1. Basic Calculus for Economists
Lecture 5
Basic Calculus for Economists

2. Analyzing a Limit
We can examine what occurs at a particular point by the limit. Using the function f (x) = 2x – 1, let’s examine what happens near x = 2 through the following chart: x
1.5
1.9
1.99
1.999 2
2.001
2.01
2.1
2.5
f (x)
2.8
2.98
2.998 ?
3.002
3.02
3.2 4
We see that as x approaches 2, f (x) approaches 3.

3. Limits In limit notation we have 3 Definition: We write 2
or as x  c, then f (x)  L,
if the functional value of f (x) is close to the single real number L whenever x is close to, but not equal to, c (on either side of c).

4. One-Sided Limits We write
and call K the limit from the left (or left-hand limit) if f (x) is close to K whenever x is close to c, but to the left of c on the real number line.
and call L the limit from the right (or right-hand limit) if f (x) is close to L whenever x is close to c, but to the right of c on the real number line.
In order for a limit to exist, the limit from the left and the limit from the right must exist and be equal.

5. Limit Properties
Let f and g be two functions, and assume that the following two limits exist and are finite:
Then
the limit of the sum of the functions is equal to the sum of the limits.
the limit of the difference of the functions is equal to the difference of the limits.

6. Limit Properties (continued)
the limit of a constant times a function is equal to the constant times the limit of the function.
the limit of the product of the functions is the product of the limits of the functions.
the limit of the quotient of the functions is the quotient of the limits of the functions, provided M  0.
the limit of the nth root of a function is the nth root of the limit of that function.

7. Examples From these examples we conclude that
f any polynomial function
r any rational function with a nonzero denominator at x = c

8. Indeterminate Forms
It is important to note that there are restrictions on some of the limit properties. In particular if
then finding may present difficulties, since the denominator is 0.
If and , then
is said to be indeterminate.
The term “indeterminate” is used because the limit may or may not exist.

9. Example
This example illustrates some techniques that can be useful for indeterminate forms.
Algebraic simplification is often useful when the numerator and denominator are both approaching 0.

10. Difference Quotients
Let f (x) = 3x - 1. Find
Solution:

11. Definition of Continuity
A function f is continuous at a point x = c if 1.
2. f (c) exists 3.
A function f is continuous on the open interval (a,b) if it is continuous at each point on the interval.
If a function is not continuous, it is discontinuous.

12. Example 1 f (x) = x – 1 at x = 2. 1. The limit exists! 2. f(2) = 1 3.
Therefore this function is continuous at x = 2. 1 2

13. Example 2
f (x) = (x2 – 9)/(x + 3) at x = The limit exists (reduce the fraction).
2. f (-3) = 0/0 is undefined! 3.
The function is not continuous at x = -3. (Graph should have an open circle there.)

14. Example 3 f (x) = |x|/x at x = 0 and at x = 1. 1. Does not exist!
2. f (0) = 0/ Undefined! 3.
The function is not continuous at x = 0.
This function is continuous at x = 1.

15. Limits in Mathematica

16. Continuity Properties
If two functions are continuous on the same interval, then their sum, difference, product, and quotient are continuous on the same interval, except for values of x that make the denominator 0.

17. Examples of Continuous Functions
A constant function is continuous for all x.
For integer n > 0, f (x) = xn is continuous for all x.
A polynomial function is continuous for all x.
A rational function is continuous for all x, except those values that make the denominator 0.
For n an odd positive integer, is continuous wherever f (x) is continuous.
For n an even positive integer, is continuous wherever f (x) is continuous and nonnegative.

18. Infinite Limits There are various possibilities under which
does not exist. For example, if the one-sided limits are different at x = a, then the limit does not exist.
Another situation where a limit may fail to exist involves functions whose values become very large as x approaches a. The special symbol  (infinity) is used to describe this type of behavior.

19. To illustrate this case, consider the function f (x) = 1/(x-1), which is discontinuous at x = 1. As x approaches 1 from the right, the values of f (x) are positive and become larger and larger. That is, f (x) increases without bound. We write this symbolically as
Since  is not a real number, the limit above does not actually exist. We are using the symbol  (infinity) to describe the manner in which the limit fails to exist, and we call this an infinite limit.

20. Example (continued)
As x approaches 1 from the left, the values of f (x) are negative and become larger and larger in absolute value. That is, f (x) decreases through negative values without bound. We write this symbolically as
The graph of this function is as shown:
Note that does not exist.

21. Infinite Limits and Vertical Asymptotes
Definition:
The vertical line x = a is a vertical asymptote for the graph of y = f (x) if f (x)   or f (x)  - as x  a+ or x  a–.
That is, f (x) either increases or decreases without bound as x approaches a from the right or from the left.
Note: If any one of the four possibilities is satisfied, this makes x = a a vertical asymptote. Most of the time, the limit will be infinite (+ or -) on both sides, but it does not have to be.

22. Vertical Asymptotes of Polynomials
How do we locate vertical asymptotes? If a function f is continuous at x = a, then
Since all of the above limits exist and are finite, f cannot have a vertical asymptote at x = a. In order for f to have a vertical asymptote at x = a, at least one of the limits above must be an infinite limit, and f must be discontinuous at x = a. We know that polynomial functions are continuous for all real numbers, so a polynomial has no vertical asymptotes.

23. Vertical Asymptotes of Rational Functions
Since a rational function is discontinuous only at the zeros of its denominator, a vertical asymptote of a rational function can occur only at a zero of its denominator. The following is a simple procedure for locating the vertical asymptotes of a rational function:
If f (x) = n(x)/d(x) is a rational function, d(c) = 0 and n(c)  0, then the line x = c is a vertical asymptote of the graph of f.
However, if both d(c) = 0 and n(c) = 0, there may or may not be a vertical asymptote at x = c.

24. Example
Let
Describe the behavior of f at each point of discontinuity. Use  and - when appropriate. Identify all vertical asymptotes.

25. Example (continued) Let
Describe the behavior of f at each point of discontinuity. Use  and - when appropriate. Identify all vertical asymptotes.
Solution: Let n(x) = x2 + x - 2 and d(x) = x Factoring the denominator, we see that d(x) = x2 - 1 = (x+1)(x-1) has two zeros, x = -1 and x = 1. These are the points of discontinuity of f.

26. Example (continued)
Since d(-1) = 0 and n(-1) = -2  0, the theorem tells us that the line x = -1 is a vertical asymptote.
Now we consider the other zero of d(x), x = 1. This time n(1) = 0 and the theorem does not apply. We use algebraic simplification to investigate the behavior of the function at x = 1:
Since the limit exists as x approaches 1, f does not have a vertical asymptote at x = 1. The graph of f is shown on the next slide.

27. Example (continued)
Vertical Asymptote
Point of discontinuity

28. Limits at Infinity of Power Functions
We begin our consideration of limits at infinity by considering power functions of the form x p and 1/x p, where p is a positive real number.
If p is a positive real number, then x p increases as x increases, and it can be shown that there is no upper bound on the values of x p. We indicate this by writing or

29. Power Functions (continued)
Since the reciprocals of very large numbers are very small numbers, it follows that 1/x p approaches 0 as x increases without bound. We indicate this behavior by writing or
This figure illustrates this behavior for f (x) = x2 and g(x) = 1/x2.

30. Power Functions (continued)
In general, if p is a positive real number and k is a nonzero real number, then
Note: k and p determine whether the limit at  is  or -.
The last limit is only defined if the pth power of a negative number is defined. This means that p has to be an integer, or a rational number with odd denominator.

31. Limits at Infinity of Polynomial Functions
What about limits at infinity for polynomial functions?
As x increases without bound in either the positive or the negative direction, the behavior of the polynomial graph will be determined by the behavior of the leading term (the highest degree term). The leading term will either become very large in the positive sense or in the negative sense (assuming that the polynomial has degree at least 1). In the first case the function will approach  and in the second case the function will approach -.
In mathematical shorthand, we write this as This covers all possibilities.

32. Limits at Infinity and Horizontal Asymptotes
A line y = b is a horizontal asymptote for the graph of y = f (x) if f (x) approaches b as either x increases without bound or decreases without bound. Symbolically, y = b is a horizontal asymptote if
In the first case, the graph of f will be close to the horizontal line y = b for large (in absolute value) negative x. In the second case, the graph will be close to the horizontal line y = b for large positive x.
Note: It is enough if one of these conditions is satisfied, but frequently they both are.

33. Example
This figure shows the graph of a function with two horizontal asymptotes, y = 1 and y = -1.

34. Horizontal Asymptotes of Rational FunctionsIf
then
There are three possible cases for these limits.
If m < n, then The line y = 0 (x axis) is a horizontal asymptote for f (x).
2. If m = n, then The line y = am/bn is a horizontal asymptote for f (x) .
3. If m > n, f (x) does not have a horizontal asymptote.

35. Horizontal Asymptotes of Rational Functions (continued)
Notice that in cases 1 and 2 on the previous slide that the limit is the same if x approaches  or -. Thus a rational function can have at most one horizontal asymptote. (See figure). Notice that the numerator and denominator have the same degree in this example, so the horizontal asymptote is the ratio of the leading coefficients of the numerator and denominator.
y = 1.5

36. Example
Find the horizontal asymptotes of each function.

37. Example Solution Find the horizontal asymptotes of each function.
Since the degree of the numerator is less than the degree of the denominator in this example, the horizontal asymptote is y = 0 (the x axis).
Since the degree of the numerator is greater than the degree of the denominator in this example, there is no horizontal asymptote.

38. Summary
An infinite limit is a limit of the form (y goes to infinity). It is the same as a vertical asymptote (as long as a is a finite number).
A limit at infinity is a limit of the form (x goes to infinity). It is the same as a horizontal asymptote (as long as L is a finite number).

39. The Rate of Change
For y = f (x), the average rate of change from x = a to x = a + h is
The above expression is also called a difference quotient. See Chiang It can be interpreted as the slope of a secant.
See the picture on the next slide for illustration.

40. Visual InterpretationQ
slope
f (a + h) – f (a)
Average rate of change = slope of the secant line through P and Q P h

41. Example 1
The revenue generated by producing and selling widgets is given by R(x) = x (75 – 3x) for 0  x  20.
What is the change in revenue if production changes from 9 to 12?

42. Example 1
The revenue generated by producing and selling widgets is given by R(x) = x (75 – 3x) for 0  x  20.
What is the change in revenue if production changes from 9 to 12?
R(12) – R(9) = $468 – $432 = $36.
Increasing production from 9 to 12 will increase revenue by $36.

43. Example 1 (continued)
The revenue is R(x) = x (75 – 3x) for 0  x  20.
What is the average rate of change in revenue (per unit change in x) if production changes from 9 to 12?

44. Example 1 (continued)
The revenue is R(x) = x (75 – 3x) for 0  x  20.
What is the average rate of change in revenue (per unit change in x) if production changes from 9 to 12?
To find the average rate of change we divide the change in revenue by the change in production:
Thus the average change in revenue is $12 when production is increased from 9 to 12.
Like Change in Y over change in X.

45. The Instantaneous Rate of Change
Consider the function y = f (x) only near the point P = (a, f (a)).
The difference quotient
gives the average rate of change of f over the interval [a, a+h].
If we make h smaller and smaller, in the limit we obtain the instantaneous rate of change of the function at the point P:

46. Visual InterpretationQ
Tangent
Slope of tangent = instantaneous rate of change.
f (a + h) – f (a) P
Let h approach 0 h

47. Instantaneous Rate of Change
Given y = f (x), the instantaneous rate of change at x = a is
provided that the limit exists. It can be interpreted as the slope of the tangent at the point (a, f (a)). See illustration on previous slide.

48. The Derivative
For y = f (x), we define the derivative of f at x, denoted f ’ (x), to be
if the limit exists.
If f ’(a) exists, we call f differentiable at a.
If f ’(x) exist for each x in the open interval (a, b), then f is said to be differentiable over (a, b).

49. Interpretations of the Derivative
If f is a function, then f ’ is a new function with the following interpretations:
For each x in the domain of f ’, f ’ (x) is the slope of the line tangent to the graph of f at the point (x, f (x)).
For each x in the domain of f ’, f ’ (x) is the instantaneous rate of change of y = f (x) with respect to x.
If f (x) is the position of a moving object at time x, then v = f ’ (x) is the velocity of the object at that time.

50. Finding the Derivative
To find f ‘ (x), we use a four-step process:
Step 1. Find f (x + h)
Step 2. Find f (x + h) – f (x)
Step 3. Find
Step 4. Find

51. Example 2 Find the derivative of f (x) = x 2 – 3x.
Step 1. f (x + h) = (x + h)2 – 3(x + h) = x2 + 2xh + h2 – 3x – 3h
Step 2. Find f (x + h) – f (x) = 2xh + h2 – 3h
Step 3. Find
Step 4. Find

52. Example 3
Find the slope of the tangent to the graph of f (x) = x 2 – 3x at x = 0, x = 2, and x = 3.
Solution: In example 2 we found the derivative of this function at x to be
f ’ (x) = 2x – 3
Hence
f ’ (0) = -3
f ’ (2) = 1, and
f ’ (3) = 3

53. Example 4
We know that the derivative of f (x) = x 2 – 3x is f ’ (x) = 2x – 3.
Verify this for x = 2 using Mathematica.
D[] is the derivative function
tangent equation

54. Example 5
Find the derivative of f (x) = 2x – 3x2 using Mathematica with a symbolic differentiation routine.

55. Example 6
Find the derivative of f (x) = 2x – 3x2 using the four-step process.
Step 1. f (x + h) = 2(x + h) – 3(x + h)2
Step 2. f (x + h) – f (x) = 2h – 6xh - 3h2
Step 3.
Step 4.

56. Nonexistence of the Derivative
The existence of a derivative at x = a depends on the existence of the limit
If the limit does not exist, we say that the function is nondifferentiable at x = a, or f ’ (a) does not exist.

57. Nonexistence of the Derivative (continued)
Some of the reasons why the derivative of a function may not exist at x = a are
The graph of f has a hole or break at x = a, or
The graph of f has a sharp corner at x = a, or
The graph of f has a vertical tangent at x = a.

58. Summary
For y = f (x), we defined the derivative of f at x, denoted f ’ (x), to be
if the limit exists.
We have seen how to find the derivative algebraically, using the four-step process.

59. Derivative Notation
In the preceding section we defined the derivative of a function. There are several widely used symbols to represent the derivative. Given y = f (x), the derivative may be represented by any of the following:
f ’(x) y’
dy/dx

60. Example 1 (continued) What is the slope of a constant function?
The graph of f (x) = C is a horizontal line with slope 0, so we would expect f ’(x) = 0.
Theorem 1. Let y = f (x) = C be a constant function, then
y’ = f ’(x) = 0.

61. THEOREM 2 IS VERY IMPORTANT. IT WILL BE USED A LOT!
Power Rule
A function of the form f (x) = xn is called a power function. This includes f (x) = x (where n = 1) and radical functions (fractional n).
Theorem 2. (Power Rule) Let y = xn be a power function, then
y’ = f ’(x) = n xn – 1.
THEOREM 2 IS VERY IMPORTANT. IT WILL BE USED A LOT!

62. Example 2 Differentiate f (x) = x5. Solution:
By the power rule, the derivative of xn is n xn–1.
In our case n = 5, so we get f ’(x) = 5 x4.

63. Example 3 Differentiate Solution:
Rewrite f (x) as a power function, and apply the power rule:

64. Constant Multiple Property
Theorem 3. Let y = f (x) = k u(x) be a constant k times a function u(x). Then
y’ = f ’(x) = k  u’(x).
In words: The derivative of a constant times a function is the constant times the derivative of the function.

65. Example 4 Differentiate f (x) = 7x4. Solution:
Apply the constant multiple property and the power rule.
f ’(x) = 7(4x3) = 28 x3.

66. Sum and Difference Properties
Theorem 5. If
y = f (x) = u(x) ± v(x),
then
y’ = f ’(x) = u’(x) ± v’(x).
In words:
The derivative of the sum of two differentiable functions is the sum of the derivatives.
The derivative of the difference of two differentiable functions is the difference of the derivatives.

67. Example 5 Differentiate f (x) = 3x5 + x4 – 2x3 + 5x2 – 7x + 4.
Solution:
Apply the sum and difference rules, as well as the constant multiple property and the power rule.
f ’(x) = 15x4 + 4x3 – 6x2 + 10x – 7.

68. Applications
Remember that the derivative gives the instantaneous rate of change of the function with respect to x. That might be:
Instantaneous velocity.
Tangent line slope at a point on the curve of the function.
Marginal Cost. If C(x) is the cost function, that is, the total cost of producing x items, then C’(x) approximates the cost of producing one more item at a production level of x items. C’(x) is called the marginal cost.

69. Tangent Line Example Let f (x) = x4 - 6x2 + 10. (a) Find f ’(x)
(b) Find the equation of the tangent line at x = 1

70. Tangent Line Example (continued)
Let f (x) = x4 - 6x
(a) Find f ’(x)
(b) Find the equation of the tangent line at x = 1
Solution:
f ’(x) = 4x3 - 12x
Slope: f ’(1) = 4(13) - 12(1) = -8. Point: If x = 1, then y = f (1) = = Point-slope form: y - y1 = m(x - x1) y - 5 = -8(x -1) y = -8x + 13

72. Application Example
The total cost (in dollars) of producing x portable radios per day is
C(x) = x – 0.5x2
for 0 ≤ x ≤ 100.
Find the marginal cost at a production level of x radios.

73. Example (continued)
The total cost (in dollars) of producing x portable radios per day is
C(x) = x – 0.5x2
for 0 ≤ x ≤ 100.
Find the marginal cost at a production level of x radios.
Solution: The marginal cost will be
C’(x) = 100 – x.

74. Example (continued)
Find the marginal cost at a production level of 80 radios and interpret the result.
Solution: C’(80) = 100 – 80 = 20.
It will cost approximately $20 to produce the 81st radio.
Find the actual cost of producing the 81st radio and compare this with the marginal cost.
Solution: The actual cost of the 81st radio will be
C(81) – C(80) = $ – $5800 = $19.50.
This is approximately equal to the marginal cost.

75. Example (continued): for you
4. If the price of the radio is $11.00, how many radios do you decide to produce?
5. Calculate the firm’s total profits.
6. Calculate the firm’s break-even point
7. Show and plot the firm’s profit function

76. Summary If f (x) = C, then f ’(x) = 0
If f (x) = xn, then f ’(x) = n xn-1
If f (x) = ku(x), then f ’ (x) = ku’(x)
If f (x) = u(x) ± v(x), then f ’(x) = u’(x) ± v’(x).

77. Increments
In a previous section we defined the derivative of f at x as the limit of the difference quotient:
Increment notation will enable us to interpret the numerator and the denominator of the difference quotient separately.

78. Example
Let y = f (x) = x3. If x changes from 2 to 2.1, then y will change from y = f (2) = 8 to y = f (2.1) =
We can write this using increment notation. The change in x is called the increment in x and is denoted by x.  is the Greek letter “delta”, which often stands for a difference or change. Similarly, the change in y is called the increment in y and is denoted by y.
In our example,
x = 2.1 – 2 = 0.1
y = f (2.1) – f (2) = – 8 =

79. Graphical Illustration of Increments
For y = f (x)
x = x2 - x1 y = y2 - y1
x2 = x1 + x = f (x2) – f (x1) = f (x1 + x) – f (x1)
(x2, f (x2))
y represents the change in y corresponding to a x change in x.
x can be either positive or negative. y
(x1, f (x1)) x1 x2 x

80. Differentials Assume that the limit exists. For small x,
Multiplying both sides of this equation by x gives us
y  f ’(x) x.
Here the increments x and y represent the actual changes in x and y.

81. Differentials (continued)
One of the notations for the derivative is
If we pretend that dx and dy are actual quantities, we get
We treat this equation as a definition, and call dx and dy differentials.

82. Interpretation of Differentials
x and dx are the same, and represent the change in x.
The increment y stands for the actual change in y resulting from the change in x.
The differential dy stands for the approximate change in y, estimated by using derivatives.
In applications, we use dy (which is easy to calculate) to estimate y (which is what we want).

83. Example 1
Find dy for f (x) = x2 + 3x and evaluate dy for x = 2 and dx = 0.1.
Solution:
dy = f ’(x) dx = (2x + 3) dx
When x = 2 and dx = 0.1, dy = [2(2) + 3] 0.1 = 0.7.

84. Example 2 Cost-Revenue
A company manufactures and sells x transistor radios per week. If the weekly cost and revenue equations are
find the approximate changes in revenue and profit if production is increased from 2,000 to 2,010 units/week.

85. Example 2 Solution
The profit is We will approximate R and P with dR and dP, respectively, using x = 2,000 and dx = 2,010 – 2,000 = 10.

86. Marginal Cost
Remember that marginal refers to an instantaneous rate of change, that is, a derivative.
Definition:
If x is the number of units of a product produced in some time interval, then
Total cost = C(x)
Marginal cost = C’(x)

87. Marginal Revenue and Marginal Profit
Definition:
If x is the number of units of a product sold in some time interval, then
Total revenue = R(x)
Marginal revenue = R’(x)
If x is the number of units of a product produced and sold in some time interval, then
Total profit = P(x) = R(x) – C(x)
Marginal profit = P’(x) = R’(x) – C’(x)

88. Marginal Cost and Exact Cost
Assume C(x) is the total cost of producing x items. Then the exact cost of producing the (x + 1)st item is
C(x + 1) – C(x).
The marginal cost is an approximation of the exact cost.
C’(x) ≈ C(x + 1) – C(x).
Similar statements are true for revenue and profit.

89. Example
The total cost of producing x electric guitars is C(x) = 1, x – 0.25x2.
Find the exact cost of producing the 51st guitar.
The exact cost is C(x + 1) – C(x).
C(51) – C(50) = 5, – 5375 = $74.75.
Use the marginal cost to approximate the cost of producing the 51st guitar.
The marginal cost is C’(x) = 100 – 0.5x
C’(50) = $75.

90. Marginal Average Cost Definition:
If x is the number of units of a product produced in some time interval, then
Average cost per unit =
Marginal average cost =

91. Marginal Average Revenue Marginal Average Profit
If x is the number of units of a product sold in some time interval, then
Average revenue per unit = Marginal average revenue =
If x is the number of units of a product produced and sold in some time interval, then
Average profit per unit = Marginal average profit =

92. Warning!
To calculate the marginal averages you must calculate the average first (divide by x), and then the derivative. If you change this order you will get no useful economic interpretations.
STOP

93. Example 2
The total cost of printing x dictionaries is $20,000 to start your business, and there are $10 in variable costs: Thus,
C(x) = 20, x
1. Find the average cost per unit if 1,000 dictionaries are produced.

94. Example 2 (continued) The total cost of printing x dictionaries is
C(x) = 20, x
1. Find the average cost per unit if 1,000 dictionaries are produced.
= $30

95. Example 2 (continued)
Find the marginal average cost at a production level of 1,000 dictionaries, and interpret the results.
Marginal average cost =
This means that if you raise production from 1,000 to 1,001 dictionaries, the price per book will fall approximately 2 cents.

96. Example 2 (continued)
3. Use the results from above to estimate the average cost per dictionary if 1,001 dictionaries are produced.
Average cost for 1000 dictionaries = $30.00
Marginal average cost =
The average cost per dictionary for 1001 dictionaries would be the average for 1000, plus the marginal average cost, or
$ $(- 0.02) = $29.98

97. Example 3
The price-demand equation and the cost function for the production of television sets are given by
where x is the number of sets that can be sold at a price of $p per set, and C(x) is the total cost of producing x sets.
Find the marginal cost.
Solution: The marginal cost is C’(x) = $30.

98. Example 3 (continued) Find the revenue function in terms of x.
The revenue function is
3. Find the marginal revenue.
The marginal revenue is
Find R’(1500) and interpret the results.
At a production rate of 1,500, each additional set increases revenue by approximately $200.

99. Example 3 (continued)
5. Graph the cost function and the revenue function on the same coordinate. Find the break-even point.
0 < x < 9,000
0 < y < 700,000

100. Example 3 (continued)
5. Graph the cost function and the revenue function on the same coordinate. Find the break-even point.
0 < x < 9,000
R(x)
0 < y < 700,000
Solution: There are two break-even points.
C(x)
(600,168,000)
(7500, 375,000)

101. Example 3 (continued) Find the profit function in terms of x.
The profit is revenue minus cost, so
Find the marginal profit.
8. Find P’(1500) and interpret the results.
At a production level of 1500 sets, profit is increasing at a rate of about $170 per set.

102. Example 3 (continued)
Find Maximum Profits… without finding MR or MC