, lim x→0+

f (x) = 2. • c = 2: The one-sided limits exist but are not equal:

lim x→2−

f (x) = 3 and lim x→2+

f (x) = 1

Therefore, lim x→2

f (x) does not exist. • c = 4: The one-sided limits exist and both have the value 2. Therefore, the two-sided

limit exists and lim x→4

f (x) = 2.

Infinite Limits For some functions, f (x) tends to ∞ or −∞ as x approaches a value c. If so, lim

x→c f (x)2

4 x − 2 1

(A)

f (x) =

x2 1f (x) =

(B)

2 4 6

−2

−2

−4

x

y

2

2 4

−2

−2−4

−4

x

y

Asymptote x = 2

Asymptote x = 0

(C)

f (x) = ln x

x 21 3 4

y

4

2

−4

−2

Asymptote x = 0

FIGURE 10

does not exist, but we say that f (x) has an infinite limit. More precisely, we write

• lim x→c f (x) = ∞ if f (x) increases without bound as x → c.

• lim x→c f (x) = −∞ if f (x) decreases without bound as x → c.

Here, “decrease without bound” means that f (x) becomes negative and |f (x)| → ∞. One-sided infinite limits are defined similarly. When using this notation, keep in mind that ∞ and −∞ are not numbers.

When f (x) approaches ∞ or −∞ as x approaches c from one or both sides, the line x = c is called a vertical asymptote. In Figure 10, the line x = 2 is a vertical asymptote in (A), and x = 0 is a vertical asymptote in both (B) and (C).

In the next example, the notation x → c± is used to indicate that the left- and right- hand limits are to be considered separately.

EXAMPLE 8 Investigate the one-sided limits graphically:

(a) lim x→2±

1 x − 2 (b) limx→0±

1 x2

(c) lim x→0+

ln x

Solution

(a) Figure 10(A) suggests that

lim x→2−

1 x − 2 = −∞, limx→2+

1 x − 2 = ∞

The vertical line x = 2 is a vertical asymptote. Why are the one-sided limits different? Because f (x) = 1

x − 2 is negative for x < 2 (so the limit from the left is −∞) and f (x) is positive for x > 2 (so the limit from the right is ∞).

(b) Figure 10(B) suggests that lim x→0

1 x2

= ∞. Indeed, f (x) = 1 x2

is positive for all x ̸= 0 and becomes arbitrarily large as x → 0 from either side. The line x = 0 is a vertical asymptote.

(c) Figure 10(C) suggests that lim x→0+

ln x = −∞ because f (x) = ln x is negative for 0 < x < 1 and tends to −∞ as x → 0+. The line x = 0 is a vertical asymptote.

S E C T I O N 2.2 Limits: A Numerical and Graphical Approach 69

CONCEPTUAL INSIGHT You should not think of an infinite limit as a true limit. The notation lim

x→c f (x) = ∞ is merely a shorthand way of saying that f (x) increases beyond all bounds as x approaches c. The limit itself does not exist. We must be careful when using this notation because ∞ and −∞ are not numbers, and contradictions can arise if we try to manipulate them as numbers. For example, if ∞ were a number, it would be larger than any finite number, and presumably, ∞ + 1 = ∞. But then

∞ + 1 = ∞ (∞ + 1) − ∞ = ∞ − ∞

1 = 0 (contradiction!)

To avoid errors, keep in mind the ∞ is not a number but rather a convenient shorthand notation.

2.2 SUMMARY

• By definition, lim x→c f (x) = L if |f (x) − L| can be made arbitrarily small by taking x

sufficiently close (but not equal) to c. We say that

– The limit of f (x) as x approaches c is L, or – f (x) approaches (or converges) to L as x approaches c.

• If f (x) approaches a limit as x → c, then the value of the limit L is unique. • If f (x) does not approach a limit as x → c, we say that lim

x→c f (x) does not exist. • The limit may exist even if f (c) is not defined. • One-sided limits:

– lim x→c−

f (x) = L if f (x) converges to L as x approaches c through values less than c.

– lim x→c+

f (x) = L if f (x) converges to L as x approaches c through values greater than c.

• The limit exists if and only if both one-sided limits exist and are equal. • Infinite limits: lim

x→c f (x) = ∞ if f (x) increases beyond bound as x approaches c, and lim x→c f (x) = −∞ if f (x) becomes arbitrarily large (in absolute value) but negative as x approaches c.

• In the case of a one- or two-sided infinite limit, the vertical line x = c is called a vertical asymptote.

2.2 EXERCISES

Preliminary Questions 1. What is the limit of f (x) = 1 as x → π? 2. What is the limit of g(t) = t as t → π? 3. Is lim

x→10 20 equal to 10 or 20?

4. Can f (x) approach a limit as x → c if f (c) is undefined? If so, give an example.

5. What does the following table suggest about lim x→1−

f (x) and

lim x→1+

f (x)?

x 0.9 0.99 0.999 1.1 1.01 1.001

f (x) 7 25 4317 3.0126 3.0047 3.00011

6. Can you tell whether lim x→5

f (x) exists from a plot of f for x > 5?

Explain.

7. If you know in advance that lim x→5

f (x) exists, can you determine its

value from a plot of f for all x > 5?

70 C H A P T E R 2 LIMITS

Exercises In Exercises 1–4, fill in the tables and guess the value of the limit.

1. lim x→1

f (x), where f (x) = x 3 − 1

x2 − 1 .

x f (x) x f (x)

1.002 0.998

1.001 0.999

1.0005 0.9995

1.00001 0.99999

2. lim t→0

h(t), where h(t) = cos t − 1 t2

. Note that h is even; that is,

h(t) = h(−t).

t ±0.002 ±0.0001 ±0.00005 ±0.00001 h(t)

3. lim y→2

f (y), where f (y) = y 2 − y − 2

y2 + y − 6 .

y f (y) y f (y)

2.002 1.998

2.001 1.999

2.0001 1.9999

4. lim x→0+

f (x), where f (x) = x ln x.

x 1 0.5 0.1 0.05 0.01 0.005 0.001

f (x)

5. Determine lim x→0.5

f (x) for f as in Figure 11.

6. Determine lim x→0.5

g(x) for g as in Figure 12.

0.5

1.5

x

y

1 y = f (x)

FIGURE 11

0.5

1.5

x

y

1 y = g(x)

FIGURE 12

In Exercises 7–8, evaluate the limit.

7. lim x→21

x 8. lim x→4.2

√ 3

In Exercises 9–16, verify each limit using the limit definition. For ex- ample, in Exercise 9, show that |3x − 12| can be made as small as desired by taking x close to 4.

9. lim x→4

3x = 12 10. lim x→5

3 = 3

11. lim x→3

(5x + 2) = 17 12. lim x→2

(7x − 4) = 10

13. lim x→0

x2 = 0 14. lim x→0

(3x2 − 9) = −9

15. lim x→0

(4x2 + 2x + 5) = 5 16. lim x→0

(x3 + 12) = 12

In Exercises 17–38, estimate the limit numerically or state that the limit does not exist. If infinite, state whether the one-sided limits are ∞ or −∞.

17. lim x→1

√ x − 1

x − 1 18. limx→−4 2x2 − 32

x + 4

19. lim x→2

x2 + x − 6 x2 − x − 2 20. limx→3

x3 − 2x2 − 9 x2 − 2x − 3

21. lim x→0

sin 2x x

22. lim x→0

sin 5x x

23. lim θ→0

cos θ − 1 θ

24. lim x→0

sin x

x2

25. lim x→4

1

(x − 4)3 26. limx→1− 3 − x x − 1

27. lim x→−3

x + 3 x2 + x − 6 28. limx→−2−

x + 1 x + 2

29. lim x→3+

x − 4 x2 − 9 30. limh→0

3h − 1 h

31. lim h→0

sin h cos 1 h

32. lim h→0

cos 1 h

33. lim x→0

|x|x 34. lim x→1+

sec−1 x√ x − 1

35. lim t→e

t − e ln t − 1 36. limr→0(1 + r)

1/r

37. lim x→1−

tan−1 x cos−1 x

38. lim x→0

tan−1 x − x sin−1 x − x

39. The greatest integer function, also known as the floor func- tion, is defined by ⌊x⌋ = n, where n is the unique integer such that n ≤ x < n + 1. Sketch the graph of y = ⌊x⌋. Calculate for c an integer: (a) lim

x→c− ⌊x⌋ (b) lim

x→c+ ⌊x⌋ (c) lim

x→2.6 ⌊x⌋

40. Determine the one-sided limits at c = 1, 2, and 4 of the function g shown in Figure 13, and state whether the limit exists at these points.

1 2 3 4 5

1

2

3

x

y

FIGURE 13

In Exercises 41–48, determine the one-sided limits numerically or graphically. If infinite, state whether the one-sided limits are ∞ or −∞, and describe the corresponding vertical asymptote. In Exercise 48, f (x) = ⌊x⌋ is the greatest integer function defined in Exercise 39.

41. lim x→0±

sin x |x| 42. limx→0±

|x|1/x

43. lim x→0±

x − sin |x| x3

44. lim x→4±

x + 1 x − 4

S E C T I O N 2.2 Limits: A Numerical and Graphical Approach 71

45. lim x→−2±

4x2 + 7 x3 + 8 46. limx→−3±

x2

x2 − 9

47. lim x→1±

x5 + x − 2 x2 + x − 2 48. limx→2±

cos (π

2 (x − ⌊x⌋)

)

49. Determine the one-sided limits at c = 2 and c = 4 of the function f in Figure 14. What are the vertical asymptotes of f ?

50. Determine the infinite one- and two-sided limits in Figure 15.

−5 42

15

5

10

x

y

FIGURE 14

x

y

−1 3 5

FIGURE 15

In Exercises 51–54, sketch the graph of a function with the given limits.

51. lim x→1

f (x) = 2, lim x→3−

f (x) = 0, lim x→3+

f (x) = 4

52. lim x→1

f (x) = ∞, lim x→3−

f (x) = 0, lim x→3+

f (x) = −∞

53. lim x→2+

f (x) = f (2) = 3, lim x→2−

f (x) = −1, lim x→4

f (x) = 2 ̸= f (4)

54. lim x→1+

f (x) = ∞, lim x→1−

f (x) = 3, lim x→4

f (x) = −∞

55. Determine the one-sided limits of the function f in Figure 16, at the points c = 1, 3, 5, 6.

−1 −2 −3 −4

1 2 3 4 5

y

x 1 2 3 4 5 6 7 8

FIGURE 16 Graph of f .

56. Does either of the two oscillating functions in Figure 17 appear to approach a limit as x → 0?

(A) (B)

xx

y y

FIGURE 17

In Exercises 57–62, plot the function and use the graph to esti- mate the value of the limit.

57. lim θ→0

sin 5θ sin 2θ

58. lim x→0

12x − 1 4x − 1

59. lim x→0

2x − cos x x

60. lim θ→0

sin2 4θ cos θ − 1

61. lim θ→0

cos 7θ − cos 5θ θ2

62. lim θ→0

sin2 2θ − θ sin 4θ θ4

63. Let n be a positive integer. For which n are the two infinite one- sided limits lim

x→0± 1/xn equal?

64. Let L(n) = lim x→1

( n

1 − xn − 1

1 − x

) for n a positive integer. In-

vestigate L(n) numerically for several values of n, and then guess the value of L(n) in general.

65. In some cases, numerical investigations can be misleading. Plot f (x) = cos πx . (a) Does lim

x→0 f (x) exist?

(b) Show, by evaluating f (x) at x = ± 12 , ± 14 , ± 16 , . . . , that you might be able to trick your friends into believing that the limit exists and is equal to L = 1. (c) Which sequence of evaluations might trick them into believing that the limit is L = −1.

Further Insights and Challenges 66. Light waves of frequency λ passing through a slit of width a pro- duce a Fraunhoferdiffraction pattern of light and dark fringes (Figure 18). The intensity as a function of the angle θ is

I (θ) = Im (

sin(R sin θ) R sin θ

)2

where R = πa/λ and Im is a constant. Show that the intensity function is not defined at θ = 0. Then choose any two values for R and check numerically that I (θ) approaches Im as θ → 0.

67. Investigate lim θ→0

sin nθ θ

numerically for several positive integer

values of n. Then guess the value in general.

a

Intensity pattern

Viewing screen

Slit

Incident light waves

θ

FIGURE 18 Fraunhofer diffraction pattern.

72 C H A P T E R 2 LIMITS

68. Show numerically that lim x→0

bx − 1 x

for b = 3 and b = 5 appears to equal ln 3 and ln 5, respectively, where ln x is the natural logarithm. Then make a conjecture (guess) for the value in general and test your conjecture for two additional values of b.

69. Investigate lim x→1

xn − 1 xm − 1 for (m, n) equal to (2, 1), (1, 2), (2, 3),

and (3, 2). Then guess the value of the limit in general and check your guess for two additional pairs.

70. Find by numerical experimentation the positive integers k such that

lim x→0

sin(sin2 x) xk

exists.

71. Plot the graph of f (x) = 2 x − 8 x − 3 .

(a) Zoom in on the graph to estimate L = lim x→3

f (x).

(b) Explain why

f (2.99999) ≤ L ≤ f (3.00001)

Use this to determine L to three decimal places.

72. The function f (x) = 2 1/x − 2−1/x

21/x + 2−1/x is defined for x ̸= 0. (a) Investigate lim

x→0+ f (x) and lim

x→0− f (x) numerically.

(b) Plot the graph of f and describe its behavior near x = 0.

2.3 Basic Limit Laws In Section 2.2, we relied on graphical and numerical approaches to investigate limits and estimate their values. In the next four sections, we go beyond this intuitive approach and develop tools for computing limits in a precise way. The next theorem provides our first set of tools.

The proof of Theorem 1 is discussed in Section 2.9 and Appendix D. To illustrate the underlying idea, consider two numbers such as 2.99 and 5.001. Observe that 2.99 is close to 3 and 5.0001 is close to 5, so certainly the sum 2.99 + 5.0001 is close to 3 + 5 and the product 2.99 × 5.0001 is close to 3 × 5. In the same way, if f (x) approaches L and g(x) approaches M as x → c, then f (x) + g(x) approaches the sum L + M, and f (x)g(x) approaches the product LM. The other laws are similar.

THEOREM 1 Basic Limit Laws If lim x→c f (x) and limx→c g(x) exist, then

(i) Sum Law: lim x→c

( f (x) + g(x)

) exists and

lim x→c

( f (x) + g(x)

) = lim

x→c f (x) + limx→c g(x)

(ii) Constant Multiple Law: For any number k, lim x→c kf (x) exists and

lim x→c kf (x) = k limx→c f (x)

(iii) Product Law: lim x→c f (x)g(x) exists and

lim x→c f (x)g(x) =

( lim x→c f (x)

) ( lim x→c g(x)

)

(iv) Quotient Law: If lim x→c g(x) ̸= 0, then limx→c

f (x)

g(x) exists and

lim x→c

f (x)

g(x) =

lim x→c f (x)

lim x→c g(x)

(v) Powers and Roots: If n is a positive integer, then

lim x→c[f (x)]

n = (

lim x→c f (x)

)n , lim

x→c n √

f (x) = n √

lim x→c f (x)

In the second limit, assume that lim x→c f (x) ≥ 0 if n is even.

If p, q are integers with q ̸= 0, then lim x→c[f (x)]

p/q exists and

lim x→c[f (x)]

p/q = (

lim x→c f (x)

)p/q

Assume that lim x→c f (x) ≥ 0 if q is even, and that limx→c f (x) ̸= 0 if p/q < 0.

Before proceeding to the examples, we make some useful remarks.

• The Sum and Product Laws are valid for any number of functions. For example,

lim x→c

( f1(x) + f2(x) + f3(x)

) = lim

x→c f1(x) + limx→c f2(x) + limx→c f3(x)

S E C T I O N 2.3 Basic Limit Laws 73

• The Sum Law has a counterpart for differences:

lim x→c

( f (x) − g(x)

) = lim

x→c f (x) − limx→c g(x)

This follows from the Sum and Constant Multiple Laws (with k = −1):

lim x→c

( f (x) − g(x)

) = lim

x→c f (x) + limx→c ( − g(x)

) = lim

x→c f (x) − limx→c g(x)

• Recall two basic limits from Theorem 1 in Section 2.2:

lim x→c k = k, limx→c x = c

Applying Law (v) to f (x) = x, we obtain

lim x→cx

p/q = cp/q 1

for integers p, q, q ̸= 0. Assume that c ≥ 0 if q is even and that c ̸= 0 if p/q < 0.

EXAMPLE 1 Use the Basic Limit Laws to evaluate:

(a) lim x→2

x3 (b) lim x→2

(x3 + 5x + 7) (c) lim x→2

√ x3 + 5x + 7

Solution

(a) By Eq. (1), lim x→2

x3 = 23 = 8. (b)

lim x→2

(x3 + 5x + 7) = lim x→2

x3 + lim x→2

5x + lim x→2

7 (Sum Law)

= lim x→2

x3 + 5 lim x→2

x + lim x→2

7 (Constant Multiple Law)

= 8 + 5(2) + 7 = 25

(c) By Law (v) for roots and (b),

lim x→2

√ x3 + 5x + 7 =

√ lim x→2

(x3 + 5x + 7) = √

25 = 5

EXAMPLE 2 Evaluate (a) lim t→−1

t + 6 2t4

and (b) lim t→3

t−1/4(t + 5)1/3. You may have noticed that each of the limits in Examples 1 and 2 could have been evaluated by a simple substitution. For example, set t = −1 to evaluate

lim t→−1

t + 6 2t4

= −1 + 6 2(−1)4 =

5 2

Substitution is valid when the function is continuous, a concept we shall study in the next section.

Solution

(a) Use the Quotient, Sum, and Constant Multiple Laws:

lim t→−1

t + 6 2t4

= lim

t→−1 (t + 6)

lim t→−1

2t4 =

lim t→−1

t + lim t→−1

6

2 lim t→−1

t4 = −1 + 6

2(−1)4 = 5 2

(b) Use the Product, Powers, and Sum Laws:

lim t→3

t−1/4(t + 5)1/3 = (

lim t→3

t−1/4 ) (

lim t→3

3√ t + 5

) =

( 3−1/4

) ( 3 √

lim t→3

t + 5 )

= 3−1/4 3 √

3 + 5 = 3−1/4(2) = 2 31/4

The next example reminds us that the Basic Limit Laws apply only when the limits of both f (x) and g(x) exist.

74 C H A P T E R 2 LIMITS

EXAMPLE 3 Assumptions Matter Show that the Product Law cannot be applied to lim x→0

f (x)g(x) if f (x) = x and g(x) = x−1.

Solution For all x ̸= 0, we have f (x)g(x) = x · x−1 = 1, so the limit of the product exists:

lim x→0

f (x)g(x) = lim x→0

1 = 1

However, lim x→0

x−1 does not exist because g(x) = x−1 approaches ∞ as x → 0+ and it approaches −∞ as x → 0−. Therefore, the Product Law cannot be applied and its conclusion does not hold:

( lim x→0

f (x) ) (

lim x→0

g(x) )

= (

lim x→0

x ) (

lim x→0

x−1 )

︸ ︷︷ ︸ Does not exist

2.3 SUMMARY

• The Basic Limit Laws: If lim x→c f (x) and limx→c g(x) both exist, then

(i) lim x→c

( f (x) + g(x)

) = lim

x→c f (x) + limx→c g(x) (ii) lim

x→c kf (x) = k limx→c f (x)

(iii) lim x→c f (x) g(x) =

( lim x→c f (x)

)( lim x→c g(x)

)

(iv) If lim x→c g(x) ̸= 0, then limx→c

f (x)

g(x) =

lim x→c f (x)

lim x→c g(x)

(v) If p, q are integers with q ̸= 0,

lim x→c[f (x)]

p/q = (

lim x→c f (x)

)p/q

For n a positive integer,

lim x→c[f (x)]

n = (

lim x→c f (x)

)n , lim

x→c n √

f (x) = n √

lim x→c f (x)

• If lim x→c f (x) or limx→c g(x) does not exist, then the Basic Limit Laws cannot be applied.

2.3 EXERCISES

Preliminary Questions 1. State the Sum Law and Quotient Law.

2. Which of the following is a verbal version of the Product Law (as- suming the limits exist)? (a) The product of two functions has a limit.

(b) The limit of the product is the product of the limits.

(c) The product of a limit is a product of functions. (d) A limit produces a product of functions.

3. Which statement is correct? The Quotient Law does not hold if: (a) the limit of the denominator is zero. (b) the limit of the numerator is zero.

Exercises In Exercises 1–24, evaluate the limit using the Basic Limit Laws and the limits lim

x→c x p/q = cp/q and lim

x→c k = k. 1. lim

x→9 x 2. lim

x→−3 14

3. lim x→ 12

x4 4. lim z→27

z2/3

5. lim t→2

t−1 6. lim x→5

x−2

7. lim x→0.2

(3x + 4) 8. lim x→ 13

(3x3 + 2x2)

9. lim x→−1

(3x4 − 2x3 + 4x) 10. lim x→8

(3x2/3 − 16x−1)

S E C T I O N 2.4 Limits and Continuity 75

11. lim x→2

(x + 1)(3x2 − 9) 12. lim x→ 12

(4x + 1)(6x − 1)

13. lim t→4

3t − 14 t + 1 14. limz→9

√ z

z − 2 15. lim

y→ 14 (16y + 1)(2y1/2 + 1) 16. lim

x→2 x(x + 1)(x + 2)

17. lim y→4

1√ 6y + 1 18. limw→7

√ w + 2 + 1√ w − 3 − 1

19. lim x→−1

x

x3 + 4x 20. limt→−1 t2 + 1

(t3 + 2)(t4 + 1)

21. lim t→25

3 √

t − 15 t (t − 20)2 22. limy→ 13

(18y2 − 4)4

23. lim t→ 32

(4t2 + 8t − 5)3/2 24. lim t→7

(t + 2)1/2 (t + 1)2/3

25. Use the Quotient Law to prove that if lim x→c f (x) exists and is

nonzero, then

lim x→c

1 f (x)

= 1 lim x→c f (x)

26. Assuming that lim x→6

f (x) = 4, compute:

(a) lim x→6

f (x)2 (b) lim x→6

1 f (x)

(c) lim x→6

x √

f (x)

In Exercises 27–30, evaluate the limit assuming that lim x→−4

f (x) = 3 and lim

x→−4 g(x) = 1.

27. lim x→−4

f (x)g(x) 28. lim x→−4

(2f (x) + 3g(x))

29. lim x→−4

g(x)

x2 30. lim

x→−4 f (x) + 1 3g(x) − 9

31. Can the Quotient Law be applied to evaluate lim x→0

sin x x

? Explain.

32. Show that the Product Law cannot be used to evaluate the limit lim

x→π/2 ( x − π2

) tan x.

33. Give an example where lim x→0

(f (x) + g(x)) exists but neither lim x→0

f (x) nor lim x→0

g(x) exists.

34. Give an example where lim x→0

(f (x) · g(x)) exists but neither lim x→0

f (x) nor lim x→0

g(x) exists.

35. Give an example where lim x→0

f (x) g(x) exists but neither limx→0

f (x) nor

lim x→0

g(x) exists.

Further Insights and Challenges 36. Show that if both lim

x→c f (x) g(x) and limx→c g(x) exist and

lim x→c g(x) ̸= 0, then limx→c f (x) exists. Hint: Write f (x) =

f (x) g(x)

g(x) .

37. Suppose that lim t→3

tg(t) = 12. Show that lim t→3

g(t) exists and equals 4.

38. Prove that if lim t→3

h(t) t = 5, then limt→3 h(t) = 15.

39. Assuming that lim x→0

f (x) x = 1, which of the following

statements is necessarily true? Why? (a) f (0) = 0 (b) lim

x→0 f (x) = 0

40. Prove that if lim x→c f (x) = L ̸= 0 and limx→c g(x) = 0, then the limit

lim x→c

f (x) g(x) does not exist.

41. Suppose that lim h→0

g(h) = L. (a) Explain why lim

h→0 g(ah) = L for any constant a ̸= 0.

(b) If we assume instead that lim h→1

g(h) = L, is it still necessarily true that lim

h→1 g(ah) = L?

(c) Illustrate (a) and (b) with the function f (x) = x2.

42. Assume that L(a) = lim x→0

ax − 1 x

exists for all a > 0.Assume also

that lim x→0

ax = 1. (a) Prove that L(ab) = L(a) + L(b) for a, b > 0. Hint: (ab)x − 1 = axbx − ax + ax − 1 = ax(bx − 1) + (ax − 1). This shows that L(a) “behaves” like a logarithm, in the sense that ln(ab) = ln a + ln b. We will see that L(a) = ln a in Section 3.9. (b) Verify numerically that L(12) = L(3) + L(4).

2.4 Limits and Continuity In everyday speech, the word “continuous” means having no breaks or interruptions. In

c

y = f (x)

f (c)

x

y

FIGURE 1 f is continuous at x = c.

calculus, continuity is used to describe functions whose graphs have no breaks. If we imagine the graph of a function f as a wavy metal wire, then f is continuous if its graph consists of a single piece of wire as in Figure 1.

Most physical phenomena are continuous in nature. Our position and velocity vary continuously with time. Temperature varies continuously with time. Viscosity of a fluid varies continuously with temperature, assuming we do not freeze or boil the liquid. Ulti- mately, when we determine the rate of change of a function as we do in the next chapter, we will need the function to be continuous.

76 C H A P T E R 2 LIMITS

A break in the wire as in Figure 2 is called a discontinuity. Observe in Figure 2 that the break in the graph occurs because the left- and right-hand limits as x approaches c are not equal and thus lim

x→c g(x) does not exist. By contrast, in Figure 1, limx→c f (x) exists and

y = g(x)

c

g(c)

x

y

FIGURE 2 Discontinuity at x = c: The left- and right-hand limits as x → c are not equal.

is equal to the function value f (c). This suggests the following definition of continuity in terms of limits.

DEFINITION Continuity at a Point Assume that f (x) is defined on an open interval containing x = c. Then f is continuous at x = c if

lim x→cf (x) = f (c)

If the limit does not exist, or if it exists but is not equal to f (c), we say that f has a discontinuity (or is discontinuous) at x = c.

Note that for f to be continuous at c, three conditions must hold:

1. f (c) is defined. 2. lim x→c f (x) exists. 3. They are equal.

A function f may be continuous at some points and discontinuous at others. If f is continuous at all points in an interval I , then f is said to be continuous on I . If I is an interval [a, b] or [a, b) that includes a as a left endpoint, we require that lim

x→a+ f (x) =

f (a). Similarly, we require that lim x→b−

f (x) = f (b) if I includes b as a right endpoint. If f is continuous at all points in its domain, then f is simply called continuous.

EXAMPLE 1 Show that the following functions are continuous:

c

k

x

y

FIGURE 3 The function f (x) = k is continuous.

(a) f (x) = k (k any constant) (b) g(x) = xn (n a whole number)

Solution

(a) We have lim x→c f (x) = limx→c k = k and f (c) = k. The limit exists and is equal to the

function value for all c, so f is continuous (Figure 3).

(b) By Eq. (1) in Section 2.3, lim x→c g(x) = limx→c x

n = cn for all c.Also g(c) = cn, so again,

c

c

x

y

FIGURE 4 The function g(x) = x is continuous.

the limit exists and is equal to the function value. Therefore, g is continuous. (Figure 4 illustrates the case n = 1.)

Examples of Discontinuities To understand continuity better, let’s consider some ways in which a function can fail to be continuous. Keep in mind that continuity at a point x = c requires that:

1. f (c) is defined. 2. lim x→c f (x) exists. 3. They are equal.

If lim x→c f (x) exists but is not equal to f (c), we say that f has a removable discon-

tinuity at x = c. The function in Figure 5(A) has a removable discontinuity at c = 2 because

f (2) = 10 but lim x→2

f (x) = 5 ︸ ︷︷ ︸

Limit exists but is not equal to function value

Removable discontinuities are “mild” in the following sense: We can make f contin- uous at x = c by redefining f (c). In Figure 5(B), f (2) has been redefined as f (2) = 5, and this makes f continuous at x = 2.

S E C T I O N 2.4 Limits and Continuity 77

2

5

10

2

5

10

x

y

x

y

(A) Removable discontinuity at x = 2 (B) Function redefined at x = 2

FIGURE 5 Removable discontinuity: The discontinuity can be removed by redefining f (2).

A“worse” type of discontinuity is a jump discontinuity, which occurs if the one-sided limits lim

x→c− f (x) and lim

x→c+ f (x) exist but are not equal. Figure 6 shows two functions with

jump discontinuities at c = 2. Unlike the removable case, we cannot make f continuous by redefining f (c).

(A) Left-continuous at x = 2 2

(B) Neither left- nor right-continuous at x = 2 2

x

y

x

y

FIGURE 6 Jump discontinuities.

In connection with jump discontinuities, it is convenient to define one-sided continuity.

DEFINITION One-Sided Continuity A function f is called:

• Left-continuous at x = c if lim x→c−

f (x) = f (c) • Right-continuous at x = c if lim

x→c+ f (x) = f (c)

In Figure 6 above, the function in (A) is left-continuous but the function in (B) is neither left- nor right-continuous. The next example explores one-sided continuity using a piecewise-defined function—that is, a function defined by different formulas on different intervals.

EXAMPLE 2 Piecewise-Defined Function Discuss the continuity of

F(x) =

⎧ ⎪⎨

⎪⎩

x for x < 1 3 for 1 ≤ x ≤ 3 x for x > 3

Solution The functions f (x) = x and g(x) = 3 are continuous, so F is also continuous,

21 4 53

1

2

3

4

5

x

y

FIGURE 7 Piecewise-defined function F in Example 2.

except possibly at the transition points x = 1 and x = 3, where the formula for F(x) changes (Figure 7).

• At x = 1, the one-sided limits exist but are not equal:

lim x→1−

F(x) = lim x→1−

x = 1, lim x→1+

F(x) = lim x→1+

3 = 3

Thus, F has a jump discontinuity at x = 1. However, the right-hand limit is equal to the function value F(1) = 3, so F is right-continuous at x = 1.

• At x = 3, the left- and right-hand limits exist and both are equal to F(3), so F is CAUTION Piecewise-defined functions may OR may not be continuous at points where they are pieced together.

continuous at x = 3:

lim x→3−

F(x) = lim x→3−

3 = 3, lim x→3+

F(x) = lim x→3+

x = 3

78 C H A P T E R 2 LIMITS

We say that f has an infinite discontinuity at x = c if one or both of the one-sided limits are infinite [even if f (x) itself is not defined at x = c]. Figure 8 illustrates three types of infinite discontinuities occurring at x = 2. Notice that x = 2 does not belong to the domain of the function in cases (A) and (B).

1

2

(B) (C)(A)

22 x x x

y y y

FIGURE 8 Functions with an infinite discontinuity at x = 2.

Finally, we note that some functions have more “severe” types of discontinuity than those discussed above. For example, f (x) = sin 1x oscillates infinitely often between +1 and −1 as x → 0 (Figure 9). Neither the left- nor the right-hand limit exists at x = 0, so this discontinuity is not a jump discontinuity. See Exercises 88 and 89 for even stranger examples.Although of interest from a theoretical point of view, these discontinuities rarely arise in practice.1−1 −

x

y

1

−1

2 π

2 π

FIGURE 9 Graph of y = sin 1x . The discontinuity at x = 0 is not a jump, removable, or infinite discontinuity.

Building Continuous Functions Having studied some examples of discontinuities, we focus again on continuous functions. How can we show that a function is continuous? One way is to use the Laws of Continuity, which state, roughly speaking, that a function is continuous if it is built out of functions that are known to be continuous.

THEOREM 1 Basic Laws of Continuity If f and g are continuous at x = c, then the following functions are also continuous at x = c:

(i) f + g and f − g (ii) kf for any constant k

(iii) fg (iv) f/g if g(c) ̸= 0

Proof These laws follow directly from the corresponding Basic Limit Laws (Theorem 1, Section 2.3). We illustrate by proving the first part of (i) in detail. The remaining laws are proved similarly. By definition, we must show that lim

x→c(f (x) + g(x)) = f (c) + g(c). Because f and g are both continuous at x = c, we have

lim x→c f (x) = f (c), limx→c g(x) = g(c)

The Sum Law for limits yields the desired result:

lim x→c(f (x) + g(x)) = limx→c f (x) + limx→c g(x) = f (c) + g(c)

In Section 2.3, we noted that the Basic Limit Laws for Sums and Products are valid for an arbitrary number of functions. The same is true for continuity; that is, if f1, . . . , fn are continuous, then so are the functions

f1 + f2 + · · · + fn, f1 · f2 · · · fn

The basic functions are continuous on their domains. Recall (Section 1.3) that the termWhen a function f is defined and continuous for all values of x, we say that f is continuous on the real line.

basic function refers to polynomials, rational functions, nth-root and algebraic functions, trigonometric functions and their inverses, and exponential and logarithmic functions.

S E C T I O N 2.4 Limits and Continuity 79

THEOREM 2 Continuity of Polynomial and Rational Functions Let P and Q be polynomials. Then:

• P and Q are continuous on the real line. • P/Q is continuous on its domain [at all values x = c such that Q(c) ̸= 0].

REMINDER A rational function is a quotient of two polynomials P/Q.

Proof The function f (x) = xm is continuous for all whole numbers m by Example 1. By Continuity Law (ii), f (x) = axm is continuous for every constant a. A polynomial

P(x) = anxn + an−1xn−1 + · · · + a1x + a0 is a sum of continuous functions, so it too is continuous. By Continuity Law (iv), a quotient function P/Q is continuous at x = c, provided that Q(c) ̸= 0.

This result shows, for example, that f (x) = 3x4 − 2x3 + 8x is continuous for all x and that

g(x) = x + 3 x2 − 1

is continuous forx ̸= ±1. Note that ifn is a positive integer, thenf (x) = x−n is continuous for x ̸= 0 because f (x) = x−n = 1/xn is a rational function.

The continuity of the nth-root, sine, cosine, exponential, and logarithmic functions should not be surprising because their graphs have no visible breaks (Figure 10). However, complete proofs of continuity are somewhat technical and are omitted.

REMINDER The domain of y = x1/n is the real line if n is odd and the half-line [0, ∞) if n is even.

THEOREM 3 Continuity of Some Basic Functions

• y = x1/n is continuous on its domain for n a natural number. • y = sin x and y = cos x are continuous on the real line. • y = bx is continuous on the real line (for b > 0, b ̸= 1). • y = logb x is continuous for x > 0 (for b > 0, b ̸= 1).

2

2

−2

4 8−8

y = x1/2

y = 2 x y = x1/3

3−3

8

x x x

y y

y

1 y = sin x

x

y

π 2

FIGURE 10 As the graphs suggest, these functions are continuous on their domains.

Because f (x) = sin x and f (x) = cos x are continuous, the Continuity Law (iv) for Quotients implies that the other standard trigonometric functions are continuous on their domains, consisting of the values of x where their denominators are nonzero:

tan x = sin x cos x

, cot x = cos x sin x

, sec x = 1 cos x

, csc x = 1 sin x

They have infinite discontinuities at points where their denominators are zero. For example, −π

2 π 2

3π 2

x

y

FIGURE 11 Graph of y = tan x.

f (x) = tan x has infinite discontinuities at the points (Figure 11)

x = ±π 2

, ±3π 2

, ±5π 2

, . . .

80 C H A P T E R 2 LIMITS

The next theorem states that the inverse f −1 of a continuous function f is continuous. This is to be expected because the graph of f −1 is the reflection of the graph of f through the line y = x. If the graph of f has “no breaks,” the same ought to be true of the graph of f −1 (see the proof of Theorem 6 in Appendix D).

THEOREM 4 Continuity of the Inverse Function If f is continuous on an interval I with range R, and if f −1 exists, then f −1 is continuous with domain R.

One consequence of this theorem is that the logarithms, nth roots, and inverse trigono- metric functions f (x) = sin−1 x, f (x) = cos−1 x, f (x) = tan−1 x, and so on are all con- tinuous on their domains.

Finally, it is important to know that a composition of continuous functions is again continuous. The following theorem is proved in Appendix D.

THEOREM 5 Continuity of Composite Functions If g is continuous at x = c, and f is continuous at x = g(c), then the composite function F(x) = f (g(x)) is continuous at x = c.

For example, F(x) = (x2 + 9)1/3 is continuous because it is the composite of the continuous functions f (x) = x1/3 and g(x) = x2 + 9. Similarly, F(x) = cos(x−1) is con- tinuous for all x ̸= 0, and F(x) = 2sin x is continuous for all x.

More generally, an elementary function is a function that is constructed out of basic functions using the operations of addition, subtraction, multiplication, division, and composition. Since the basic functions are continuous (on their domains), an elementary function is also continuous on its domain by the Laws of Continuity. An example of an elementary function is

F(x) = tan−1 (

x2 + cos(2x + 9) x − 8

)

This function is continuous on its domain {x : x ̸= 8}.

Substitution: Evaluating Limits Using Continuity It is easy to evaluate a limit when the function in question is known to be continuous. In this case, by definition, the limit is equal to the function value:

lim x→c f (x) = f (c)

We call this the Substitution Method because the limit is evaluated by “plugging in” x = c.

EXAMPLE 3 Evaluate (a) lim y→ π3

sin y and (b) lim x→−1

3x√ x + 5

.

Solution

(a) We can use substitution because f (y) = sin y is continuous.

lim y→ π3

sin y = sin π 3

= √

3 2

(b) The function f (x) = 3x/√x + 5 is continuous at x = −1 because the numerator and denominator are both continuous at x = −1 and the denominator √x + 5 is nonzero at x = −1. Therefore, we can use substitution:

lim x→−1

3x√ x + 5

= 3 −1

√ −1 + 5

= 1 6

S E C T I O N 2.4 Limits and Continuity 81

The greatest integer function f (x) = ⌊x⌋ is the function defined by ⌊x⌋ = n, where

1

2

3

−3

1 32−2 x

y

FIGURE 12 Graph of f (x) = ⌊x⌋.

n is the unique integer such that n ≤ x < n + 1 (Figure 12). For example, ⌊4.7⌋ = 4 and ⌊−2.3⌋ = −3.

EXAMPLE 4 Assumptions Matter Can we evaluate lim x→2

⌊x⌋ using substitution?

Solution Substitution cannot be applied because f (x) = ⌊x⌋ is not continuous at x = 2. Although f (2) = 2, lim

x→2 ⌊x⌋ does not exist because the one-sided limits are not equal:

lim x→2+

⌊x⌋ = 2 and lim x→2−

⌊x⌋ = 1

CONCEPTUAL INSIGHT Real-World Modeling by Continuous Functions Continuous func- tions are used often to represent physical quantities such as velocity, temperature, and voltage. This reflects our everyday experience that change in the physical world tends to occur continuously rather than through abrupt transitions. However, mathematical models are at best approximations to reality, and it is important to be aware of their limitations.

In Figure 13, atmospheric temperature is represented as a continuous function of al- titude. This is justified for large-scale objects such as the earth’s atmosphere because the reading on a thermometer appears to vary continuously as altitude changes. However, temperature is a measure of the average kinetic energy of molecules. At the micro- scopic level, it would not be meaningful to treat temperature as a quantity that varies continuously from point to point.

Similarly, the size P(t) of a population is usually treated as a continuous function of time t . Strictly speaking, P(t) is a whole number that changes by ±1 when an individual is born or dies, so it is not continuous, but if the population is large, the effect of an individual birth or death is small, and it is both reasonable and convenient to treat P as a continuous function.

World population (millions)

6,000

4,000

2,000

0 17501700 1800 1850 1900 1950 2000

Tr op

os ph

er e

St ra

to sp

he re

M es

os ph

er e

T he

rm os

ph er

e

Temperature (°C)

200

100

0

−100

10 50 100 150

FIGURE 13 Atmospheric temperature and world population are represented by continuous graphs.

2.4 SUMMARY

• Definition: f is continuous at x = c if lim x→c f (x) = f (c). This means that f (c) exists,

lim x→c f (x) exists, and they are equal.

• If lim x→c f (x) does not exist, or if it exists but does not equal f (c), then f is discontinuous

at x = c. • If f is continuous at all points in its domain, f is simply called continuous. • Right-continuous at x = c: lim

x→c+ f (x) = f (c).

• Left-continuous at x = c: lim x→c−

f (x) = f (c).

82 C H A P T E R 2 LIMITS

• Three common types of discontinuities: removable discontinuity [

lim x→c f (x) exists but

does not equal f (c) ] , jump discontinuity (the one-sided limits both exist but are not

equal), and infinite discontinuity (the limit is infinite as x approaches c from one or both sides).

• Laws of Continuity: Sums, products, multiples, inverses, and composites of continuous functions are again continuous. The same holds for a quotient f/g at points where g(x) ̸= 0.

• Basic functions: Polynomials, rational functions, nth-root and algebraic functions, trigonometric functions and their inverses, exponential and logarithmic functions. Basic functions are continuous on their domains.

• Substitution Method: If f is known to be continuous at x = c, then the value of the limit lim x→c f (x) is f (c).

2.4 EXERCISES

Preliminary Questions 1. Which property of f (x) = x3 allows us to conclude that lim x→2

x3 = 8?

2. What can be said about f (3) if f is continuous and lim x→3

f (x) = 12 ?

3. Suppose that f (x) < 0 if x is positive and f (x) > 1 if x is negative. Can f be continuous at x = 0? 4. Is it possible to determine f (7) if f (x) = 3 for all x < 7 and f is right-continuous at x = 7? What if f is left-continuous? 5. Are the following true or false? If false, then draw or give a coun- terexample, and state a correct version.

(a) f is continuous at x = a if the left- and right-hand limits of f (x) as x → a exist and are equal.

(b) f is continuous at x = a if the left- and right-hand limits of f (x) as x → a exist and equal f (a).

(c) If the left- and right-hand limits of f (x) as x → a exist, then f has a removable discontinuity at x = a.

(d) If f and g are continuous at x = a, then f + g is continuous at x = a.

(e) If f and g are continuous at x = a, then f/g is continuous at x = a.

Exercises 1. Referring to Figure 14, state whether f is left- or right-continuous

(or neither) at each point of discontinuity. Does f have any removable discontinuities?

Exercises 2–4 refer to the function g whose graph appears in Figure 15.

2. State whether g is left- or right-continuous (or neither) at each of its points of discontinuity.

3. At which point c does g have a removable discontinuity? How should g(c) be redefined to make g continuous at x = c?

4. Find the point c1 at which g has a jump discontinuity but is left- continuous. How should g(c1) be redefined to make g right-continuous at x = c1?

1 2 3 4 5 6 x

5

4

3

2

1

y

FIGURE 14 Graph of y = f (x). 1 2 3 4 5 6

x

5

4

3

2

1

y

FIGURE 15 Graph of y = g(x).

5. In Figure 16, determine the one-sided limits at the points of dis- continuity. Which discontinuity is removable and how should f be redefined to make it continuous at this point?

42−2

6

2

x

y

FIGURE 16

6. Suppose that f (x) = 2 for x < 3 and f (x) = −4 for x > 3. (a) What is f (3) if f is left-continuous at x = 3? (b) What is f (3) if f is right-continuous at x = 3? In Exercises 7–16, use the Laws of Continuity and Theorems 2 and 3 to show that the function is continuous.

7. f (x) = x + sin x 8. f (x) = x sin x 9. f (x) = 3x + 4 sin x 10. f (x) = 3x3 + 8x2 − 20x

11. f (x) = 1 x2 + 1 12. f (x) =

x2 − cos x 3 + cos x

13. f (x) = cos(x2) 14. f (x) = tan−1(4x) 15. f (x) = ex cos 3x 16. f (x) = ln(x4 + 1) In Exercises 17–34, determine the points of discontinuity. State the type of discontinuity (removable, jump, infinite, or none of these) and whether the function is left- or right-continuous.

S E C T I O N 2.4 Limits and Continuity 83

17. f (x) = 1 x

18. f (x) = |x|

19. f (x) = x − 2|x − 1| 20. f (x) = ⌊x⌋

21. f (x) = ⌊x

2

⌋ 22. g(t) = 1

t2 − 1

23. f (x) = x + 1 4x − 2 24. h(z) =

1 − 2z z2 − z − 6

25. f (x) = 3x2/3 − 9x3 26. g(t) = 3t−2/3 − 9t3

27. f (x) =

⎧ ⎨

⎩

x − 2 |x − 2| x ̸= 2

−1 x = 2 28. f (x) =

{ cos

1 x

x ̸= 0 1 x = 0

29. g(t) = tan 2t 30. f (x) = csc(x2) 31. f (x) = tan(sin x) 32. f (x) = cos(π⌊x⌋)

33. f (x) = 1 ex − e−x 34. f (x) = ln |x − 4|

In Exercises 35–48, determine the domain of the function and prove that it is continuous on its domain using the Laws of Continuity and the facts quoted in this section.

35. f (x) = 2 sin x + 3 cos x 36. f (x) = √

x2 + 9

37. f (x) = √x sin x 38. f (x) = x 2

x + x1/4

39. f (x) = x2/32x 40. f (x) = x1/3 + x3/4

41. f (x) = x−4/3 42. f (x) = ln(9 − x2)

43. f (x) = tan2 x 44. f (x) = cos(2x)

45. f (x) = (x4 + 1)3/2 46. f (x) = e−x2

47. f (x) = cos(x 2)

x2 − 1 48. f (x) = 9 tan x

49. Show that the function

f (x) =

⎧ ⎪⎨

⎪⎩

x2 + 3 for x < 1 10 − x for 1 ≤ x ≤ 2 6x − x2 for x > 2

is continuous for x ̸= 1, 2. Then compute the right- and left-hand lim- its at x = 1, 2, and determine whether f is left-continuous, right- continuous, or continuous at these points (Figure 17).

621

9

y = 10 − x

y = 6x − x2

y = x2 + 3 x

y

FIGURE 17

50. Sawtooth Function Draw the graph of f (x) = x − ⌊x⌋. At which points is f discontinuous? Is it left- or right-continuous at those points?

In Exercises 51–54, sketch the graph of f . At each point of discontinuity, state whether f is left- or right-continuous.

51. f (x) = {

x2 for x ≤ 1 2 − x for x > 1

52. f (x) =

⎧ ⎨

⎩

x + 1 for x < 1 1 x

for x ≥ 1

53. f (x) =

⎧ ⎨

⎩

x2 − 3x + 2 |x − 2| x ̸= 2

0 x = 2

54. f (x) =

⎧ ⎪⎨

⎪⎩

x3 + 1 for −∞ < x ≤ 0 −x + 1 for 0 < x < 2 −x2 + 10x − 15 for x ≥ 2

55. Show that the function

f (x) =

⎧ ⎨

⎩

x2 − 16 x − 4 x ̸= 4

10 x = 4

has a removable discontinuity at x = 4.

56. Define f (x) = x sin 1x + 2 for x ̸= 0. Plot f . How should f (0) be defined so that f is continuous at x = 0? In Exercises 57–59, find the value of the constant (a, b, or c) that makes the function continuous.

57. f (x) = {

x2 − c for x < 5 4x + 2c for x ≥ 5

58. f (x) = {

2x + 9x−1 for x ≤ 3 −4x + c for x > 3

59. f (x) =

⎧ ⎪⎨

⎪⎩

x−1 for x < −1 ax + b for − 1 ≤ x ≤ 12 x−1 for x > 12

60. Define

g(x) =

⎧ ⎪⎨

⎪⎩

x + 3 for x < −1 cx for − 1 ≤ x ≤ 2 x + 2 for x > 2

Find a value of c such that g is (a) left-continuous (b) right-continuous In each case, sketch the graph of g.

61. Define g(t) = tan−1 (

1 t − 1

) for t ̸= 1. Answer the following

questions, using a plot if necessary. (a) Can g(1) be defined so that g is continuous at t = 1? (b) How should g(1) be defined so that g is left-continuous at t = 1?

62. Each of the following statements is false. For each statement, sketch the graph of a function that provides a counterexample. (a) If lim

x→a f (x) exists, then f is continuous at x = a. (b) If f has a jump discontinuity at x = a, then f (a) is equal to either

lim x→a−

f (x) or lim x→a+

f (x).

84 C H A P T E R 2 LIMITS

In Exercises 63–66, draw the graph of a function on [0, 5] with the given properties.

63. f is not continuous at x = 1, but lim x→1+

f (x) and lim x→1−

f (x) exist

and are equal.

64. f is left-continuous but not continuous at x = 2, and right- continuous but not continuous at x = 3. 65. f has a removable discontinuity at x = 1, a jump discontinuity at x = 2, and

lim x→3−

f (x) = −∞, lim x→3+

f (x) = 2

66. f is right- but not left-continuous at x = 1, left- but not right- continuous at x = 2, and neither left- nor right-continuous at x = 3. In Exercises 67–80, evaluate using substitution.

67. lim x→−1

(2x3 − 4) 68. lim x→2

(5x − 12x−2)

69. lim x→3

x + 2 x2 + 2x 70. limx→π sin

(x 2

− π )

71. lim x→ π4

tan(3x) 72. lim x→π

1 cos x

73. lim x→4

x−5/2 74. lim x→2

√ x3 + 4x

75. lim x→−1

(1 − 8x3)3/2 76. lim x→2

(7x + 2 4 − x

)2/3

77. lim x→3

10x 2−2x 78. lim

x→− π2 3sin x

79. lim x→4

sin−1 (x

4

) 80. lim

x→0 tan−1(ex)

81. Suppose that f and g are discontinuous at x = c. Does it follow that f + g is discontinuous at x = c? If not, give a counterexample. Does this contradict Theorem 1(i)?

82. Prove that f (x) = |x| is continuous for all x. Hint: To prove con- tinuity at x = 0, consider the one-sided limits.

83. Use the result of Exercise 82 to prove that if g is continuous, then f (x) = |g(x)| is also continuous.

84. Which of the following quantities would be represented by contin- uous functions of time and which would have one or more discontinu- ities? (a) Velocity of an airplane during a flight (b) Temperature in a room under ordinary conditions (c) Value of a bank account with interest paid yearly (d) The salary of a teacher (e) The population of the world

85. In 2009, the federal income tax T on income of x dollars (up to $82,250) was determined by the formula

T (x) =

⎧ ⎪⎨

⎪⎩

0.10x for 0 ≤ x < 8350 0.15x − 417.50 for 8350 ≤ x < 33,950 0.25x − 3812.50 for 33,950 ≤ x < 82,250

Sketch the graph of T . Does T have any discontinuities? Explain why, if T had a jump discontinuity, it might be advantageous in some situations to earn less money.

Further Insights and Challenges 86. If f has a removable discontinuity at x = c, then it is pos- sible to redefine f (c) so that f is continuous at x = c. Can this be done in more than one way?

87. Give an example of functions f and g such that f (g(x)) is con- tinuous but g has at least one discontinuity.

88. Continuous at Only One Point Show that the following func- tion is continuous only at x = 0:

f (x) = {

x for x rational −x for x irrational

89. Show that f is a discontinuous function for all x, where f (x) is defined as follows:

f (x) = {

1 for x rational −1 for x irrational

Show that f 2 is continuous for all x.

2.5 Evaluating Limits Algebraically Substitution can be used to evaluate limits when the function in question is known to be continuous. For example, f (x) = x−2 is continuous at x = 3, and therefore,

lim x→3

x−2 = 3−2 = 1 9

When we study derivatives in Chapter 3, we will be faced with limits lim x→c f (x), where

f (c) is not defined. In such cases, substitution cannot be used directly. However, many of these limits can be evaluated if we use algebra to rewrite the formula for f (x).

To illustrate, consider the limit (Figure 1): 4 8

4

8

12

x

y

FIGURE 1 Graph of f (x) = x 2 − 16 x − 4 . This

function is undefined at x = 4, but the limit as x → 4 exists. limx→4

x2 − 16 x − 4

S E C T I O N 2.5 Evaluating Limits Algebraically 85

The function f (x) = x 2 − 16 x − 4 is not defined at x = 4 because the formula for f (4) pro-

duces the undefined expression 0/0. However, the numerator of f (x) factors:

x2 − 16 x − 4 =

(x + 4)(x − 4) x − 4 = x + 4 (valid for x ̸= 4)

This shows that f coincides with the continuous function y = x + 4 for all x ̸= 4. Since the limit depends only on the values of f (x) for x ̸= 4, we have

lim x→4

x2 − 16 x − 4 = limx→4(x + 4) = 8︸ ︷︷ ︸

Evaluate by substitution

We say that f (x) has an indeterminate form (or is indeterminate) at x = c if the formula for f (c) yields an undefined expression of the type 00 ,

∞ ∞ , ∞ · 0, ∞ − ∞.

Indeterminate forms are warning signs that tell us more work needs to be done to evaluate the limit. Our strategy, when this occurs, is to transform f (x) algebraically, if possible, into a new expression that is defined and continuous at x = c, and then evaluate the limit by substitution (“plugging in”). As you study the following examples, notice that the critical step is to cancel a common factor from the numerator and denominator at the appropriate moment, thereby removing the indeterminacy.

EXAMPLE 1 Calculate lim x→3

x2 − 4x + 3 x2 + x − 12 .

Other indeterminate forms are 1∞, ∞0, and 00. These are treated in Section 4.5.

Solution The function has the indeterminate form 0/0 at x = 3 because Numerator at x = 3: x2 − 4x + 3 = 32 − 4(3) + 3 = 0 Denominator at x = 3: x2 + x − 12 = 32 + 3 − 12 = 0

Step 1. Transform algebraically and cancel.

x2 − 4x + 3 x2 + x − 12 =

(x − 3)(x − 1) (x − 3)(x + 4)︸ ︷︷ ︸

Cancel common factor

= x − 1 x + 4︸ ︷︷ ︸

Continuous at x = 3

(if x ̸= 3) 1

Step 2. Substitute (evaluate using continuity). Because the expression on the right in Eq. (1) is continuous at x = 3,

lim x→3

x2 − 4x + 3 x2 + x − 12 = limx→3

x − 1 x + 4 =

2 7︸ ︷︷ ︸

Evaluate by substitution

EXAMPLE 2 The Form ∞ ∞ Calculate limx→ π2

tan x sec x

.

Solution As we see in Figure 2, both f (x) = tan x and f (x) = sec x have infinite dis- continuities at x = π2 , so this limit has the indeterminate form ∞/∞ at x = π2 .

1 −1

y = tan x

y = sec x

y

x π 2

FIGURE 2

86 C H A P T E R 2 LIMITS

Step 1. Transform algebraically and cancel.

tan x sec x

= (sin x)

( 1

cos x

)

1 cos x

= sin x (if cos x ̸= 0)

Step 2. Substitute (evaluate using continuity). Because f (x) = sin x is continuous,

lim x→ π2

tan x sec x

= lim x→ π2

sin x = sin π 2

= 1

The next example illustrates the algebraic technique of “multiplying by the conju- gate,” which can be used to treat some indeterminate forms involving square roots.

EXAMPLE 3 Multiplying by the Conjugate Evaluate lim x→4

√ x − 2

x − 4 .

Solution We check that f (x) = √

x − 2 x − 4 has the indeterminate form 0/0 at x = 4:

Numerator at x = 4: √x − 2 = √

4 − 2 = 0 Denominator at x = 4: x − 4 = 4 − 4 = 0

Step 1. Multiply by the conjugate and cancel. Note, in Step 1, that the conjugate of√

x − 2 is √x + 2, so ( √

x − 2)(√x + 2) = x − 4.

(√ x − 2

x − 4

) (√ x + 2√ x + 2

) = x − 4

(x − 4)(√x + 2) = 1√

x + 2 (if x ̸= 4)

Step 2. Substitute (evaluate using continuity). Because f (x) = 1/(√x + 2) is continuous at x = 4,

lim x→4

√ x − 2

x − 4 = limx→4 1√

x + 2 = 1 4

EXAMPLE 4 Evaluate lim h→5

h − 5√ h + 4 − 3

.

Solution We note that f (h) = h − 5√ h + 4 − 3

yields 0/0 at h = 5:

Numerator at h = 5: h − 5 = 5 − 5 = 0 Denominator at h = 5:

√ h + 4 − 3 =

√ 5 + 4 − 3 = 0

The conjugate of √

h + 4 − 3 is √

h + 4 + 3, and

h − 5√ h + 4 − 3

= (

h − 5√ h + 4 − 3

) (√ h + 4 + 3√ h + 4 + 3

)

= (h − 5) (√

h + 4 + 3 )

(√ h + 4 − 3

)(√ h + 4 + 3

)

The denominator is equal to (√

h + 4 − 3 )(√

h + 4 + 3 )

= (√

h + 4 )2 − 9 = h − 5

Thus, for h ̸= 5,

f (h) = h − 5√ h + 4 − 3

= (h − 5) (√

h + 4 + 3 )

h − 5 = √

h + 4 + 3

We obtain

lim h→5

h − 5√ h + 4 − 3

= lim h→5

(√ h + 4 + 3

) =

√ 9 + 3 = 6

S E C T I O N 2.5 Evaluating Limits Algebraically 87

EXAMPLE 5 The Form ∞ − ∞ Calculate lim x→1

( 1

x − 1 − 2

x2 − 1

) .

Solution As we see in Figure 3, y = 1 x − 1 and y =

2 x2 − 1 both have infinite disconti-

nuities at x = 1, so this limit has the indeterminate form ∞ − ∞. 1

−2 −4

−6

2

4

6

x2 − 1 y = 2

x − 1 y = 1

y

x

FIGURE 3

Step 1. Transform algebraically and cancel. Combine the fractions and simplify (for x ̸= 1):

1 x − 1 −

2 x2 − 1 =

x + 1 x2 − 1 −

2 x2 − 1 =

x − 1 x2 − 1 =

x − 1 (x − 1)(x + 1) =

1 x + 1

Step 2. Substitute (evaluate using continuity).

lim x→1

( 1

x − 1 − 2

x2 − 1

) = lim

x→1 1

x + 1 = 1

1 + 1 = 1 2

In the next example, the function has the undefined form a/0 with a nonzero. This isCAUTION The form a0 with nonzero a is NOT an indeterminate form. In this case, the limit does not exist, and we have a vertical asymptote.

not an indeterminate form (it is not of the form 0/0, ∞/∞, etc.).

EXAMPLE 6 Infinite But Not Indeterminate Evaluate lim x→2

x2 − x + 5 x − 2 .

Solution The function f (x) = x 2 − x + 5 x − 2 is undefined at x = 2 because the formula

for f (2) yields 7/0:

Numerator at x = 2: x2 − x + 5 = 22 − 2 + 5 = 7 Denominator at x = 2: x − 2 = 2 − 2 = 0

But f (x) is not indeterminate at x = 2 because 7/0 is not an indeterminate form. As in Figure 4, the one-sided limits are infinite:

−20

20

2 x

y

FIGURE 4 Graph of f (x) = x 2 − x + 5 x − 2 .

lim x→2−

x2 − x + 5 x − 2 = −∞, limx→2+

x2 − x + 5 x − 2 = ∞

The limit itself does not exist since the numerator approaches a fixed nonzero value and the denominator approaches 0.

As preparation for the derivative in Chapter 3, we evaluate a limit involving a sym- bolic constant.

EXAMPLE 7 Symbolic Constant Calculate lim h→0

(h + a)2 − a2 h

, where a is a constant.

Solution We have the indeterminate form 0/0 at h = 0 because Numerator at h = 0: (h + a)2 − a2 = (0 + a)2 − a2 = 0 Denominator at h = 0: h = 0

Expand the numerator and simplify (for h ̸= 0): (h + a)2 − a2

h = (h

2 + 2ah + a2) − a2 h

= h 2 + 2ah

h = h(h + 2a)

h = h + 2a

The function y = h + 2a is continuous (for any constant a), so

lim h→0

(h + a)2 − a2 h

= lim h→0

(h + 2a) = 2a

Given an expression that yields an indeterminate form, it will not always be the case

that algebraic manipulation yields the limit. For instance, consider lim x→0

sin x x

. Although

this is an indeterminate form of type 0/0, algebraic manipulation does not provide an

88 C H A P T E R 2 LIMITS

answer. In the next section, we use the Squeeze Theorem to evaluate this limit. It can also be evaluated by L’Hôpital’s Rule, which will allow us to evaluate many indeterminate forms and which we will discuss in Section 4.5.

2.5 SUMMARY

• When f is known to be continuous at x = c, the limit can be evaluated by substitu- tion: lim

x→c f (x) = f (c). • We say that f (x) is indeterminate (or has an indeterminate form) at x = c if the formula

for f (c) yields an undefined expression of the type

0 0 ,

∞ ∞ , ∞ · 0, ∞ − ∞

• If f (x) is indeterminate at x = c, try to transform f (x) algebraically into a new expres- sion that is defined and continuous at x = c. Then evaluate by substitution.

2.5 EXERCISES

Preliminary Questions 1. Which of the following is indeterminate at x = 1?

x2 + 1 x − 1 ,

x2 − 1 x + 2 ,

x2 − 1√ x + 3 − 2

, x2 + 1√ x + 3 − 2

2. Give counterexamples to show that these statements are false: (a) If f (c) is indeterminate, then the right- and left-hand limits as x → c are not equal.

(b) If lim x→c f (x) exists, then f (c) is not indeterminate.

(c) If f (x) is undefined at x = c, then f (x) has an indeterminate form at x = c.

3. The method for evaluating limits discussed in this section is some- times called “simplify and plug in.” Explain how it actually relies on the property of continuity.

Exercises In Exercises 1–4, show that the limit leads to an indeterminate form. Then carry out the two-step procedure: Transform the function alge- braically and evaluate using continuity.

1. lim x→6

x2 − 36 x − 6 2. limh→3

9 − h2 h − 3

3. lim x→−1

x2 + 2x + 1 x + 1 4. limt→9

2t − 18 5t − 45

In Exercises 5–34, evaluate the limit, if it exists. If not, determine whether the one-sided limits exist (finite or infinite).

5. lim x→7

x − 7 x2 − 49 6. limx→8

x2 − 64 x − 9

7. lim x→−2

x2 + 3x + 2 x + 2 8. limx→8

x3 − 64x x − 8

9. lim x→5

2x2 − 9x − 5 x2 − 25 10. limh→0

(1 + h)3 − 1 h

11. lim x→− 12

2x + 1 2x2 + 3x + 1 12. limx→3

x2 − x x2 − 9

13. lim x→2

3x2 − 4x − 4 2x2 − 8 14. limh→0

(3 + h)3 − 27 h

15. lim t→0

42t − 1 4t − 1 16. limh→4

(h + 2)2 − 9h h − 4

17. lim x→16

√ x − 4

x − 16 18. limt→−2 2t + 4

12 − 3t2

19. lim h→0

1

(h + 2)2 − 1 4

h 20. lim

y→3 y2 + y − 12 y3 − 10y + 3

21. lim h→0

√ 2 + h − 2

h 22. lim

x→8

√ x − 4 − 2 x − 8

23. lim x→4

x − 4√ x −

√ 8 − x

24. lim x→4

√ 5 − x − 1 2 − √x

25. lim x→4

( 1√

x − 2 − 4

x − 4

) 26. lim

x→0+

( 1√ x

− 1√ x2 + x

)

27. lim x→0

cot x csc x

28. lim θ→ π2

cot θ csc θ

29. lim x→1

( 1

1 − x − 2

1 − x2 )

30. lim x→ π4

sin x − cos x tan x − 1

31. lim t→2

22t + 2t − 20 2t − 4 32. limθ→ π2

( sec θ − tan θ

)

33. lim θ→ π4

( 1

tan θ − 1 − 2

tan2 θ − 1

)

34. lim x→ π3

2 cos2 x + 3 cos x − 2 2 cos x − 1

S E C T I O N 2.6 Trigonometric Limits 89

35. Use a plot of f (x) = x − 4√ x −

√ 8 − x

to estimate lim x→4

f (x)

to two decimal places. Compare with the answer obtained algebraically in Exercise 23.

36. Use a plot of f (x) = 1√ x − 2 −

4 x − 4 to estimate

lim x→4

f (x) numerically. Compare with the answer obtained alge-

braically in Exercise 25.

In Exercises 37–42, evaluate using the identity

a3 − b3 = (a − b)(a2 + ab + b2)

37. lim x→2

x3 − 8 x − 2 38. limx→3

x3 − 27 x2 − 9

39. lim x→1

x2 − 5x + 4 x3 − 1 40. limx→−2

x3 + 8 x2 + 6x + 8

41. lim x→1

x4 − 1 x3 − 1 42. limx→27

x − 27 x1/3 − 3

43. Evaluate lim h→0

4√1 + h − 1 h

. Hint: Set x = 4 √

1 + h, express h as a function of x, and rewrite as a limit as x → 1.

44. Evaluate lim h→0

3√1 + h − 1 2√1 + h − 1

. Hint: Set x = 6 √

1 + h, express h as a function of x, and rewrite as a limit as x → 1. In Exercises 45–54, evaluate in terms of the constant a.

45. lim x→0

(2a + x) 46. lim h→−2

(4ah + 7a)

47. lim t→−1

(4t − 2at + 3a) 48. lim h→0

(3a + h)2 − 9a2 h

49. lim h→0

2(a + h)2 − 2a2 h

50. lim x→a

(x + a)2 − 4x2 x − a

51. lim x→a

√ x − √a x − a 52. limh→0

√ a + 2h − √a

h

53. lim x→0

(x + a)3 − a3 x

54. lim h→a

1 h

− 1 a

h − a

Further Insights and Challenges In Exercises 55–58, find all values of c such that the limit exists.

55. lim x→c

x2 − 5x − 6 x − c 56. limx→1

x2 + 3x + c x − 1

57. lim x→1

( 1

x − 1 − c

x3 − 1

) 58. lim

x→0 1 + cx2 −

√ 1 + x2

x4

59. For which sign, + or −, does the following limit exist?

lim x→0

( 1 x

± 1 x(x − 1)

)

2.6 Trigonometric Limits In our study of the derivative, we will need to evaluate certain limits involving transcen- dental functions such as sine and cosine. The algebraic techniques of the previous section are often ineffective for such functions, and other tools are required. In this section, we discuss one such tool—the Squeeze Theorem—and use it to evaluate the trigonometric limits needed in Section 3.6.

The Squeeze Theorem Consider a function f that is “trapped” between two functions l, for lower bound, and u,

y " u(x)

c x

y

y " f (x)

y " l (x)

FIGURE 1 f is trapped between l and u (but not squeezed at x = c).

for upper bound, on an interval I . In other words,

l(x) ≤ f (x) ≤ u(x) for all x ∈ I Thus, the graph of f lies between the graphs of l and u (Figure 1).

The Squeeze Theorem applies when f is not just trapped but squeezed at a point x = c (Figure 2). By this we mean that for all x ̸= c in some open interval containing c,

c

L

x

y

y " u(x)

y " l(x)

y " f (x)

FIGURE 2 f is squeezed by l and u at x = c.

l(x) ≤ f (x) ≤ u(x) and lim x→c l(x) = limx→c u(x) = L

We do not require that f (x) be defined at x = c, but it is clear graphically that f (x) must approach the limit L, as stated in the next theorem. See Appendix D for a proof.

THEOREM 1 Squeeze Theorem Assume that for x ̸= c (in some open interval con- taining c),

l(x) ≤ f (x) ≤ u(x) and lim x→c l(x) = limx→c u(x) = L

Then lim x→c f (x) exists and limx→c f (x) = L.

90 C H A P T E R 2 LIMITS

EXAMPLE 1 Show that lim x→0

x sin 1x = 0.

Solution Although f (x) = x sin 1x is a product of two functions, we cannot use the Prod- uct Law because lim

x→0 sin 1x does not exist. However, the sine function takes on values

between 1 and −1, and therefore ∣∣sin 1x

∣∣ ≤ 1 for all x ̸= 0. Multiplying by |x|, we obtain∣∣x sin 1x ∣∣ ≤ |x| and conclude that (Figure 3)

−|x| ≤ x sin 1 x

≤ |x|

Because

y = x sin 1x

−0.4

−0.1

0.4

0.1

y = −|x|

y = |x|

x

y

FIGURE 3

lim x→0

|x| = 0 and lim x→0

(−|x|) = − lim x→0

|x| = 0

we can apply the Squeeze Theorem to conclude that lim x→0

x sin 1x = 0.

In Section 2.2, we found numerical and graphical evidence suggesting that the limit

lim θ→0

sin θ θ

is equal to 1. The Squeeze Theorem will allow us to prove this fact.

THEOREM 2 Important Trigonometric Limits

lim θ→0

sin θ θ

= 1 and lim θ→0

1 − cos θ θ

= 0

To apply the Squeeze Theorem, we must find functions that squeeze sin θ

θ at θ = 0.

Note that both sin θ θ

and cos θ−1 θ

are indeterminate at θ = 0, so Theorem 2 cannot be proved by substitution.

These are provided by the next theorem (Figure 4).

−

−1

1 y = 1

y = cos θ

y = sin θ θ

−π ππ 2

θ

y

π 2

FIGURE 4 Graph illustrating the inequalities of Theorem 3.

THEOREM 3

cos θ ≤ sin θ θ

≤ 1 for −π 2

< θ < π

2 , θ ̸= 0 1

Proof Assume first that 0 < θ < π2 . Our proof is based on the following relation between the areas in Figure 5:

Area of △OAB < area of sector BOA < area of △OAC 2 Let’s compute these three areas. First, △OAB has base 1 and height sin θ , so its area is

REMINDER Let’s recall why a sector of angle θ in a circle of radius r has area 1 2 r

2θ . A sector of angle θ represents a fraction θ2π of the entire circle. The circle has area πr2, so the sector has area(

θ 2π

) πr2 = 12 r2θ . In the unit circle

(r = 1), the sector has area 12 θ .

NOTE Our proof of Theorem 3 uses the formula 12 θ for the area of a sector, but this formula is based on the formula πr2 for the area of a circle, a complete proof of which requires integral calculus.

1 2 sin θ . Next, recall that a sector of angle θ has area

1 2θ . Finally, to compute the area of

△OAC, we observe that

tan θ = opposite side adjacent side

= AC OA

= AC 1

= AC

Thus, △OAC has base 1, height tan θ , and area 12 tan θ . We have shown, therefore, that 1 2

sin θ ︸ ︷︷ ︸

Area △OAB

≤ 1 2 θ

︸︷︷︸ Area of sector

≤ 1 2

sin θ cos θ︸ ︷︷ ︸

Area △OAC

3

S E C T I O N 2.6 Trigonometric Limits 91

O A

B = (cos θ, sin θ)

Area of triangle = sin θ Area of triangle =Area of sector =

tan θ

θ

O A11 1

C

B

θ

O A

B

θ xxx

y y y

1 2

tan θθ

11

1 2

1 2

FIGURE 5

The first inequality yields sin θ ≤ θ , and because θ > 0, we obtain sin θ

θ ≤ 1 4

Next, multiply the second inequality in (3) by 2 cos θ

θ to obtain

cos θ ≤ sin θ θ

5

The combination of (4) and (5) gives us (1) when 0 < θ < π2 . However, the functions in (1) do not change value when θ is replaced by −θ because both f (x) = cos θ and f (x) = sin θ

θ are even functions. Indeed, cos(−θ) = cos θ and

sin(−θ) −θ =

− sin θ −θ =

sin θ θ

Therefore, (1) holds for −π2 < θ < 0 as well. This completes the proof of Theorem 3.

Proof of Theorem 2 According to Theorem 3,

cos θ ≤ sin θ θ

≤ 1

Since lim θ→0

cos θ = cos 0 = 1 and lim θ→0

1 = 1, the Squeeze Theorem yields lim θ→0

sin θ θ

= 1, as required. It then follows that

lim θ→0

1 − cos θ θ

= lim θ→0

( 1 + cos θ 1 + cos θ

) 1 − cos θ

θ = lim

θ→0 1 − cos2 θ

(1 + cos θ)θ

= lim θ→0

1 1 + cos θ

sin2 θ θ

= lim x→0

sin θ 1 + cos θ

sin θ θ

= 0 2

· 1 = 0

In the next example, we evaluate another trigonometric limit. The key idea is to rewrite the function of h in terms of the new variable θ = 4h.

EXAMPLE 2 Evaluating a Limit by Changing Variables Investigate lim h→0

sin 4h hnumerically and then evaluate it exactly.

Solution The table of values at the left suggests that the limit is equal to 4. To evaluate

h sin 4h

h

±1.0 −0.75680 ±0.5 1.81859 ±0.2 3.58678 ±0.1 3.8 9 418 ±0.05 3.9 7 339 ±0.01 3.99 893 ±0.005 3.999 73

the limit exactly, we rewrite it in terms of the limit of sin θ

θ so that Theorem 2 can be

applied. Thus, we set θ = 4h and write sin 4h

h = 4

( sin 4h 4h

) = 4sin θ

θ

92 C H A P T E R 2 LIMITS

The new variable θ tends to zero as h → 0 because θ is a multiple of h. Therefore, we may change the limit as h → 0 into a limit as θ → 0 to obtain

lim h→0

sin 4h h

= lim θ→0

4 sin θ

θ = 4

( lim θ→0

sin θ θ

) = 4(1) = 4 6

Note that the change of variables demonstrates that sin(kx)

kx approaches 1 as x → 0.

We can use this to our advantage in the next example.

EXAMPLE 3 Find lim x→0

tan(3x) tan(2x)

.

Solution

lim x→0

tan(3x) tan(2x)

= lim x→0

sin(3x) cos(3x)

· cos(2x) sin(2x)

= lim x→0

sin(3x) cos(3x)

· cos(2x) sin(2x)

· x x

7

= lim x→0

3 2

( sin(3x)

3x

) ( 2x

sin(2x)

) cos(2x) cos(3x)

= 3 2

· 1 · 1 · 1 1

= 3 2

8

2.6 SUMMARY

• We say that f is squeezed at x = c if there exist functions l and u such that l(x) ≤ f (x) ≤ u(x) for all x ̸= c in an open interval I containing c, and

lim x→c l(x) = limx→c u(x) = L

The Squeeze Theorem states that in this case, lim x→c f (x) = L.

• Two important trigonometric limits:

lim θ→0

sin θ θ

= 1 and lim θ→0

1 − cos θ θ

= 0

2.6 EXERCISES

Preliminary Questions 1. Assume that −x4 ≤ f (x) ≤ x2. What is lim

x→0 f (x)? Is there

enough information to evaluate lim x→ 12

f (x)? Explain.

2. State the Squeeze Theorem carefully.

3. If you want to evaluate lim h→0

sin 5h 3h

, it is a good idea to rewrite the

limit in terms of the variable (choose one):

(a) θ = 5h (b) θ = 3h (c) θ = 5h 3

Exercises 1. State precisely the hypothesis and conclusions of the Squeeze The-

orem for the situation in Figure 6.

1 2

2

x

y y ! u(x)

y ! l(x)

y ! f (x)

FIGURE 6

2. In Figure 7, is f squeezed by u and l at x = 3? At x = 2?

1 2 3 4

1.5

x

y y ! u(x)

y ! l (x)

y ! f (x)

FIGURE 7

S E C T I O N 2.6 Trigonometric Limits 93

3. What does the Squeeze Theorem say about lim x→7

f (x) if the limits

lim x→7

l(x) = lim x→7

u(x) = 6 and f , u, and l are related as in Figure 8? The inequality f (x) ≤ u(x) is not satisfied for all x. Does this affect the validity of your conclusion?

7

6

x

y y ! u(x)

y ! l(x)

y ! f (x)

FIGURE 8

4. Determine lim x→0

f (x) assuming that cos x ≤ f (x) ≤ 1.

5. State whether the inequality provides sufficient information to de- termine lim

x→1 f (x), and if so, find the limit.

(a) 4x − 5 ≤ f (x) ≤ x2 (b) 2x − 1 ≤ f (x) ≤ x2 (c) 4x − x2 ≤ f (x) ≤ x2 + 2 6. Plot the graphs of u(x) = 1 +

∣∣x − π2 ∣∣ and l(x) = sin x on

the same set of axes. What can you say about lim x→ π2

f (x) if f is squeezed by l and u at x = π2 ? In Exercises 7–16, evaluate using the Squeeze Theorem.

7. lim x→0

x2 cos 1 x

8. lim x→0

x sin 1

x2

9. lim x→1

(x − 1) sin π x − 1 10. limx→3(x

2 − 9) x − 3|x − 3|

11. lim t→0

(2t − 1) cos 1 t

12. lim x→0+

√ x ecos(π/x)

13. lim t→2

(t2 − 4) cos 1 t − 2 14. limx→0 tan x cos

( sin

1 x

)

15. lim θ→ π2

cos θ cos(tan θ) 16. lim t→0+

sin t tan−1(ln t)

In Exercises 17–26, evaluate using Theorem 2 as necessary.

17. lim x→0

tan x x

18. lim x→0

sin x sec x x

19. lim t→0

√ t3 + 9 sin t

t 20. lim

t→0 sin2 t

t

21. lim x→0

x2

sin2 x 22. lim

t→ π2

1 − cos t t

23. lim θ→0

sec θ − 1 θ

24. lim θ→0

1 − cos θ sin θ

25. lim t→ π4

sin t t

26. lim t→0

cos t − cos2 t t

27. Let L = lim x→0

sin 14x x

.

(a) Show, by letting θ = 14x, that L = lim θ→0

14 sin θ

θ .

(b) Compute L.

28. Evaluate lim h→0

sin 9h sin 7h

. Hint: sin 9h sin 7h

= (

9 7

) ( sin 9h

9h

) ( 7h

sin 7h

) .

In Exercises 29–48, evaluate the limit.

29. lim h→0

sin 9h h

30. lim h→0

sin 4h 4h

31. lim h→0

sin h 5h

32. lim x→ π6

x

sin 3x

33. lim θ→0

sin 7θ sin 3θ

34. lim x→0

tan 4x 9x

35. lim x→0

x csc 25x 36. lim t→0

tan 4t t sec t

37. lim h→0

sin 2h sin 3h

h2 38. lim

z→0 sin(z/3)

sin z

39. lim θ→0

sin(−3θ) sin(4θ)

40. lim x→0

tan 4x tan 9x

41. lim t→0

csc 8t csc 4t

42. lim x→0

sin 5x sin 2x sin 3x sin 5x

43. lim x→0

sin 3x sin 2x x sin 5x

44. lim h→0

1 − cos 2h h

45. lim h→0

sin(2h)(1 − cos h) h2

46. lim t→0

1 − cos 2t sin2 3t

47. lim θ→0

cos 2θ − cos θ θ

48. lim h→ π2

1 − cos 3h h

49. Calculate lim x→0−

sin x |x| .

50. Use the identity sin 3θ = 3 sin θ − 4 sin3 θ to evaluate the limit lim θ→0

sin 3θ − 3 sin θ θ3

.

51. Prove the following result:

lim θ→0

csc θ − cot θ = 0 9

52. Investigate lim h→0

1 − cos h h2

numerically (and graphically if

you have a graphing utility). Then prove that the limit is equal to 12 . Hint: See the proof of Theorem 2.

In Exercises 53–55, evaluate using the result of Exercise 52.

53. lim h→0

cos 3h − 1 h2

54. lim h→0

cos 3h − 1 cos 2h − 1

55. lim t→0

√ 1 − cos t

t

56. Use the Squeeze Theorem to prove that if lim x→c |f (x)| = 0, then

lim x→c f (x) = 0.

94 C H A P T E R 2 LIMITS

Further Insights and Challenges 57. Use the result of Exercise 52 to prove that for m ̸= 0,

lim x→0

cos mx − 1 x2

= −m 2

2

58. Using a diagram of the unit circle and the Pythagorean Theorem, show that

sin2 θ ≤ (1 − cos θ)2 + sin2 θ ≤ θ2

Conclude that sin2 θ ≤ 2(1 − cos θ) ≤ θ2 and use this to give an alter- native proof of Eq. (9) in Exercise 51. Then give an alternative proof of the result in Exercise 52.

59. (a) Investigate lim x→c

sin x − sin c x − c numerically for the five values

c = 0, π6 , π4 , π3 , π2 . (b) Can you guess the answer for general c? (c) Check numerically that your answer to (b) works for two other values of c.

2.7 Limits at Infinity

5 10 15 20 25

283.00 283.05 283.10 283.15 283.20 283.25

T (K)

t (years)

(N A

SA )

FIGURE 1 The earth’s average temperature (according to a simple climate model) in response to an 0.25% increase in solar radiation. According to this model, lim

t→∞ T (t) = 283.255.

So far we have considered limits as x approaches a number c. It is also important to consider limits where x approaches ∞ or −∞, which we refer to as limits at infinity. In applications, limits at infinity arise naturally when we describe the “long-term” behavior of a system as in Figure 1.

The notation x → ∞ indicates that x increases without bound, and x → −∞ indi- cates that x decreases (through negative values) without bound. We write

• lim x→∞ f (x) = L if f (x) gets closer and closer to L as x → ∞.

• lim x→−∞

f (x) = L if f (x) gets closer and closer to L as x → −∞.

As before, “closer and closer” means that |f (x) − L| becomes arbitrarily small. In either case, the line y = L is called a horizontal asymptote. We use the notation x → ±∞ to indicate that we are considering both infinite limits, as x → ∞ and as x → −∞.

Infinite limits describe the asymptotic behavior of a function, which is determined by the behavior of the graph as we move out indefinitely to the right or the left.

EXAMPLE 1 Discuss the asymptotic behavior in Figure 2.

Solution The function g approaches L = 7 as we move to the right and it approaches L = 3 as we move to left, so

lim x→∞ g(x) = 7, limx→−∞ g(x) = 3

Accordingly, the lines y = 7 and y = 3 are horizontal asymptotes of g.

A function may approach an infinite limit as x → ±∞. We write

−400 −200 200 400

3

7

x

y y = g(x)

FIGURE 2 The lines y = 7 and y = 3 are horizontal asymptotes of g.

lim x→∞ f (x) = ∞ or limx→−∞ f (x) = ∞

if f (x) becomes arbitrarily large as x → ∞ or −∞. Similar notation is used if f (x) approaches −∞ as x → ±∞. For example, we see in Figure 3(A) that

lim x→∞ e

x = ∞ and lim x→−∞

ex = 0

x

1

yy

x

y = ex y = sin x

(A) (B)

−1

1

2π−2π

FIGURE 3

S E C T I O N 2.7 Limits at Infinity 95

However, limits at infinity do not always exist. For example, f (x) = sin x oscillatesWhen a limit equals either ∞ or −∞, the limit does not exist, since the function is not approaching a finite number. But the fact that the limit is ∞ or −∞ is useful information nonetheless, and we record that fact, rather than just saying the limit does not exist.

indefinitely [Figure 3(B)], so

lim x→∞ sin x and limx→−∞

sin x

do not exist. The limits at infinity of the power functions f (x) = xn are easily determined. If

n > 0, then xn certainly increases without bound as x → ∞, so (Figure 4)

lim x→∞ x

n = ∞ and lim x→∞ x

−n = lim x→∞

1 xn

= 0

To describe the limits as x → −∞, assume that n is a whole number so that xn is defined for x < 0. If n is even, then xn becomes large and positive as x → −∞, and if n is odd, it becomes large and negative. In summary,CAUTION lim

x→−∞ x1/2 does not exist, since

the square root of a negative number is not a real number. THEOREM 1 For all n > 0,

lim x→∞ x

n = ∞ and lim x→∞ x

−n = lim x→∞

1 xn

= 0

If n is a positive whole number,

lim x→−∞

xn = {∞ if n is even −∞ if n is odd and limx→−∞ x

−n = lim x→−∞

1 xn

= 0

x

y

y = x2

y = x4

x

y

y = x3

y = x5

(B) n odd: lim xn = ∞, lim xn = −∞

x

y

y = 1x

x→∞ x→−∞(A) n even: lim x n = lim xn = ∞

x→∞ x→−∞ (C) lim = 0 = lim x→∞ x→−∞ 1 x

1 x

FIGURE 4

Note also that if p and q are positive integers, then

lim x→−∞

xp/q = {−∞ if q is odd

undefined if q is even

The Basic Limit Laws (Theorem 1 in Section 2.3) are valid for limits at infinity. For example, the Sum and Constant Multiple Laws yield

lim x→∞

( 3 − 4x−3 + 5x−5

) = lim

x→∞ 3 − 4 limx→∞ x −3 + 5 lim

x→∞ x −5

= 3 + 0 + 0 = 3

EXAMPLE 2 Calculate lim x→∞

20x2 − 3x 3x5 − 4x2 + 5 .

Solution It would be nice if we could apply the Quotient Law directly, but this law is valid only if the denominator has a finite, nonzero limit. Our limit has the indeterminate form ∞/∞ because

lim x→∞ (20x

2 − 3x) = ∞ and lim x→∞ (3x

5 − 4x2 + 5) = ∞

The way around this difficulty is to divide the numerator and denominator by x5 (the highest power of x in the denominator):

20x2 − 3x 3x5 − 4x2 + 5 =

x−5(20x2 − 3x) x−5(3x5 − 4x2 + 5) =

20x−3 − 3x−4 3 − 4x−3 + 5x−5

96 C H A P T E R 2 LIMITS

Now we can use the Quotient Law:

lim x→∞

20x2 − 3x 3x5 − 4x2 + 5 =

lim x→∞

( 20x−3 − 3x−4

)

lim x→∞

( 3 − 4x−3 + 5x−5

) = 0 3

= 0

In general, if

f (x) = anx n + an−1xn−1 + · · · + a0

bmxm + bm−1xm−1 + · · · + b0 where an ̸= 0 and bm ̸= 0, divide the numerator and denominator by xm:

f (x) = anx n−m + an−1xn−1−m + · · · + a0x−m bm + bm−1x−1 + · · · + b0x−m

= xn−m (

an + an−1x−1 + · · · + a0x−n bm + bm−1x−1 + · · · + b0x−m

)

The quotient in parentheses approaches the finite limit an/bm because

lim x→∞(an + an−1x

−1 + · · · + a0x−n) = an

lim x→∞(bm + bm−1x

−1 + · · · + b0x−m) = bm This also holds true for x → −∞, and therefore,

lim x→±∞

f (x) = lim x→±∞

xn−m lim x→±∞

an + an−1x−1 + · · · + a0x−n bm + bm−1x−1 + · · · + b0x−m

= an bm

lim x→±∞

xn−m

THEOREM 2 Limits at Infinity of a Rational Function The asymptotic behavior of a rational function depends only on the leading terms of its numerator and denominator. If an, bm ̸= 0, then

lim x→±∞

anx n + an−1xn−1 + · · · + a0

bmxm + bm−1xm−1 + · · · + b0 = an

bm lim

x→±∞ xn−m

Here are some examples:

• n = m: lim x→∞

3x4 − 7x + 9 7x4 − 4 =

3 7

lim x→∞ x

0 = 3 7

• n < m: lim x→∞

3x3 − 7x + 9 7x4 − 4 =

3 7

lim x→∞ x

−1 = 0

• n > m, n − m odd: lim x→−∞

3x8 − 7x + 9 7x3 − 4 =

3 7

lim x→−∞

x5 = −∞

• n > m, n − m even: lim x→−∞

3x7 − 7x + 9 7x3 − 4 =

3 7

lim x→−∞

x4 = ∞

Our method can be adapted to noninteger exponents and algebraic functions.

EXAMPLE 3 Calculate the limits (a) lim x→∞

3x7/2 + 7x−1/2 x2 − x1/2 (b) limx→∞

x2√ x3 + 1

.

Solution

(a) As before, divide the numerator and denominator by x2, which is the highest powerThe Quotient Law is valid if limx→c f (x) = ∞ and lim

x→c g(x) = L, where L ̸= 0:

lim x→c

f (x)

g(x) =

lim x→c

f (x)

lim x→c

g(x) =

{∞ if L > 0 −∞ if L < 0

A similar result holds when lim x→c

f (x) = −∞.

of x occurring in the denominator (this means multiply by x−2):

3x7/2 + 7x−1/2 x2 − x1/2 =

( x−2

x−2

) 3x7/2 + 7x−1/2

x2 − x1/2 = 3x3/2 + 7x−5/2

1 − x−3/2

lim x→∞

3x7/2 + 7x−1/2 x2 − x1/2 =

lim x→∞(3x

3/2 + 7x−5/2) lim

x→∞(1 − x −3/2)

= ∞ 1

= ∞

S E C T I O N 2.7 Limits at Infinity 97

(b) The key is to observe that the denominator of x2√

x3 + 1 “behaves” like x3/2:

√ x3 + 1 =

√ x3(1 + x−3) = x3/2

√ 1 + x−3 (for x > 0)

This suggests that we divide the numerator and denominator by x3/2:

x2√ x3 + 1

= (

x−3/2

x−3/2

) x2

x3/2 √

1 + x−3 = x

1/2 √

1 + x−3

Then apply the Quotient Law:

lim x→∞

x2√ x3 + 1

= lim x→∞

x1/2√ 1 + x−3

= lim

x→∞ x 1/2

lim x→∞

√ 1 + x−3

= ∞ 1

= ∞

EXAMPLE 4 Calculate the limits at infinity of f (x) = 12x + 25√ 16x2 + 100x + 500

.

Solution Divide the numerator and denominator by x (multiply by x−1), but notice the difference between x positive and x negative. For x > 0,

x−1 √

16x2 + 100x + 500 = √

x−2 √

16x2 + 100x + 500 = √

16 + 100 x

+ 500 x2

lim x→∞

12x + 25√ 16x2 + 100x + 500

= lim

x→∞

( 12 + 25x

)

lim x→∞

√ 16 + 100x + 500x2

= 12√ 16

= 3

However, if x < 0, then x = − √

x2 and

x−1 √

16x2 + 100x + 500 = − √

x−2 √

16x2 + 100x + 500 = − √

16 + 100 x

+ 500 x2

So the limit as x → −∞ is −3 instead of 3 (Figure 5): 25 50−25−50

−3

3

x

y

FIGURE 5 Graph of

f (x) = 12x + 25√ 16x2 + 100x + 500

.

lim x→−∞

12x + 25√ 16x2 + 100x + 500

= lim

x→−∞

( 12 + 25x

)

− lim x→−∞

√ 16 + 100x + 500x2

= 12 −

√ 16

= −3

2.7 SUMMARY

• Limits as infinity:

– lim x→∞ f (x) = L if |f (x) − L| becomes arbitrarily small as x increases without bound.

– lim x→−∞

f (x) = L if |f (x) − L| becomes arbitrarily small as x decreases without bound.

– lim x→∞ e

x = ∞ and lim x→−∞

ex = 0

• A horizontal line y = L is a horizontal asymptote if

lim x→∞ f (x) = L and/or limx→−∞ f (x) = L

A function can have 0, 1 or 2 horizontal asymptotes.

98 C H A P T E R 2 LIMITS

• If n > 0, then lim x→∞ x

n = ∞ and lim x→±∞

x−n = 0. If n > 0 is a whole number, then

lim x→−∞

xn = {∞ if n is even −∞ if n is odd and limx→−∞ x

−n = 0

• If f (x) = anx n + an−1xn−1 + · · · + a0

bmxm + bm−1xm−1 + · · · + b0 with an, bm ̸= 0, then

lim x→±∞

f (x) = an bm

lim x→±∞

xn−m

2.7 EXERCISES

Preliminary Questions 1. Assume that

lim x→∞ f (x) = L and limx→L g(x) = ∞

Which of the following statements are correct? (a) x = L is a vertical asymptote of g. (b) y = L is a horizontal asymptote of g. (c) x = L is a vertical asymptote of f . (d) y = L is a horizontal asymptote of f . 2. What are the following limits?

(a) lim x→∞ x

3 (b) lim x→−∞ x

3 (c) lim x→−∞ x

4

3. Sketch the graph of a function that approaches a limit as x → ∞ but does not approach a limit (either finite or infinite) as x → −∞.

4. What is the sign of a if f (x) = ax3 + x + 1 satisfies lim

x→−∞ f (x) = ∞?

5. What is the sign of the coefficient multiplying x7 if f is a polyno- mial of degree 7 such that lim

x→−∞ f (x) = ∞?

6. Explain why lim x→∞ sin

1 x exists but limx→0

sin 1x does not exist. What

is lim x→∞ sin

1 x ?

Exercises 1. What are the horizontal asymptotes of the function in Figure 6?

−20 20 40 60 80 x

1

2

y

y = f (x)

FIGURE 6

2. Sketch the graph of a function f that has both y = −1 and y = 5 as horizontal asymptotes.

3. Sketch the graph of a function f with a single horizontal asymptote y = 3.

4. Sketch the graphs of two functions f and g that have both y = −2 and y = 4 as horizontal asymptotes but lim

x→∞ f (x) ̸= limx→∞ g(x).

5. Investigate the asymptotic behavior of f (x) = x 3

x3 + x numerically and graphically: (a) Make a table of values of f (x) for x = ±50, ±100, ±500, ±1000. (b) Plot the graph of f . (c) What are the horizontal asymptotes of f ?

6. Investigate lim x→±∞

12x + 1 √

4x2 + 9 numerically and graphically:

(a) Make a table of values of f (x) = 12x + 1√ 4x2 + 9

for x = ±100, ±500, ±1000, ±10,000. (b) Plot the graph of f . (c) What are the horizontal asymptotes of f ?

In Exercises 7–16, evaluate the limit.

7. lim x→∞

x

x + 9 8. limx→∞ 3x2 + 20x

4x2 + 9

9. lim x→∞

3x2 + 20x 2x4 + 3x3 − 29 10. limx→∞

4 x + 5

11. lim x→∞

7x − 9 4x + 3 12. limx→∞

9x2 − 2 6 − 29x

13. lim x→−∞

7x2 − 9 4x + 3 14. limx→−∞

5x − 9 4x3 + 2x + 7

15. lim x→−∞

3x3 − 10 x + 4 16. limx→−∞

2x5 + 3x4 − 31x 8x4 − 31x2 + 12

In Exercises 17–22, find the horizontal asymptotes.

17. f (x) = 2x 2 − 3x

8x2 + 8 18. f (x) = 8x3 − x2

7 + 11x − 4x4

19. f (x) = √

36x2 + 7 9x + 4 20. f (x) =

√ 36x4 + 7 9x2 + 4

21. f (t) = e t

1 + e−t 22. f (t) = t1/3

(64t2 + 9)1/6 In Exercises 23–30, evaluate the limit.

23. lim x→∞

√ 9x4 + 3x + 2

4x3 + 1 24. limx→∞

√ x3 + 20x 10x − 2

S E C T I O N 2.7 Limits at Infinity 99

25. lim x→−∞

8x2 + 7x1/3 √

16x4 + 6 26. lim

x→−∞ 4x − 3

√ 25x2 + 4x

27. lim t→∞

t4/3 + t1/3 (4t2/3 + 1)2 28. limt→∞

t4/3 − 9t1/3 (8t4 + 2)1/3

29. lim x→−∞

|x| + x x + 1 30. limt→−∞

4 + 6e2t 5 − 9e3t

31. Determine lim x→∞ tan

−1 x. Explain geometrically.

32. Show that lim x→∞(

√ x2 + 1 − x) = 0. Hint: Observe that

√ x2 + 1 − x = 1√

x2 + 1 + x 33. According to the Michaelis–Menten equation (Figure 7), when an enzyme is combined with a substrate of concentration s (in millimo- lars), the reaction rate (in micromolars/min) is

R(s) = As K + s (A, K constants)

(a) Show, by computing lim s→∞ R(s), that A is the limiting reaction rate

as the concentration s approaches ∞. (b) Show that the reaction rate R(s) attains one-half of the limiting value A when s = K . (c) For a certain reaction, K = 1.25 mM and A = 0.1. For which con- centration s is R(s) equal to 75% of its limiting value?

Leonor Michaelis 1875–1949 (Rockefeller Archive Center)

Maud Menten 1879–1960 (University Archives, University of Pittsburgh)

FIGURE 7 Canadian-born biochemist Maud Menten is best known for her fundamental work on enzyme kinetics with German scientist Leonor Michaelis. She was also an accomplished painter, clarinetist, mountain climber, and master of numerous languages.

34. Suppose that the average temperature of Earth is T (t) = 283 + 3(1 − e−0.03t ) kelvins, where t is the number of years since 2000.

(a) Calculate the long-term average L = lim t→∞ T (t).

(b) At what time is T (t) within one-half a degree of its limiting value?

In Exercises 35–42, calculate the limit.

35. lim x→∞(

√ 4x4 + 9x − 2x2) 36. lim

x→∞( √

9x3 + x − x3/2)

37. lim x→∞(2

√ x −

√ x + 2) 38. lim

x→∞

( 1 x

− 1 x + 2

)

39. lim x→∞ (ln(3x + 1) − ln(2x + 1))

40. lim x→∞

( ln(

√ 5x2 + 2) − ln x

)

41. lim x→∞ tan

−1 (

x2 + 9 9 − x

)

42. lim x→∞ tan

−1 (

1 + x 1 − x

)

43. Let P(n) be the perimeter of an n-gon inscribed in a unit circle (Figure 8). (a) Explain, intuitively, why P(n) approaches 2π as n → ∞. (b) Show that P(n) = 2n sin

(π n

) .

(c) Combine (a) and (b) to conclude that lim n→∞

n π sin

(π n

) = 1.

(d) Use this to give another argument that lim θ→0

sin θ θ

= 1.

n = 6 n = 9 n = 12 FIGURE 8

44. Physicists have observed that Einstein’s theory of special relativ- ity reduces to Newtonian mechanics in the limit as c → ∞, where c is the speed of light. This is illustrated by a stone tossed up vertically from ground level so that it returns to Earth 1 s later. Using Newton’s Laws, we find that the stone’s maximum height is h = g/8 meters (g = 9.8 m/s2). According to special relativity, the stone’s mass depends on its velocity divided by c, and the maximum height is

h(c) = c √

c2/g2 + 1/4 − c2/g

Prove that lim c→∞ h(c) = g/8.

Further Insights and Challenges 45. Every limit as x → ∞ can be rewritten as a one-sided limit as t → 0+, where t = x−1. Setting g(t) = f (t−1), we have

lim x→∞ f (x) = limt→0+

g(t)

Show that lim x→∞

3x2 − x 2x2 + 5 = limt→0+

3 − t 2 + 5t2 , and evaluate using the

Quotient Law.

46. Rewrite the following as one-sided limits as in Exercise 45 and evaluate.

(a) lim x→∞

3 − 12x3 4x3 + 3x + 1 (b) limx→∞ e

1/x

(c) lim x→∞ x sin

1 x

(d) lim x→∞ ln

( x + 1 x − 1

)

47. Let G(b) = lim x→∞(1 + b

x)1/x for b ≥ 0. Investigate G(b) numer- ically and graphically for b = 0.2, 0.8, 2, 3, 5 (and additional values if necessary). Then make a conjecture for the value of G(b) as a function of b. Draw a graph of y = G(b). Does G appear to be continuous? We will evaluate G(b) using L’Hôpital’s Rule in Section 4.5 (see Ex- ercise 69 there).

100 C H A P T E R 2 LIMITS

2.8 Intermediate Value Theorem The Intermediate Value Theorem (IVT) says, roughly speaking, that a continuous func- tion cannot skip values. Consider a plane that takes off and climbs from 0 to 10,000 m in 20 min. The plane must reach every altitude between 0 and 10,000 m during this 20-min interval. Thus, at some moment, the plane’s altitude must have been exactly 8371 m. Of course, this assumes that the plane’s motion is continuous, so its altitude cannot jump abruptly from, say, 8000 to 9000 m.

To state this conclusion formally, let A(t) be the plane’s altitude at time t . The IVT asserts that for every altitude M between 0 and 10,000, there is a time t0 between 0 and 20 such that A(t0) = M . In other words, the graph of A must intersect the horizontal line y = M [Figure 1(A)].

−1 1 2 3

1

2

3

4

20t0 Time (min)

(A) Altitude of plane A(t) (B) Graph of f (x) = ⌊x⌋

Altitude (m)

5000

10,000

M

x

y

FIGURE 1

By contrast, a discontinuous function can skip values. The greatest integer function f (x) = ⌊x⌋ in Figure 1(B) satisfies ⌊1⌋ = 1 and ⌊2⌋ = 2, but it does not take on the value 1.5 (or any other value between 1 and 2).

THEOREM 1 Intermediate Value Theorem If f is continuous on a closed interval [a, b], then for every value M between f (a) and f (b), there exists at least one value c ∈ (a, b) such that f (c) = M .

Graphically, as in Figure 2, the result appears obvious. For a continuous function, a bc

f (c) = M

x

y

f (a)

f (b) 0

FIGURE 2 For every height M , there is a c in (a, b) such that f (c) = M .

every horizontal line at height M between f (a) and f (b) is forced to hit the continuous graph and therefore there must be at least one value c in (a, b) such that f (c) = M . The proof appears in Appendix B.

EXAMPLE 1 Prove that the equation sin x = 0.3 has at least one solution.

0.3

1 y = sin x

y = 0.3

c π 2

x

y

FIGURE 3

Solution We may apply the IVT because f (x) = sin x is continuous. We choose an interval where we suspect that a solution exists. The desired value 0.3 lies between the two values of the function

sin 0 = 0 and sin π 2

= 1

so the interval [ 0, π2

] will work (Figure 3). The IVT tells us that sin x = 0.3 has at least

one solution in ( 0, π2

) .

The IVT can be used to show the existence of zeros of functions. If f is continuous

A zero or root of a function is a value c such that f (c) = 0. Sometimes the word “root” is reserved to refer specifically to the zero of a polynomial.

and takes on both positive and negative values—say, f (a) < 0 and f (b) > 0—then the IVT guarantees that f (c) = 0 for some c between a and b. This is extremely useful in a case where we cannot explicitly solve for the zero but would like to know that it is there.

COROLLARY 2 Existence of Zeros If f is continuous on [a, b] and if f (a) and f (b) are nonzero and have opposite signs, then f has a zero in (a, b).

We can locate zeros of functions to arbitrary accuracy using the Bisection Method. The idea is to find an interval [a, b] such that the function has opposite signs at the end points. Then Corollary 2 tells us that there is a zero on this interval. To find its location

S E C T I O N 2.8 Intermediate Value Theorem 101

more precisely, we cut the interval into two equal subintervals. Then, check the signs at the end points of each of these intervals to see which one Corollary 2 tells us has a zero. (But keep in mind that there may be more than one zero, so both could contain a zero). Next we repeat the process on this smaller interval. Eventually, we narrow down on the zero. This is illustrated in the next example.

EXAMPLE 2 The Bisection Method Show that f (x) = cos2 x − 2 sin x4 has a zero in (0, 2). Then locate the zero more accurately using the Bisection Method.

1 2

1

−1

1

0 2

x

y

f (x) > 0

f (x) < 0

Zero of f

3 4

FIGURE 4 Graph of f (x) = cos2 x − 2 sin x4 .

Solution Using a calculator, we find that f (0) and f (2) have opposite signs:

f (0) = 1 > 0, f (2) ≈ −0.786 < 0

Corollary 2 guarantees that f (x) = 0 has a solution in (0, 2) (Figure 4). To locate a zero more accurately, divide [0, 2] into two intervals [0, 1] and [1, 2]. At

Computer algebra systems have built-in commands for finding roots of a function or solving an equation numerically. These systems use a variety of methods, including more sophisticated versions of the Bisection Method. Notice that to use the Bisection Method, we must first find an interval containing a root.

least one of these intervals must contain a zero of f . To determine which, evaluate f at the midpoint m = 1. A calculator gives f (1) ≈ −0.203 < 0, and since f (0) = 1, we see that

f (x) takes on opposite signs at the endpoints of [0, 1]

Therefore, (0, 1) must contain a zero. We discard [1, 2] because both f (1) and f (2) are negative.

The Bisection Method consists of continuing this process until we narrow down the location of the zero to any desired accuracy. In the following table, the process is carried out three times:

Interval Midpoint of interval Function values Conclusion

[0, 1] 12 f ( 1

2 )

≈ 0.521 f (1) ≈ −0.203

Zero lies in ( 1

2 , 1 )

[ 1 2 , 1

] 3 4 f

( 3 4 )

≈ 0.163 f (1) ≈ −0.203

Zero lies in ( 3

4 , 1 )

[ 3 4 , 1

] 7 8 f

( 7 8 )

≈ −0.0231 f

( 3 4 )

≈ 0.163 Zero lies in

( 3 4 ,

7 8 )

We conclude that f has a zero c satisfying 0.75 < c < 0.875.

CONCEPTUAL INSIGHT The IVT seems to state the obvious, namely that a continuous function cannot skip values. Yet its proof (given in Appendix B) is subtle because it depends on the completeness property of real numbers. To highlight the subtlety, observe that the IVT is false for functions defined only on the rational numbers. For example, f (x) = x2 is continuous, but it does not have the intermediate value property if we restrict its domain to the rational numbers. Indeed, f (0) = 0 and f (2) = 4, but f (c) = 2 has no solution for c rational. The solution c =

√ 2 is “missing” from the set

of rational numbers because it is irrational. No doubt the IVT was always regarded as obvious, but it was not possible to give a correct proof until the completeness property was clarified in the second half of the nineteenth century.

2.8 SUMMARY

• The Intermediate Value Theorem (IVT) says that a continuous function cannot skip values.

• More precisely, if f is continuous on [a, b] with f (a) ̸= f (b), and if M is a number between f (a) and f (b), then f (c) = M for some c ∈ (a, b).

102 C H A P T E R 2 LIMITS

• Existence of zeros: If f is continuous on [a, b] and if f (a) and f (b) take opposite signs (one is positive and the other negative), then f (c) = 0 for some c ∈ (a, b).

• Bisection Method: Assume f is continuous and that f (a) and f (b) have opposite signs, so that f has a zero in (a, b). Then f has a zero in [a, m] or [m, b], where m = (a + b)/2 is the midpoint of [a, b]. A zero lies in (a, m) if f (a) and f (m) have opposite signs and in (m, b) if f (m) and f (b) have opposite signs. Continuing the process, we can locate a zero with arbitrary accuracy.

2.8 EXERCISES

Preliminary Questions 1. Prove that f (x) = x2 takes on the value 0.5 in the interval [0, 1].

2. The temperature in Vancouver was 8◦C at 6 am and rose to 20◦C at noon. Which assumption about temperature allows us to conclude that the temperature was 15◦C at some moment of time between 6 am and noon?

3. What is the graphical interpretation of the IVT?

4. Show that the following statement is false by drawing a graph that provides a counterexample:

If f is continuous and has a root in [a, b], then f (a) and f (b) have opposite signs.

5. Assume that f is continuous on [1, 5] and that f (1) = 20, f (5) = 100. Determine whether each of the following statements is always true, never true, or sometimes true. (a) f (c) = 3 has a solution with c ∈ [1, 5]. (b) f (c) = 75 has a solution with c ∈ [1, 5]. (c) f (c) = 50 has no solution with c ∈ [1, 5]. (d) f (c) = 30 has exactly one solution with c ∈ [1, 5].

Exercises 1. Use the IVT to show that f (x) = x3 + x takes on the value 9 for

some x in [1, 2].

2. Show that g(t) = t t + 1 takes on the value 0.499 for some t in

[0, 1].

3. Show that g(t) = t2 tan t takes on the value 12 for some t in [ 0, π4

] .

4. Show that f (x) = x 2

x7 + 1 takes on the value 0.4.

5. Show that cos x = x has a solution in the interval [0, 1]. Hint: Show that f (x) = x − cos x has a zero in [0, 1].

6. Use the IVT to find an interval of length 12 containing a root of f (x) = x3 + 2x + 1.

In Exercises 7–16, prove using the IVT.

7. √

c + √

c + 2 = 3 has a solution.

8. For all integers n, sin nx = cos x for some x ∈ [0, π ].

9. √

2 exists. Hint: Consider f (x) = x2.

10. A positive number c has an nth root for all positive integers n.

11. For all positive integers k, cos x = xk has a solution.

12. 2x = bx has a solution if b > 2.

13. 2x + 3x = 4x has a solution.

14. cos x = cos−1 x has a solution in (0, 1).

15. ex + ln x = 0 has a solution.

16. tan−1 x = cos−1 x has a solution.

17. Use the Intermediate Value Theorem to show that the equation x6 − 8x4 + 10x2 − 1 = 0 has at least six distinct solutions.

In Exercises 18–20, determine whether or not the IVT applies to show that the given function takes on all values between f (a) and f (b) for x ∈ (a, b). If it does not apply, determine any values between f (a) and f (b) that the function does not take on for x ∈ (a, b). 18.

f (x) = { x for x < 0 x2 for x ≥ 0

for the interval [−1, 1]. 19.

g(x) = {−x for x < 0 x3 + 1 for x ≥ 0

for the interval [−1, 1]. 20.

j (x) =

⎧ ⎨

⎩

−x2 for x < 0 1 for x = 0 x for x > 0

for the interval [−2, 2]. 21. Carry out three steps of the Bisection Method for f (x) = 2x − x3 as follows: (a) Show that f has a zero in [1, 1.5]. (b) Show that f has a zero in [1.25, 1.5]. (c) Determine whether [1.25, 1.375] or [1.375, 1.5] contains a zero. 22. Figure 5 shows that f (x) = x3 − 8x − 1 has a root in the inter- val [2.75, 3]. Apply the Bisection Method twice to find an interval of length 116 containing this root.

1 2 3 x

y

FIGURE 5 Graph of y = x3 − 8x − 1.

S E C T I O N 2.9 The Formal Definition of a Limit 103

23. Find an interval of length 14 in [1, 2] containing a root of the equa- tion x7 + 3x − 10 = 0.

24. Show that tan3 θ − 8 tan2 θ + 17 tan θ − 8 = 0 has a root in [0.5, 0.6]. Apply the Bisection Method twice to find an interval of length 0.025 containing this root.

In Exercises 25–28, draw the graph of a function f on [0, 4] with the given property.

25. Jump discontinuity at x = 2 and does not satisfy the conclusion of the IVT

26. Jump discontinuity at x = 2 and satisfies the conclusion of the IVT on [0, 4]

27. Infinite one-sided limits at x = 2 and does not satisfy the conclu- sion of the IVT

28. Infinite one-sided limits at x = 2 and satisfies the conclusion of the IVT on [0, 4]

29. Can Corollary 2 be applied to f (x) = x−1 on [−1, 1]? Does f have any roots?

30. x6 − 8x4 + 10x2 − 1 = 0 has at least six distinct solutions.

Further Insights and Challenges 31. Take any map and draw a circle on it anywhere (Figure 6). Prove that at any moment in time there exists a pair of diametrically opposite points A and B on that circle corresponding to locations where the tem- peratures at that moment are equal. Hint: Let θ be an angular coordinate along the circle and let f (θ) be the difference in temperatures at the locations corresponding to θ and θ + π .

θ

B

A

FIGURE 6 f (θ) is the difference between the temperatures at A and B.

32. Show that if f is continuous and 0 ≤ f (x) ≤ 1 for 0 ≤ x ≤ 1, then f (c) = c for some c in [0, 1] (Figure 7).

1

1

y = f (x)

y = x

c x

y

FIGURE 7 A function satisfying 0 ≤ f (x) ≤ 1 for 0 ≤ x ≤ 1.

33. Use the IVT to show that if f is continuous and one-to-one on an interval [a, b], then f is either an increasing or a decreasing function.

34. Ham Sandwich Theorem Figure 8(A) shows a slice of ham. Prove that for any angle θ (0 ≤ θ ≤ π ), it is possible to cut the slice in half with a cut of incline θ . Hint: The lines of inclination θ are given by the equations y = (tan θ)x + b, where b varies from −∞ to ∞. Each such line divides the slice into two pieces (one of which may be empty). Let A(b) be the amount of ham to the left of the line minus the amount to the right, and let A be the total area of the ham. Show that A(b) = −A if b is sufficiently large and A(b) = A if b is sufficiently negative. Then use the IVT. This works if θ ̸= 0 or π2 . If θ = 0, define A(b) as the amount of ham above the line y = b minus the amount below. How can you modify the argument to work when θ = π2 (in which case tan θ = ∞)?

35. Figure 8(B) shows a slice of ham on a piece of bread. Prove that it is possible to slice this open-faced sandwich so that each part has equal amounts of ham and bread. Hint: By Exercise 34, for all 0 ≤ θ ≤ π there is a line L(θ) of incline θ (which we assume is unique) that divides the ham into two equal pieces. Let B(θ) denote the amount of bread to the left of (or above) L(θ) minus the amount to the right (or below). Notice that L(π) and L(0) are the same line, but B(π) = −B(0) since left and right get interchanged as the angle moves from 0 to π . Assume that B is continuous and apply the IVT. (By a further extension of this argument, one can prove the full “Ham Sandwich Theorem,” which states that if you allow the knife to cut at a slant, then it is possible to cut a sandwich consisting of a slice of ham and two slices of bread so that all three layers are divided in half.)

L (0) = L(π)

L(θ)L ( )π2

θ

(A) Cutting a slice of ham at an angle θ

(B) A slice of ham on top of a slice of bread

x

y

x

y

FIGURE 8

2.9 The Formal Definition of a Limit In this section, we reexamine the definition of a limit in order to state it in a more rigorous and precise fashion. Why is this necessary? In Section 2.2, we defined limits by saying that lim

x→c f (x) = L if |f (x) − L| becomes arbitrarily small when x is sufficiently close (but not equal) to c. The problem with this definition lies in the phrases “arbitrarily small” and “sufficiently close.” We must find a way to specify just how close is sufficiently close.

104 C H A P T E R 2 LIMITS

The Size of the Gap Recall that the distance from f (x) to L is |f (x) − L|. It is convenient to refer to theA “rigorous proof” in mathematics is a

proof based on a complete chain of logic without any gaps or ambiguity. The formal definition of a limit is a key ingredient of rigorous proofs in calculus. A few such proofs are included in Appendix D. More complete developments can be found in textbooks on the branch of mathematics called “analysis.”

quantity |f (x) − L| as the gap between the value f (x) and the limit L. Let’s reexamine the trigonometric limit

lim x→0

sin x x

= 1 1

In this example, f (x) = sin x x

and L = 1, so Eq. (1) tells us that the gap |f (x) − 1| gets arbitrarily small when x is sufficiently close, but not equal, to 0 [Figure 1(A)].

The gap | f (x) − 1| is less than 0.004.

The gap | f (x) − 1| is less than 0.2.

The interval |x | < 1 2π 3π

1

−π−2π 00

0.8

1

1

−1 1−3π π

(A () B)

The interval |x | < 0.15 0.150−0.15

(C)

0.996

Enlarged view of graph near x = 0

x

y

xx

y

y

0.5

FIGURE 1 Graphs of y = sin x x

. To shrink the gap from 0.2 to 0.004, we require that x lie within 0.15 of 0.

Suppose we want the gap |f (x) − 1| to be less than 0.2. How close to 0 must x be? Figure 1(B) shows that f (x) lies within 0.2 of L = 1 for all values of x in the interval [−1, 1]. In other words, the following statement is true:

If 0 < |x| < 1, then ∣∣∣∣ sin x

x − 1

∣∣∣∣ < 0.2

If we insist instead that the gap be smaller than 0.004, we can check by zooming in on the graph, as in Figure 1(C), that

If 0 < |x| < 0.15, then ∣∣∣∣ sin x

x − 1

∣∣∣∣ < 0.004

It would seem that this process can be continued: By zooming in on the graph, we can find a small interval around c = 0, where the gap |f (x) − 1| is smaller than any prescribed positive number.

To express this in a precise fashion, we follow time-honored tradition in using the Greek letters ϵ (epsilon) and δ (delta) to denote small numbers specifying the sizes of the gap and the quantity |x − c|, respectively. In our case, c = 0 and |x − c| = |x|. The precise meaning of Eq. (1) is that for every choice of ϵ > 0, there exists some δ (depending on ϵ) such that

If 0 < |x| < δ, then ∣∣∣∣ sin x

x − 1

∣∣∣∣ < ϵ

The number δ pins down just how close is “sufficiently close” for a given ϵ. With this motivation, we are ready to state the formal definition of the limit.

S E C T I O N 2.9 The Formal Definition of a Limit 105

FORMAL DEFINITION OF A LIMIT Suppose that f (x) is defined for all x in an open interval containing c (but not necessarily at x = c). Then

lim x→c f (x) = L

if for all ϵ > 0, there exists δ > 0 such that

if 0 < |x − c| < δ, then |f (x) − L| < ϵ

The condition 0 < |x − c| < δ in this definition excludes x = c. In other words, the limit

The formal definition of a limit is often called the ϵ-δ definition. The tradition of using the symbols ϵ and δ originated in the writings of Augustin-Louis Cauchy on calculus and analysis in the 1820s.

If the symbols ϵ and δ seem to make this definition too abstract, keep in mind that we can take ϵ = 10−n and δ = 10−m. Thus, lim

x→c f (x) = L if, for any n, there

exist m > 0 such that if 0 < |x − c| < 10−m, then |f (x) − L| < 10−n.

depends only on values of f (x) near c but not on f (c) itself. As we have seen in previous sections, the limit may exist even when f (c) is not defined.

EXAMPLE 1 Let f (x) = 8x + 3. (a) Prove that lim

x→3 f (x) = 27 using the formal definition of the limit.

(b) Find values of δ that work for ϵ = 0.2 and 0.001. Solution

(a) We break the proof into two steps.

Step 1. Relate the gap to |x − c|. We must find a relation between two absolute values: |f (x) − L| for L = 27 and |x − c| for c = 3. Observe that

|f (x) − 27|︸ ︷︷ ︸ Size of gap

= |(8x + 3) − 27| = |8x − 24| = 8|x − 3|

Thus, the gap is 8 times as large as |x − 3|. Step 2. Choose δ (in terms of ϵ).

We can now see how to make the gap small: If |x − 3| < ϵ8 , then the gap is less than 8

( ϵ 8

) = ϵ. Therefore, for any ϵ > 0, we choose δ = ϵ8 . With this choice, the following

statement holds:

If 0 < |x − 3| < δ, then |f (x) − 27| < ϵ, where δ = ϵ 8

Since we have specified δ for all ϵ > 0, we have fulfilled the requirements of the formal definition, thus proving rigorously that lim

x→3 (8x + 3) = 27.

(b) For the particular choice ϵ = 0.2, we may take δ = ϵ8 = 0.28 = 0.025:

If 0 < |x − 3| < 0.025, then |f (x) − 27| < 0.2

This statement is illustrated in Figure 2. But note that any positive δ smaller than 0.025

27

27.2

26.8

3 3.0252.975

y = 8x + 3

x

y

FIGURE 2 To make the gap less than 0.2, we may take δ = 0.025 (not drawn to scale).

will also work. For example, the following statement is also true, although it places an unnecessary restriction on x:

If 0 < |x − 3| < 0.019, then |f (x) − 27| < 0.2

Similarly, to make the gap less than ϵ = 0.001, we may take

δ = ϵ 8

= 0.001 8

= 0.000125

The difficulty in applying the limit definition lies in trying to relate |f (x) − L| to |x − c|. The next two examples illustrate how this can be done in special cases.

106 C H A P T E R 2 LIMITS

EXAMPLE 2 Prove that lim x→2

x2 = 4.

Solution Let f (x) = x2. Step 1. Relate the gap to |x − c|.

In this case, we must relate the gap |f (x) − 4| = |x2 − 4| to the quantity |x − 2| (Figure 3). This is more difficult than in the previous example because the gap is not a constant multiple of |x − 2|. To proceed, consider the factorization

|x2 − 4| = |x + 2| |x − 2| Because we are going to require that |x − 2| be small, we may as well assume from the outset that |x − 2| < 1, which means that 1 < x < 3. In this case, |x + 2| is less than 5 and the gap satisfies

If |x − 2| < 1, then |x2 − 4| = |x + 2| |x − 2| < 5 |x − 2| 2

22 − δ

4 − ϵ

4 + ϵ

2 + δ

4 Function values

within ϵ of 4

x

y

FIGURE 3 Graph of f (x) = x2. We may choose δ so that f (x) lies within ϵ of 4 for all x in [2 − δ, 2 + δ].

Step 2. Choose δ (in terms of ϵ). We see from Eq. (2) that if |x − 2| is smaller than both ϵ5 and 1, then the gap satisfies

|x2 − 4| < 5|x − 2| < 5 (ϵ

5

) = ϵ

Therefore, the following statement holds for all ϵ > 0:

If 0 < |x − 2| < δ, then |x2 − 4| < ϵ, where δ is the smaller of ϵ5 and 1

We have specified δ for all ϵ > 0, so we have fulfilled the requirements of the formal limit definition, thus proving that lim

x→2 x2 = 4.

EXAMPLE 3 Prove that lim x→3

1 x

= 1 3

.

Solution

Step 1. Relate the gap to |x − c|. The gap is equal to

∣∣∣∣ 1 x

− 1 3

∣∣∣∣ = ∣∣∣∣ 3 − x

3x

∣∣∣∣ = |x − 3| ∣∣∣∣

1 3x

∣∣∣∣

Because we are going to require that |x − 3| be small, we may as well assume from the outset that |x − 3| < 1, or equivalently, 2 < x < 4. Now observe that if x > 2, then 3x > 6 and 13x <

1 6 , so the following inequality is valid if |x − 3| < 1:

REMINDER If a > b > 0, then 1 a

< 1 b .

Thus, if 3x > 6, then 13x < 1 6 .

∣∣∣∣f (x) − 1 3

∣∣∣∣ = ∣∣∣∣ 3 − x

3x

∣∣∣∣ = ∣∣∣∣

1 3x

∣∣∣∣ |x − 3| < 1 6

|x − 3| 3

Step 2. Choose δ (in terms of ϵ). By Eq. (3), if |x − 3| < 1 and |x − 3| < 6ϵ, then

∣∣∣∣ 1 x

− 1 3

∣∣∣∣ < 1 6

|x − 3| < 1 6 (6ϵ) = ϵ

S E C T I O N 2.9 The Formal Definition of a Limit 107

Therefore, given any ϵ > 0, we let δ be the smaller of the numbers 6ϵ and 1. Then we have:

If 0 < |x − 3| < δ, then ∣∣∣∣ 1 x

− 1 3

∣∣∣∣ < ϵ, where δ is the smaller of 6ϵ and 1

Again, we have fulfilled the requirements of the formal limit definition, thus proving rigorously that lim

x→3 1 x = 13 .

GRAPHICAL INSIGHT Keep the graphical interpretation of limits in mind. In Figure 4(A), f (x) approaches L as x → c because for any ϵ > 0, we can make the gap less than ϵ by taking δ sufficiently small. By contrast, the function in Figure 4(B) has a jump discontinuity at x = c. The gap cannot be made small, no matter how small δ is taken. Therefore, the limit does not exist.

c − δ c + δ

L

L + ϵ

L − ϵ

The function is continuous at x = c. By taking δ sufficiently small, we can make the gap smaller than ϵ.

(A) (B) The function is not continuous at x = c. The gap is always larger than (b − a)/2, no matter how small δ is.

Width 2ϵ

Width at least b − a

c c − δ c + δc

b

a

x x

y y

FIGURE 4

Proving Limit Theorems In practice, the formal definition of the limit is rarely used to evaluate limits. Most limits are evaluated using the Basic Limit Laws or other techniques such as the Squeeze Theorem. However, the formal definition allows us to prove these laws in a rigorous fashion and thereby ensure that calculus is built on a solid foundation. We illustrate by proving the Sum Law. Other proofs are given in Appendix D.

Proof of the Sum Law Assume that

lim x→c f (x) = L and limx→c g(x) = M

We must prove that lim x→c(f (x) + g(x)) = L + M .

Apply the Triangle Inequality (see margin) with a = f (x) − L and b = g(x) − M:REMINDER The Triangle Inequality [Eq. (1) in Section 1.1] states

|a + b| ≤ |a| + |b|

for all a and b.

|(f (x) + g(x)) − (L + M)| ≤ |f (x) − L| + |g(x) − M| 4 Each term on the right in (4) can be made small by the limit definition. More precisely, given ϵ > 0, we can choose δ such that |f (x) − L| < ϵ2 and |g(x) − M| < ϵ2 if 0 < |x − c| < δ (in principle, we might choose different δ’s for f and g, but we may then use the smaller of the two δ’s). Thus, Eq. (4) gives

If 0 < |x − c| < δ, then |f (x) + g(x) − (L + M)| < ϵ 2

+ ϵ 2

= ϵ 5

This proves that

lim x→c

( f (x) + g(x)

) = L + M = lim

x→c f (x) + limx→c g(x)

108 C H A P T E R 2 LIMITS

2.9 SUMMARY

• Informally speaking, the statement lim x→c f (x) = L means that the gap |f (x) − L| tends

to 0 as x approaches c. • The formal definition (called the ϵ-δ definition): lim

x→c f (x) = L if, for all ϵ > 0, there exists a δ > 0 such that

if 0 < |x − c| < δ, then |f (x) − L| < ϵ

2.9 EXERCISES

Preliminary Questions 1. Given that lim

x→0 cos x = 1, which of the following statements is true?

(a) If |cos x − 1| is very small, then x is close to 0.

(b) There is an ϵ > 0 such that if if 0 < |cos x − 1| < ϵ, then |x| < 10−5 .

(c) There is a δ > 0 such that if 0 < |x| < δ, then |cos x − 1| < 10−5.

(d) There is a δ > 0 such that if 0 < |x − 1| < δ, then |cos x| < 10−5.

2. Suppose it is known that for a given ϵ and δ, if 0 < |x − 3| < δ, then |f (x) − 2| < ϵ . Which of the following statements must also be true? (a) If 0 < |x − 3| < 2δ, then |f (x) − 2| < ϵ. (b) If 0 < |x − 3| < δ, then|f (x) − 2| < 2ϵ. (c) If 0 < |x − 3| < δ

2 , then |f (x) − 2| < ϵ

2 .

(d) If 0 < |x − 3| < δ 2

, then |f (x) − 2| < ϵ.

Exercises 1. Based on the information conveyed in Figure 5(A), find values of

L, ϵ, and δ > 0 such that the following statement holds: If |x| < δ, then |f (x) − L| < ϵ.

2. Based on the information conveyed in Figure 5(B), find values of c, L, ϵ, and δ > 0 such that the following statement holds: If 0 < |x − c| < δ, then |f (x) − L| < ϵ.

3 3.12.9

10 10.4

9.8

x

y

y = f (x) y = f (x)

(A) (B)

0.1−0.1

4

4.8

3.5

x

y

FIGURE 5

3. Consider lim x→4

f (x), where f (x) = 8x + 3. (a) Show that |f (x) − 35| = 8|x − 4|. (b) Show that for any ϵ > 0, if 0 < |x − 4| < δ, then |f (x) − 35| < ϵ, where δ = ϵ8 . Explain how this proves rigorously that limx→4 f (x) = 35.

4. Consider lim x→2

f (x), where f (x) = 4x − 1. (a) Show that if 0 < |x − 2| < δ, then |f (x) − 7| < 4δ. (b) Find a δ such that

If 0 < |x − 2| < δ, then |f (x) − 7| < 0.01 (c) Prove rigorously that lim

x→2 f (x) = 7.

5. Consider lim x→2

x2 = 4 (refer to Example 2).

(a) Show that if 0 < |x − 2| < 0.01, then |x2 − 4| < 0.05.

(b) Show that if 0 < |x − 2| < 0.0002, then |x2 − 4| < 0.0009. (c) Find a value of δ such that if 0 < |x − 2| < δ, then |x2 − 4| is less than 10−4.

6. With regard to the limit lim x→5

x2 = 25,

(a) show that if 4 < x < 6, then |x2 − 25| < 11|x − 5|. Hint: Write |x2 − 25| = |x + 5| · |x − 5|. (b) find a δ such that if 0 < |x − 5| < δ, then |x2 − 25| < 10−3. (c) give a rigorous proof of the limit by showing that if 0 < |x − 5| < δ, then |x2 − 25| < ϵ , where δ is the smaller of ϵ11 and 1.

7. Refer to Example 3 to find a value of δ > 0 such that

If 0 < |x − 3| < δ, then ∣∣∣∣ 1 x

− 1 3

∣∣∣∣ < 10 −4

8. Use Figure 6 to find a value of δ > 0 such that the following state- ment holds: If 0 < |x − 2| < δ, then

∣∣1/x2 − 14 ∣∣ < ϵ for ϵ = 0.03.

Then find a value of δ that works for ϵ = 0.01.

0.05

0.10

0.15

0.20

0.25

0.30

1.9 2.0 2.1

y

x

1 x2

y =

FIGURE 6

9. Plot f (x) = √

2x − 1 together with the horizontal lines y = 2.9 and y = 3.1. Use this plot to find a value of δ > 0 such that if 0 < |x − 5| < δ, then |

√ 2x − 1 − 3| < 0.1.

S E C T I O N 2.9 The Formal Definition of a Limit 109

10. Plot f (x) = tan x together with the horizontal lines y = 0.99 and y = 1.01. Use this plot to find a value of δ > 0 such that if 0 <

∣∣x − π4 ∣∣ < δ, then |tan x − 1| < 0.01.

11. The number e has the following property: lim x→0

ex − 1 x

= 1.

Use a plot of f (x) = e x − 1 x

to find a value of δ > 0 such that if

0 < |x − 1| < δ, then |f (x) − 1| < 0.01.

12. Let f (x) = 4 x2 + 1 and ϵ = 0.5. Using a plot of f , find

a value of δ > 0 such that if 0 < ∣∣∣x − 12

∣∣∣ < δ, then ∣∣∣f (x) − 165

∣∣∣ < ϵ. Repeat for ϵ = 0.2 and 0.1.

13. Consider lim x→2

1 x

.

(a) Show that if |x − 2| < 1, then ∣∣∣∣ 1 x

− 1 2

∣∣∣∣ < 1 2 |x − 2|

(b) Let δ be the smaller of 1 and 2ϵ. Prove the following:

If 0 < |x − 2| < δ, then ∣∣∣∣ 1 x

− 1 2

∣∣∣∣ < ϵ

(c) Find a δ > 0 such that if 0 < |x − 2| < δ, then ∣∣∣ 1x − 12

∣∣∣ < 0.01.

(d) Prove rigorously that lim x→2

1 x

= 1 2

.

14. Consider lim x→1

√ x + 3.

(a) Show that if |x − 1| < 4, then | √

x + 3 − 2| < 12 |x − 1|. Hint: Multiply the inequality by |

√ x + 3 + 2| and observe that |

√ x + 3 +

2| > 2. (b) Find δ > 0 such that if 0 < |x − 1| < δ, then |

√ x + 3 − 2| <10−4.

(c) Prove rigorously that the limit is equal to 2.

15. Let f (x) = sin x. Using a calculator, we find

f (π

4 − 0.1

) ≈ 0.633, f

(π 4

) ≈ 0.707, f

(π 4

+ 0.1 )

≈ 0.774

Use these values and the fact that f is increasing on [ 0, π2

] to justify

the statement

If 0 < ∣∣∣x − π

4

∣∣∣ < 0.1, then ∣∣∣f (x) − f

(π 4

)∣∣∣ < 0.08

Then draw a figure like Figure 3 to illustrate this statement.

16. Adapt the argument in Example 1 to prove rigorously that lim x→c(ax + b) = ac + b, where a, b, c are arbitrary.

17. Adapt the argument in Example 2 to prove rigorously that lim x→c x

2 = c2 for all c.

18. Adapt the argument in Example 3 to prove rigorously that lim x→c x

−1 = 1c for all c ̸= 0.

In Exercises 19–24, use the formal definition of the limit to prove the statement rigorously.

19. lim x→4

√ x = 2 20. lim

x→1 (3x2 + x) = 4

21. lim x→1

x3 = 1 22. lim x→0

(x2 + x3) = 0

23. lim x→2

x−2 = 1 4

24. lim x→0

x sin 1 x

= 0

25. Let f (x) = x|x| . Prove rigorously that limx→0 f (x) does not exist. Hint: Show that for any L, there always exists some x such that |x| < δ but |f (x) − L| ≥ 12 , no matter how small δ is taken.

26. Prove rigorously that lim x→0

|x| = 0.

27. Let f (x) = min(x, x2), where min(a, b) is the minimum of a and b. Prove rigorously that lim

x→1 f (x) = 1.

28. Prove rigorously that lim x→0

sin 1x does not exist.

29. First, use the identity

sin x + sin y = 2 sin (

x + y 2

) cos

( x − y

2

)

to verify the relation

sin(a + h) − sin a = h sin(h/2) h/2

cos (

a + h 2

) 6

Then use the inequality ∣∣∣∣ sin x

x

∣∣∣∣ ≤ 1 for x ̸= 0 to show that

|sin(a + h) − sin a| < |h| for all a. Finally, prove rigorously that lim x→a sin x = sin a.

Further Insights and Challenges 30. Uniqueness of the Limit Prove that a function converges to at most one limiting value. In other words, use the limit definition to prove that if lim

x→c f (x) = L1 and limx→c f (x) = L2, then L1 = L2.

In Exercises 31–33, prove the statement using the formal limit defini- tion.

31. The Constant Multiple Law [Theorem 1, part (ii) in Section 2.3].

32. The Squeeze Theorem (Theorem 1 in Section 2.6).

33. The Product Law [Theorem 1, part (iii) in Section 2.3]. Hint: Use the identity

f (x)g(x) − LM = (f (x) − L) g(x) + L(g(x) − M)

34. Let f (x) = 1 if x is rational and f (x) = 0 if x is irrational. Prove that lim

x→c f (x) does not exist for any c. Hint: There exist rational and irrational numbers arbitrarily close to any c.

35. Here is a function with strange continuity properties:

f (x) =

⎧ ⎪⎨

⎪⎩

1 q

if x is the rational number p/q in lowest terms

0 if x is an irrational number

(a) Show that f is discontinuous at c if c is rational. Hint: There exist irrational numbers arbitrarily close to c.

110 C H A P T E R 2 LIMITS

(b) Show that f is continuous at c if c is irrational. Hint: Let I be the interval {x : |x − c| < 1}. Show that for any Q > 0, I contains at most finitely many fractions p/q with q < Q. Conclude that there is a δ such that all fractions in {x : |x − c| < δ} have a denominator larger than Q.

36. Write a formal definition of the following:

lim x→∞ f (x) = L

37. Write a formal definition of the following:

lim x→a f (x) = ∞

CHAPTER REVIEW EXERCISES

1. The position of a particle at time t (s) is s(t) = √

t2 + 1 m. Com- pute its average velocity over [2, 5] and estimate its instantaneous ve- locity at t = 2.

2. The “wellhead” price p of natural gas in the United States (in dol- lars per 1000 ft3) on the first day of each month in 2012 is listed in the table below.

J F M A M J

2.89 2.46 2.25 1.89 1.94 2.54

J A S O N D

2.59 2.86 2.71 3.03 3.35 3.35

Compute the average rate of change of p (in dollars per 1000 ft3 per month) over the quarterly periods January–March, April–June, and July–September.

3. For a positive integer n, let P(n) be the number of partitions of n, that is, the number of ways of writing n as a sum of one or more positive integers. For example, P(4) = 5 since the number 4 can be partitioned in five different ways: 4, 3 + 1, 2 + 2, 2 + 1 + 1, and 1 + 1 + 1 + 1. Suppose P is a continuous function whose values at positive integers are known. Use Figure 1 to estimate the rate of change of P at n = 12.

n

P(n)

14121086420 0

40

80

120

160

FIGURE 1 Graph of P .

4. The average velocity v (m/s) of an oxygen molecule in the air at temperature T (◦C) is v = 25.7

√ 273.15 + T . What is the average

speed at T = 25◦ (room temperature)? Estimate the rate of change of average velocity with respect to temperature at T = 25◦. What are the units of this rate?

In Exercises 5–10, estimate the limit numerically to two decimal places or state that the limit does not exist.

5. lim x→0

1 − cos3(x) x2

6. lim x→1

x1/(x−1)

7. lim x→2

xx − 4 x2 − 4 8. limx→2

x − 2 ln(3x − 5)

9. lim x→1

( 7

1 − x7 − 3

1 − x3 )

10. lim x→2

3x − 9 5x − 25

In Exercises 11–50, evaluate the limit if it exists. If not, determine whether the one-sided limits exist (finite or infinite).

11. lim x→4

(3 + x1/2) 12. lim x→1

5 − x2 4x + 7

13. lim x→−2

4

x3 14. lim

x→−1 3x2 + 4x + 1

x + 1

15. lim t→9

√ t − 3

t − 9 16. limx→3

√ x + 1 − 2 x − 3

17. lim x→1

x3 − x x − 1 18. limh→0

2(a + h)2 − 2a2 h

19. lim t→9

t − 6√ t − 3 20. lims→0

1 − √

s2 + 1 s2

21. lim x→−1+

1 x + 1 22. limy→ 13

3y2 + 5y − 2 6y2 − 5y + 1

23. lim x→1

x3 − 2x x − 1 24. lima→b

a2 − 3ab + 2b2 a − b

25. lim x→0

e3x − ex ex − 1 26. limθ→0

sin 5θ θ

27. lim x→1.5

⌊ 1 x

⌋ 28. lim

θ→ π4 sec θ

29. lim z→−3

z + 3 z2 + 4z + 3 30. limx→1

x3 − ax2 + ax − 1 x − 1

31. lim x→b

x3 − b3 x − b 32. limx→0

sin 4x sin 3x

33. lim x→0

( 1

3x − 1

x(x + 3)

) 34. lim

θ→ 14 3tan(πθ)

35. lim x→0−

⌊x⌋ x

36. lim x→0+

⌊x⌋ x

37. lim θ→ π2

θ sec θ 38. lim y→2

ln (

sin π

y

)

39. lim θ→0

cos θ − 2 θ

40. lim x→4.3

1 x − ⌊x⌋

41. lim x→2−

x − 3 x − 2 42. limt→0

sin2 t

t3

43. lim x→1+

( 1√

x − 1 − 1√

x2 − 1

)

44. lim t→e

√ t(ln t − 1)

Chapter Review Exercises 111

45. lim x→ π2

tan x 46. lim t→0

cos 1 t

47. lim t→0+

√ t cos

1 t

48. lim x→5+

x2 − 24 x2 − 25

49. lim x→0

cos x − 1 sin x

50. lim θ→0

tan θ − sin θ sin3 θ

51. Find the left- and right-hand limits of the function f in Figure 2 at x = 0, 2, 4. State whether f is left- or right-continuous (or both) at these points.

x

y

1 3 52 4

1

2

FIGURE 2

52. Sketch the graph of a function f such that (a) lim

x→2− f (x) = 1, lim

x→2+ f (x) = 3

(b) lim x→4

f (x) exists but does not equal f (4).

53. Graph h and describe the discontinuity:

h(x) = {

ex for x ≤ 0 ln x for x > 0

Is h left- or right-continuous?

54. Sketch the graph of a function g such that

lim x→−3−

g(x) = ∞, lim x→−3+

g(x) = −∞, lim x→4

g(x) = ∞

55. Find the points of discontinuity of

g(x) =

⎧ ⎪⎨

⎪⎩

cos (πx

2

) for |x| < 1

|x − 1| for |x| ≥ 1

Determine the type of discontinuity and whether g is left- or right- continuous.

56. Show that f (x) = xesin x is continuous on its domain.

57. Find a constant b such that h is continuous at x = 2, where

h(x) = {

x + 1 for |x| < 2 b − x2 for |x| ≥ 2

With this choice of b, find all points of discontinuity.

In Exercises 58–63, find the horizontal asymptotes of the function by computing the limits at infinity.

58. f (x) = 9x 2 − 4

2x2 − x 59. f (x) = x2 − 3x4

x − 1

60. f (u) = 8u − 3√ 16u2 + 6

61. f (u) = 2u 2 − 1

√ 6 + u4

62. f (x) = 3x 2/3 + 9x3/7

7x4/5 − 4x−1/3 63. f (t) = t1/3 − t−1/3 (t − t−1)1/3

64. Calculate (a)–(d), assuming that

lim x→3

f (x) = 6, lim x→3

g(x) = 4

(a) lim x→3

(f (x) − 2g(x)) (b) lim x→3

x2f (x)

(c) lim x→3

f (x)

g(x) + x (d) limx→3(2g(x) 3 − g(x)3/2)

65. Assume that the following limits exist:

A = lim x→a f (x), B = limx→a g(x), L = limx→a

f (x)

g(x)

Prove that if L = 1, then A = B. Hint: You cannot use the Quotient Law if B = 0, so apply the Product Law to L and B instead. 66. Define g(t) = (1 + 21/t )−1 for t ̸= 0. How should g(0) be defined to make g left-continuous at t = 0?

67. In the notation of Exercise 65, give an example where L exists but neither A nor B exists.

68. True or false? (a) If lim

x→3 f (x) exists, then lim

x→3 f (x) = f (3).

(b) If lim x→0

f (x)

x = 1, then f (0) = 0.

(c) If lim x→−7

f (x) = 8, then lim x→−7

1 f (x)

= 1 8

.

(d) If lim x→5+

f (x) = 4 and lim x→5−

f (x) = 8, then lim x→5

f (x) = 6.

(e) If lim x→0

f (x)

x = 1, then lim

x→0 f (x) = 0.

(f) If lim x→5

f (x) = 2, then lim x→5

f (x)3 = 8.

69. Let f (x) = ⌊

1 x

⌋ , where ⌊x⌋ is the greatest integer func-

tion. Show that for x ̸= 0,

1 x

− 1 < ⌊

1 x

⌋ ≤ 1

x

Then use the Squeeze Theorem to prove that

lim x→0

x

⌊ 1 x

⌋ = 1

Hint: Treat the one-sided limits separately.

70. Let r1 and r2 be the roots of f (x) = ax2 − 2x + 20. Observe that f “approaches” the linear function L(x) = −2x + 20 as a → 0. Be- cause r = 10 is the unique root of L, we might expect one of the roots of f to approach 10 as a → 0 (Figure 3). Prove that the roots can be labeled so that lim

a→0 r1 = 10 and lim

a→0 r2 = ∞.

x

y

100 200

Root tends to ∞ as a → 0

Root near 10

300 400

200

−200 y = −2x + 20

a = 0.002 a = 0.008

FIGURE 3 Graphs of f (x) = ax2 − 2x + 20.

112 C H A P T E R 2 LIMITS

71. Use the IVT to prove that the curves y = x2 and y = cos x inter- sect.

72. Use the IVT to prove that f (x) = x3 − x 2 + 2

cos x + 2 has a root in the interval [0, 2].

73. Use the IVT to show that e−x2 = x has a solution on (0, 1). 74. Use the Bisection Method to locate a solution of x2 − 7 = 0 to two decimal places.

75. Give an example of a (discontinuous) function that does not satisfy the conclusion of the IVT on [−1, 1]. Then show that the function

f (x) =

⎧ ⎨

⎩ sin

1 x

x ̸= 0

0 x = 0

satisfies the conclusion of the IVT on every interval [−a, a].

76. Let f (x) = 1 x + 2 .

(a) Show that if |x − 2| < 1, then ∣∣∣f (x) − 14

∣∣∣ < |x − 2|

12 . Hint: Ob-

serve that if |x − 2| < 1, then |4(x + 2)| > 12. (b) Find δ > 0 such that if |x − 2| < δ, then

∣∣∣f (x) − 14 ∣∣∣ < 0.01.

(c) Prove rigorously that lim x→2

f (x) = 14 .

77. Plot the function f (x) = x1/3. Use the zoom feature to find a δ > 0 such that if |x − 8| < δ, then |x1/3 − 2| < 0.05.

78. Use the fact that f (x) = 2x is increasing to find a value of δ such that |2x − 8| < 0.001 if |x − 2| < δ. Hint: Find c1 and c2 such that 7.999 < f (c1) < f (c2) < 8.001.

79. Prove rigorously that lim x→−1

(4 + 8x) = −4.

80. Prove rigorously that lim x→3

(x2 − x) = 6.

The velocity at any given moment in free fall is

given by evaluating the derivative of the position

function at that time. (Germanskydiver/Shutterstock)

3 DIFFERENTIATION

D ifferential calculus is the study of the derivative, and differentiation is the process ofcomputing derivatives. What is a derivative? There are three equally important an- swers: A derivative is a rate of change, it is the slope of a tangent line, and (more formally), it is the limit of a difference quotient, as we will explain shortly. In this chapter, we explore all three facets of the derivative and develop the basic rules of differentiation. When you master these techniques, you will possess one of the most useful and flexible tools that mathematics has to offer.

3.1 Definition of the Derivative We begin with two questions: What is the precise definition of a tangent line? And how can we compute its slope? To answer these questions, let’s return to the relationship between tangent and secant lines first mentioned in Section 2.1.

The secant line through distinct points P = (a, f (a)) and Q = (x, f (x)) on the graph of a function f has slope [Figure 1(A)]

REMINDER A secant line is any line through two points on a curve or graph.

!f

!x = f (x) − f (a)

x − a

where

!f = f (x) − f (a) and !x = x − a

The expression f (x) − f (a)

x − a is called the difference quotient.

xa

(A)

a

(B)

Tangent

y

x

y

x

P = (a, f (a))

Q = (x, f (x))

!x = x − a

∆ f = f (x) − f (a) P = (a, f (a))

FIGURE 1 The secant line has slope !f/!x. Our goal is to compute the slope of the tangent line at (a, f (a)).

Now observe what happens as Q approaches P or, equivalently, as x approaches a. Figure 2 suggests that the secant lines get progressively closer to the tangent line. If we imagine Q moving toward P , then the secant line appears to rotate into the tangent line as in (D). Therefore, we may expect the slopes of the secant lines to approach the slope of the tangent line.

(C)(B)

P

xa

Q

a (D)

y

x

y

x xa

(A)

y

x xa

y

x

P

Q

P Q

P

x

Q

FIGURE 2 The secant lines approach the tangent line as Q approaches P .

113

114 C H A P T E R 3 DIFFERENTIATION

Based on this intuition, we define the derivative f ′(a) (which is read “f prime of a”) at a point x = a as the limit

f ′(a)︸ ︷︷ ︸ Slope of the tangent line

= lim x→a

f (x) − f (a) x − a︸ ︷︷ ︸

Limit of slopes of secant lines

There is another way of writing the difference quotient using a new variable h:

h = x − a

We have x = a + h and, for x ̸= a (Figure 3),P

Q

y

x a

h

f (a + h) − f (a)

x = a + h

FIGURE 3 The difference quotient can be written in terms of h.

f (x) − f (a) x − a =

f (a + h) − f (a) h

The variable h approaches 0 as x → a, so we can rewrite the derivative as

f ′(a) = lim h→0

f (a + h) − f (a) h

Each way of writing the derivative is useful. The version using h is often more convenient in computations.

DEFINITION The Derivative The derivative of f at a point a is the limit of the dif- ference quotients (if it exists):

f ′(a) = lim h→0

f (a + h) − f (a) h

1

When the limit exists, we say that f is differentiable at a. An equivalent definition of the derivative at a point a is

f ′(a) = lim x→a

f (x) − f (a) x − a 2

We can now define the tangent line in a precise way, as the line of slope f ′(a) through P = (a, f (a)).

DEFINITION Tangent Line Assume that f is differentiable at a. The tangent line to the graph of y = f (x) at P = (a, f (a)) is the line through P of slope f ′(a). The equation of the tangent line in point-slope form is

y − f (a) = f ′(a)(x − a) 3

REMINDER The equation of the line through P = (a, b) of slope m in point-slope form:

y − b = m(x − a)

EXAMPLE 1 Equation of a Tangent Line Find an equation of the tangent line to the graph of f (x) = x2 at x = 5.

y = 10x − 25

y = x2

3 5 7

(5, 25)

75

50

25

x

y

FIGURE 4 Tangent line to y = x2 at x = 5.

Solution First, we must compute f ′(5). We are free to use either Eq. (1) or Eq. (2). Using Eq. (2), we have

f ′(5) = lim x→5

f (x) − f (5) x − 5 = limx→5

x2 − 25 x − 5 = limx→5

(x − 5)(x + 5) x − 5

= lim x→5

(x + 5) = 10

Next, we apply Eq. (3) with a = 5. Because f (5) = 25, an equation of the tangent line is y − 25 = 10(x − 5), or in slope-intercept form, y = 10x − 25 (Figure 4).

S E C T I O N 3.1 Definition of the Derivative 115

If instead, we had used Eq. (1) to find f ′(5), we obtain the same slope:

f ′(5) = lim h→0

f (5 + h) − f (5) h

= lim h→0

(5 + h)2 − 52 h

= lim h→0

(25 + 10h + h2) − 25 h

= lim h→0

(10 + h) = 10

In the next two examples, we perform the differentiation (the process of computing the derivative) using Eq. (1). For clarity, we break up the computations into three steps.Isaac Newton referred to calculus as the

“method of fluxions” (from the Latin word for “flow”), but the term “differential calculus,” introduced in its Latin form “calculus differentialis” by Gottfried Wilhelm Leibniz, eventually won out and was adopted universally.

EXAMPLE 2 Compute f ′(3), where f (x) = x2 − 8x.

Solution Using Eq. (1), we write the difference quotient at a = 3 as

f (a + h) − f (a) h

= f (3 + h) − f (3) h

(h ̸= 0)

Step 1. Write out the numerator of the difference quotient.

f (3 + h) − f (3) = ( (3 + h)2 − 8(3 + h)

) −

( 32 − 8(3)

)

= ( (9 + 6h + h2) − (24 + 8h)

) − (9 − 24)

= h2 − 2h

Step 2. Divide by h and simplify.

f (3 + h) − f (3) h

= h 2 − 2h

h = h(h − 2)

h = h − 2

︸ ︷︷ ︸ Cancel h

Step 3. Compute the limit.

f ′(3) = lim h→0

f (3 + h) − f (3) h

= lim h→0

(h − 2) = −2

EXAMPLE 3 Sketch the graph of f (x) = 1 x

and the tangent line at x = 2.

(a) Based on the sketch, do you expect f ′(2) to be positive or negative? (b) Find an equation of the tangent line at x = 2.

Solution The graph and tangent line at x = 2 are shown in Figure 5.

54321

3

2

1

x

y

FIGURE 5 Graph of f (x) = 1x . The tangent line at x = 2 has equation y = − 14x + 1.

(a) We see that the tangent line has negative slope, so f ′(2) must be negative. (b) We compute f ′(2) in three steps as before.

Step 1. Write out the numerator of the difference quotient.

f (2 + h) − f (2) = 1 2 + h −

1 2

= 2 2(2 + h) −

2 + h 2(2 + h) = −

h

2(2 + h)

Step 2. Divide by h and simplify.

f (2 + h) − f (2) h

= 1 h

· (

− h 2(2 + h)

) = − 1

2(2 + h)

116 C H A P T E R 3 DIFFERENTIATION

Step 3. Compute the limit.

f ′(2) = lim h→0

f (2 + h) − f (2) h

= lim h→0

−1 2(2 + h) = −

1 4

The function value is f (2) = 12 , so the tangent line passes through ( 2, 12

) and has equation

y − 1 2

= −1 4 (x − 2)

In slope-intercept form, y = − 14x + 1.

The graph of a linear function f (x) = mx + b (where m and b are constants) is a line of slope m. The tangent line at any point coincides with the line itself (Figure 6), so we should expect that f ′(a) = m for all a. Let’s check this by computing the derivative:

f ′(a) = lim h→0

f (a + h) − f (a) h

= lim h→0

(m(a + h) + b) − (ma + b) h

= lim h→0

mh

h = lim

h→0 m = m

If m = 0, then f (x) = b is constant and f ′(a) = 0 (Figure 7). In summary,

4

2

42

f (x) = mx + b

−1

y

x

FIGURE 6 The derivative of f (x) = mx + b is f ′(a) = m for all a.

4

2

42 x

y

f (x) = b

FIGURE 7 The derivative of a constant function f (x) = b is f ′(a) = 0 for all a.

THEOREM 1 Derivative of Linear and Constant Functions

• If f (x) = mx + b is a linear function, then f ′(a) = m for all a. • If f (x) = b is a constant function, then f ′(a) = 0 for all a.

EXAMPLE 4 Find the derivative of f (x) = 9x − 5 at x = 2 and x = 5. Solution We have f ′(a) = 9 for all a. Hence, f ′(2) = f ′(5) = 9.

EXAMPLE 5 Determine the derivative of the function f whose graph appears in Figure 8 at x = 2, 3, 4.3

2

1

0 54321 x

y

FIGURE 8 Determining the derivative at x = 2, 3, 4.

Solution At x = 2, the graph is a line with slope given by

m = f (3) − f (0) 3 − 0 =

3 − 0 3 − 0 = 1

Hence, f ′(2) = 1. At x = 3, there is no well-defined tangent line, since the limit of the slopes of the

secant lines as x approaches 3 from the left is different from the limit of the slopes of the secant lines as x approaches 3 from the right. So the derivative f ′(3) does not exist.

At x = 4, the tangent line passes through the two points (3, 2) and (5, 0). Hence, it has slope m = 2−03−5 = −1. Therefore, f ′(4) = −1.

Estimating the Derivative Approximations to the derivative are useful in situations where we cannot evaluate f ′(a) exactly. Since the derivative is the limit of difference quotients, the difference quotient should give a good numerical approximation when h is sufficiently small:

f ′(a) ≈ f (a + h) − f (a) h

if h is small

Graphically, this says that for small h, the slope of the secant line is nearly equal to the slope of the tangent line (Figure 9).

a + ha

y

x

P Q

Tangent

Secant

FIGURE 9 When h is small, the secant line has nearly the same slope as the tangent line.

S E C T I O N 3.1 Definition of the Derivative 117

EXAMPLE 6 Estimate the derivative of f (x) = sin x at x = π6 . Solution We calculate the difference quotient for several small values of h:

sin (

π 6 + h

) − sin π6

h = sin

( π 6 + h

) − 0.5

h

Table 1 suggests that the limit has a decimal expansion beginning 0.866. In other words, f ′

( π 6

) ≈ 0.866.

TABLE 1 Values of the Difference Quotient for Small h

h > 0 sin

(π 6 + h

) − 0.5

h h < 0

sin (π

6 + h ) − 0.5

h

0.01 0.863511 −0.01 0.868511 0.001 0.865775 −0.001 0.866275 0.0001 0.8660 00 −0.0001 0.866 050 0.00001 0.8660 229 −0.00001 0.8660 279

In the next example, we use graphical reasoning to determine the accuracy of the estimates obtained in Example 6.

EXAMPLE 7 Determining Accuracy Graphically Let f (x) = sin x. Show that the approximation f ′

( π 6

) ≈ 0.8660 is accurate to four decimal places.π

6

Tangent

Secant (h > 0)

y = sin x

Secant (h < 0)

x

y

FIGURE 10 The tangent line is squeezed in between the secant lines with h > 0 and h < 0.

Solution Observe in Figure 10 that the position of the secant line relative to the tangent line depends on whether h is positive or negative. When h > 0, the slope of the secant line is smaller than the slope of the tangent line, but it is larger when h < 0. This tells us that the difference quotients in the second column of Table 1 are smaller than f ′

( π 6

) and

those in the fourth column are greater than f ′ (

π 6

) . From the last line in Table 1 we may

conclude thatThis technique of estimating an unknown quantity by showing that it lies between two known values (“squeezing it”) is used frequently in calculus.

0.866022 ≤ f ′ (π

6

) ≤ 0.866028

It follows that the estimate f ′ (

π 6

) ≈ 0.8660 is accurate to four decimal places. In Sec-

tion 3.6, we will see that the exact value is f ′ (

π 6

) = cos

( π 6

) =

√ 3

2 ≈ 0.8660254, just about midway between 0.866022 and 0.866028.

CONCEPTUAL INSIGHT Are Limits Really Necessary? It is natural to ask whether limits are really necessary. The tangent line is easy to visualize. Is there perhaps a better or simpler way to find its equation? History gives one answer: The methods of calculus based on limits have stood the test of time and are used more widely today than ever before.

History aside, we can see directly why limits play such a crucial role. The slope of a line can be computed if the coordinates of two points P = (x1, y1) and Q = (x2, y2) on the line are known:

Slope of line = y2 − y1 x2 − x1

This formula cannot be applied to the tangent line because we know only that it passes through the single point P = (a, f (a)). Limits provide an ingenious way around this obstacle. We choose a point Q = (a + h, f (a + h)) on the graph near P and form the secant line. The slope of this secant line is just an approximation to the slope of the tangent line:

Slope of secant line = f (a + h) − f (a) h

≈ slope of tangent line

But this approximation improves as h → 0, and by taking the limit, we convert our approximations into the exact slope.

118 C H A P T E R 3 DIFFERENTIATION

3.1 SUMMARY

• The difference quotient:

f (a + h) − f (a) h

The difference quotient is the slope of the secant line through the points P = (a, f (a)) and Q = (a + h, f (a + h)) on the graph of f .

• The derivative f ′(a) is defined by the following equivalent limits:

f ′(a) = lim h→0

f (a + h) − f (a) h

= lim x→a

f (x) − f (a) x − a

If the limit exists, we say that f is differentiable at x = a. • By definition, the tangent line at P = (a, f (a)) is the line through P with slope f ′(a)

[assuming that f ′(a) exists]. • Equation of the tangent line in point-slope form:

y − f (a) = f ′(a)(x − a) • To calculate f ′(a) using the limit definition:

Step 1. Write out the numerator of the difference quotient. Step 2. Divide by h and simplify. Step 3. Compute the derivative by taking the limit.

• For small values of h, we have the estimate f ′(a) ≈ f (a + h) − f (a) h

.

3.1 EXERCISES

Preliminary Questions 1. Which of the lines in Figure 11 are tangent to the curve?

A

B C

D

FIGURE 11

2. What are the two ways of writing the difference quotient?

3. Find a and h such that f (a + h) − f (a)

h is equal to the slope of

the secant line between (3, f (3)) and (5, f (5)).

4. Which derivative is approximated by tan

(π 4 + 0.0001

) − 1

0.0001 ?

5. What do the following quantities represent in terms of the graph of f (x) = sin x? (a) sin 1.3 − sin 0.9 (b) sin 1.3 − sin 0.9

0.4 (c) f ′(0.9)

Exercises 1. Let f (x) = 5x2. Show that f (3 + h) = 5h2 + 30h + 45. Then

show that

f (3 + h) − f (3) h

= 5h + 30

and compute f ′(3) by taking the limit as h → 0.

2. Let f (x) = 2x2 − 3x − 5. Show that the secant line through (2, f (2)) and (2 + h, f (2 + h)) has slope 2h + 5. Then use this for- mula to compute the slope of: (a) The secant line through (2, f (2)) and (3, f (3)) (b) The tangent line at x = 2 (by taking a limit) In Exercises 3–8, compute f ′(a) in two ways, using Eq. (1) and Eq. (2).

3. f (x) = x2 + 9x, a = 0

4. f (x) = x2 + 9x, a = 2 5. f (x) = 3x2 + 4x + 2, a = −1 6. f (x) = x3, a = 2 7. f (x) = x3 + 2x, a = 1 8. f (x) = 1x , a = 2

In Exercises 9–12, refer to Figure 12.

9. Find the slope of the secant line through (2, f (2)) and (2.5, f (2.5)). Is it larger or smaller than f ′(2)? Explain.

10. Estimate f (2 + h) − f (2)

h for h = −0.5. What does this

quantity represent? Is it larger or smaller than f ′(2)? Explain.

S E C T I O N 3.1 Definition of the Derivative 119

11. Estimate f ′(1) and f ′(2).

12. Find a value of h for which f (2 + h) − f (2)

h = 0.

0.5

1.0

1.5

2.0

2.5

3.0

f (x)

1.0 2.0 3.00.5 1.5 2.5 x

y

FIGURE 12

In Exercises 13–16, refer to Figure 13.

13. Determine f ′(a) for a = 1, 2, 4, 7. 14. For which values of x is f ′(x) < 0?

15. Which is larger, f ′(5.5) or f ′(6.5)?

16. Show that f ′(3) does not exist.

1

2

3

5

4

1 2 3 4 5 6 7 8 9 x

y

FIGURE 13 Graph of f .

In Exercises 17–20, use the limit definition to calculate the derivative of the linear function.

17. f (x) = 7x − 9 18. f (x) = 12 19. g(t) = 8 − 3t 20. k(z) = 14z + 12 21. Find an equation of the tangent line at x = 3, assuming that f (3) = 5 and f ′(3) = 2. 22. Find f (3) and f ′(3), assuming that the tangent line to y = f (x) at a = 3 has equation y = 5x + 2. 23. Describe the tangent line at an arbitrary point on the “curve” y = 2x + 8. 24. Suppose that f (2 + h) − f (2) = 3h2 + 5h. Calculate: (a) The slope of the secant line through (2, f (2)) and (6, f (6)) (b) f ′(2)

25. Let f (x) = 1 x

. Does f (−2 + h) equal 1−2 + h or 1

−2 + 1 h

?

Compute the difference quotient at a = −2 with h = 0.5. 26. Let f (x) = √x. Does f (5 + h) equal √5 + h or

√ 5 +

√ h?

Compute the difference quotient at a = 5 with h = 1. 27. Let f (x) = 1/√x. Compute f ′(5) by showing that

f (5 + h) − f (5) h

= − 1√ 5 √

5 + h(√5 + h + √

5)

28. Find an equation of the tangent line to the graph of f (x) = 1/√x at x = 9.

In Exercises 29–46, use the limit definition to compute f ′(a) and find an equation of the tangent line.

29. f (x) = 2x2 + 10x, a = 3 30. f (x) = 4 − x2, a = −1

31. f (t) = t − 2t2, a = 3 32. f (x) = 8x3, a = 1

33. f (x) = x3 + x, a = 0 34. f (t) = 2t3 + 4t , a = 4

35. f (x) = x−1, a = 8 36. f (x) = x + x−1, a = 4

37. f (x) = 1 x + 3 , a = −2 38. f (t) =

2 1 − t , a = −1

39. f (x) = √

x + 4, a = 1 40. f (t) = √

3t + 5, a = −1

41. f (x) = 1√ x

, a = 4 42. f (x) = 1√ 2x + 1

, a = 4

43. f (t) = √

t2 + 1, a = 3 44. f (x) = x−2, a = −1

45. f (x) = 1 x2 + 1 , a = 0 46. f (t) = t

−3, a = 1

47. Figure 14 displays data collected by the biologist Julian Huxley (1887–1975) on the average antler weight W of male red deer as a function of age t . Estimate the derivative at t = 4. For which values of t is the slope of the tangent line equal to zero? For which values is it negative?

2 40 6 8 10 12 14 t

Age (years)

Antler weight W (kg)

0 1 2 3 4 5 6 7 8

FIGURE 14

48. Figure 15(A) shows the graph of f (x) = √x. The close-up in Fig- ure 15(B) shows that the graph is nearly a straight line near x = 16. Estimate the slope of this line and take it as an estimate for f ′(16). Then compute f ′(16) and compare with your estimate.

3.9

4.1

(B) Zoom view near (16, 4)

x

y

1 2

5 4 3

2 4 6 8 10 12 14 16 18 x

y

(A) Graph of y = x

16.1

15.9

FIGURE 15

49. Let f (x) = 4 1 + 2x .

(a) Plot f over [−2, 2]. Then zoom in near x = 0 until the graph ap- pears straight, and estimate the slope f ′(0). (b) Use (a) to find an approximate equation to the tangent line at x = 0. Plot this line and y = f (x) on the same set of axes.

120 C H A P T E R 3 DIFFERENTIATION

50. Let f (x) = cot x. Estimate f ′ (π

2 )

graphically by zooming in on a plot of f near x = π2 .

51. Determine the intervals along the x-axis on which the derivative in Figure 16 is positive.

1.0 1.5 2.0 2.5 3.0 3.5 4.00.5

1.0 0.5

1.5 2.0 2.5 3.0 3.5 4.0

x

y

FIGURE 16

52. Sketch the graph of f (x) = sin x on [0, π ] and guess the value of f ′

(π 2 ) . Then calculate the difference quotient at x = π2 for two small

positive and negative values of h.Are these calculations consistent with your guess?

In Exercises 53–58, each limit represents a derivative f ′(a). Find f (x) and a.

53. lim h→0

(5 + h)3 − 125 h

54. lim x→5

x3 − 125 x − 5

55. lim h→0

sin (π

6 + h ) − 0.5

h 56. lim

x→ 14

x−1 − 4 x − 14

57. lim h→0

52+h − 25 h

58. lim h→0

5h − 1 h

59. Apply the method of Example 7 to f (x) = sin x to determine f ′

(π 4 )

accurately to four decimal places.

60. Apply the method of Example 7 to f (x) = cos x to de- termine f ′

(π 5 )

accurately to four decimal places. Use a graph of f to explain how the method works in this case.

61. For each graph in Figure 17, determine whether f ′(1) is larger or smaller than the slope of the secant line between x = 1 and x = 1 + h for h > 0. Explain.

1 1

(A) (B)

y

x

y

x

y = f (x) y = f (x)

FIGURE 17

62. Refer to the graph of f (x) = 2x in Figure 18. (a) Explain graphically why, for h > 0,

f (−h) − f (0) −h ≤ f

′(0) ≤ f (h) − f (0) h

(b) Use (a) to show that 0.69314 ≤ f ′(0) ≤ 0.69315.

(c) Similarly, compute f ′(x) to four decimal places for x = 1, 2, 3, 4. (d) Now compute the ratios f ′(x)/f ′(0) for x = 1, 2, 3, 4. Can you guess an approximate formula for f ′(x)?

321−1

1 x

y

FIGURE 18 Graph of f (x) = 2x .

63. Sketch the graph of f (x) = x5/2 on [0, 6]. (a) Use the sketch to justify the inequalities for h > 0:

f (4) − f (4 − h) h

≤ f ′(4) ≤ f (4 + h) − f (4) h

(b) Use (a) to compute f ′(4) to four decimal places. (c) Use a graphing utility to plot y = f (x) and the tangent line at x = 4, utilizing your estimate for f ′(4).

64. Verify that P = ( 1, 12

) lies on the graphs of both

f (x) = 1/(1 + x2) and L(x) = 12 + m(x − 1) for every slope m. Plot y = f (x) and y = L(x) on the same axes for several values of m until you find a value of m for which y = L(x) appears tangent to the graph of f . What is your estimate for f ′(1)?

65. Use a plot of f (x) = xx to estimate the value c such that f ′(c) = 0. Find c to sufficient accuracy so that

∣∣∣∣ f (c + h) − f (c)

h

∣∣∣∣ ≤ 0.006 for h = ±0.001

66. Plot f (x) = xx and y = 2x + a on the same set of axes for several values of a until the line becomes tangent to the graph. Then estimate the value c such that f ′(c) = 2.

In Exercises 67–73, estimate derivatives using the symmetric differ- ence quotient (SDQ), defined as the average of the difference quotients at h and −h:

1 2

( f (a + h) − f (a)

h + f (a − h) − f (a)−h

)

= f (a + h) − f (a − h) 2h

4

The SDQ usually gives a better approximation to the derivative than the difference quotient.

67. The vapor pressure of water at temperature T (in kelvins) is the atmospheric pressure P at which no net evaporation takes place. Use the following table to estimate P ′(T ) for T = 303, 313, 323, 333, 343 by computing the SDQ given by Eq. (4) with h = 10.

T (K) 293 303 313 323 333 343 353

P (atm) 0.0278 0.0482 0.0808 0.1311 0.2067 0.3173 0.4754

68. Use the SDQ with h = 1 year to estimate P ′(T ) in the years 2005, 2007, 2009, 2011, where P(T ) is the U.S. ethanol production (Figure 19). Express your answer in the correct units.

S E C T I O N 3.2 The Derivative as a Function 121

2.12

4.00

6.20

19 98

19 97

19 99

20 00

20 01

20 02

20 03

20 04

20 05

20 06

20 07

20 08

20 09

20 10

20 11

20 12

1.63 1.77 2.81

4.89

P (billions of gallons)

9.31

10.94

13.30 13.93

13.30

1.30 1.40 1.47

3.40

FIGURE 19 U.S. ethanol production.

In Exercises 69–70, traffic speed S along a certain road (in kilometers per hour) varies as a function of traffic density q (number of cars per kilometer of road). Use the following data to answer the questions:

q (density) 60 70 80 90 100

S (speed) 72.5 67.5 63.5 60 56

69. Estimate S′(80).

70. Explain why V = qS, called traffic volume, is equal to the number of cars passing a point per hour. Use the data to estimate V ′(80).

Exercises 71–73: The current (in amperes) at time t (in seconds) flowing in the circuit in Figure 20 is given by Kirchhoff’s Law:

i(t) = Cv′(t) + R−1v(t)

where v(t) is the voltage (in volts, V ), C the capacitance (in farads, F ), and R the resistance (in ohms, #).

+

−

v R

i

C

FIGURE 20

71. Calculate the current at t = 3 if

v(t) = 0.5t + 4 V

where C = 0.01 F and R = 100 #.

72. Use the following data to estimate v′(10) (by an SDQ). Then esti- mate i(10), assuming C = 0.03 and R = 1000.

t 9.8 9.9 10 10.1 10.2

v(t) 256.52 257.32 258.11 258.9 259.69

73. Assume that R = 200 # but C is unknown. Use the following data to estimate v′(4) (by an SDQ) and deduce an approximate value for the capacitance C.

t 3.8 3.9 4 4.1 4.2

v(t) 388.8 404.2 420 436.2 452.8

i(t) 32.34 33.22 34.1 34.98 35.86

Further Insights and Challenges 74. The SDQ usually approximates the derivative much more closely than does the ordinary difference quotient. Let f (x) = 2x and a = 0. Compute the SDQ with h = 0.001 and the ordinary difference quo- tients with h = ±0.001. Compare with the actual value, which is f ′(0) = ln 2.

75. Explain how the symmetric difference quotient defined by Eq. (4) can be interpreted as the slope of a secant line.

76. Which of the two functions in Figure 21 satisfies the inequality

f (a + h) − f (a − h) 2h

≤ f (a + h) − f (a) h

for h > 0? Explain in terms of secant lines.

a x

y

a x

y

(A) (B)

FIGURE 21

77. Show that if f is a quadratic polynomial, then the SDQ at x = a (for any h ̸= 0) is equal to f ′(a). Explain the graphical meaning of this result.

78. Let f (x) = x−2. Compute f ′(1) by taking the limit of the SDQs (with a = 1) as h → 0.

3.2 The Derivative as a Function In the previous section, we computed the derivative f ′(a) for specific values of a. It is also useful to view the derivative as a function f ′ whose value at x = a is f ′(a). The function f ′ is still defined as a limit, but the fixed number a is replaced by the variable x:

f ′(x) = lim h→0

f (x + h) − f (x) h

1

If y = f (x), we also write y′ or y′(x) for f ′(x).

122 C H A P T E R 3 DIFFERENTIATION

The domain of f ′ consists of all values of x in the domain of f for which the limit inOften, the domain of f ′ is clear from the context. If so, we usually do not mention the domain explicitly.

Eq. (1) exists. We say that f is differentiable on (a, b) if f ′(x) exists for all x in (a, b). When f ′(x) exists for all x in the interval or intervals on which f (x) is defined, we say simply that f is differentiable.

EXAMPLE 1 Prove that f (x) = x3 − 12x is differentiable. Compute f ′(x) and find an equation of the tangent line at x = −3.

Solution We compute f ′(x) in three steps as in the previous section.

Step 1. Write out the numerator of the difference quotient.

f (x + h) − f (x) = ( (x + h)3 − 12(x + h)

) −

( x3 − 12x

)

= (x3 + 3x2h + 3xh2 + h3 − 12x − 12h) − (x3 − 12x) = 3x2h + 3xh2 + h3 − 12h = h(3x2 + 3xh + h2 − 12) (factor out h)

Step 2. Divide by h and simplify.

f (x + h) − f (x) h

= h(3x 2 + 3xh + h2 − 12)

h = 3x2 + 3xh + h2 − 12 (h ̸= 0)

Step 3. Compute the limit.

f ′(x) = lim h→0

f (x + h) − f (x) h

= lim h→0

(3x2 + 3xh + h2 − 12) = 3x2 − 12

In this limit, x is treated as a constant because it does not change as h → 0. We see that the limit exists for all x, so f is differentiable and f ′(x) = 3x2 − 12.

Now evaluate:

f ′(−3) = 3(−3)2 − 12 = 15

This is the slope of the tangent line. Since f (−3) = 9, the line passes through (−3, 9). So an equation of the tangent line at x = −3 is y − 9 = 15(x + 3) or y = 15x + 54

20

x

y

y = 15x + 54

f (x) = x3 − 12x

(−3, 9)

−3 2

−20

FIGURE 1 Graph of f (x) = x3 − 12x. (Figure 1).

EXAMPLE 2 Prove that y = x−2 is differentiable and calculate y′.

Solution The domain of f (x) = x−2 is {x : x ̸= 0}, so assume that x ̸= 0. We compute f ′(x) directly, without the separate steps of the previous example:

y′ = lim h→0

f (x + h) − f (x) h

= lim h→0

1 (x + h)2 −

1 x2

h

= lim h→0

x2 − (x + h)2 x2(x + h)2

h = lim

h→0 1 h

( x2 − (x + h)2 x2(x + h)2

)

= lim h→0