Transcript x - United International College

Mathematics for Business
Instructor: Prof. Ken Tsang
Room E409-R11
Email: [email protected]
1
CALCULUS
For Business, Economics, and the Social
and life Sciences
Hoffmann, L.D. & Bradley, G.L.
2
TA information
Mr Zhu Zhibin
Room E409 Tel: 3620630
[email protected]
3
Web-page for this class
Watch for announcements about this class
and
 download lecture notes from
 http://www.uic.edu.hk/~kentsang/calcu2012/c
alcu.htm
 Or from this page:
http://www.uic.edu.hk/~kentsang/
Or from Ispace

4
Tutorials





One hour each week
Time & place to be announced later (we need
your input)
More explanations
More examples
More exercises
5
How is my final grade determined?
Quizzes
 Mid-term exam
 Assignments
 Final Examination

20%
20%
10%
50%
6
UIC Score System
7
Grade Distribution Guidelines
8
What can you do to maximize your
chances for success?
Work hard, more importantly, work smart:
1.
Understand, don't memorize.
2.
Ask why, not how.
3.
See every problem as a challenge.
4.
Learn techniques, not results.
5.
Make sure you understand each topic before
going on to the next.
9
More info about this Course



Assignments must be handed in before the deadline.
There will be about 3 to 4 quizzes.
We will tell you your scores for the mid-term test
and quizzes so that you know your progress.
However, for the final examination, we cannot tell
you the score before the AR release the official
results.
10
Mathematics? Why?

Mathematics is about



numbers, space, structures, …
Mathematicians seek out patterns, formulate
new conjectures, and establish truth by
rigorous deduction from appropriately
chosen axioms and definitions.
Most important, it teaches us how to
analysis problem in an abstract form, with
logical thinking.
11
They invented Calculus!
Sir Isaac Newton
(1642-1727)
Gottfriend Wilhelm von Leibniz
(1646-1716)
12
What is Calculus all about?

Calculus is the study of changing quantities, or more
precisely, the rate of changes: e.g. velocities, interest
rate, return on an asset.

The two key areas of Calculus are Differential
Calculus and Integral Calculus.

The big surprise is that these two seemingly unrelated
areas are actually connected via the Fundamental
Theorem of Calculus.
13
Calculus has practical applications, such as understanding the true meaning of the
infinitesimals. (Image concept by Dr. Lachowska.)
14
Isaac Newton (4 January 1643 – 31 March 1727)
English physicist,
mathematician, astronomer,
natural philosopher and
theologian, one of the most
influential men in human
history.
Newton in a 1702 portrait by Godfrey Kneller
15
Newton’s contributions
Newton described universal gravitation and the three laws
of motion, laying the groundwork for classical mechanics, which
dominated the scientific view of the physical Universe for the next
three centuries and is the basis for modern engineering.
Newton showed that the motions of objects on Earth
and of celestial bodies are governed by the same set of
natural laws.
16
Newton's own first edition copy of his Philosophiae Naturalis
Principia Mathematica with his handwritten corrections for the
second edition.
The book can be seen in the Wren Library of Trinity College, Cambridge.
Cosmos1
17
Newton's 2nd law of motion
Newton's Second Law states that an applied force, on an object equals
the rate of change of its momentum, with time.
For a system with constant mass, the equation can
be written in the iconic form:
F= ma,
where a is the acceleration of an object.
Acceleration is the rate of change in velocity.
This can be rewritten as a differential equation.
Most laws of nature can be expressed as differential
equations or partial differential equations (PDE).
18
If you are a finance major

Finance is a quantitative discipline





How to calculate the return of your investment?
Asset valuation
Portfolio theory
Derivatives
Risk management
19
A simple example in asset valuation

Suppose we have a riskless asset
 r is the constant rate of return
20
If your major is finance, you will know this:



Fischer Black and Myron Scholes first articulated the
Black-Scholes formula in their 1973 paper, "The
Pricing of Options and Corporate Liabilities."
Robert C. Merton was the first to publish a paper
expanding the mathematical understanding of the
options pricing model and coined the term "BlackScholes" options pricing model.
Merton and Scholes received the 1997 Prize in
Economic Sciences in Memory of Alfred Nobel for
this and related work.
21
The Black-Scholes model

In the Black-Scholes model, we assume that the
underlying security (typically the stock) follows a
geometric Brownian motion. That is,
where S is the price of the stock at time t,
μ is the drift rate of S, annualized,
σ is the volatility of the stock,
the dW term here stands in for any and all sources of uncertainty in
the price history of a stock, modeled by a Brownian motion.
22
If you are a science major

Science is
Quantitative
 Logical

23
Ecology: Population dynamics

The basic accounting relation for population
dynamics is:
N1 = N0 + B − D + I − E
where N1 is the number of individuals at time
1, N0 is the number of individuals at time 0, B
is the number of individuals born, D the
number that died, I the number that
immigrated, and E the number that emigrated
between time 0 and time 1.
24
The Lotka–Volterra (predator–prey)
equations
are a pair of first-order, non-linear, differential
equations frequently used to describe the
dynamics of biological systems in which two
species interact, one a predator and one its
prey.
25
where
* y is the number of some predator (for example, wolves);
* x is the number of its prey (for example, rabbits);
* dy/dt and dx/dt represents the growth of the two populations against
time;
* t represents the time; and
* α, β, γ and δ are parameters representing the interaction of the two
species.
26
Suppose there are two species of animals, a baboon (prey) and a cheetah
(predator). If the initial conditions are 80 baboons and 40 cheetahs, one can plot the
progression of the two species over time.
27
OK, any question?


That’s all for introduction.
Let’s begin the real thing!
28
Chapter 1
Functions, Graphs and Limits
In this Chapter, we will encounter some
important concepts.
 Functions (函数)
 Limits （极限）
 One-sided Limits （单边极限） and
Continuity （连续）
29
Section 1.1 Functions （函数）

A function is a rule that assigns to each object in a set
A exactly one object in a set B.

The set A is called the domain （定义域）of the
function, and the set of assigned objects in B is called the
range. （值域）
30
Function, or not?
f
A
B
YES
f
A
f
B
NO
B
A
NO
31

To be convenient, we represent a functional relationship by
an equation y  f (x)

In this context, x and y are called variables, furthermore,
we refer to y as the dependent variable (因变量) and to x
as the independent variable （自变量）. For instant, the
function representation y  f ( x)  x 2  4

Noted that x and y can be substituted by other letters. For
example, the above function can be represented by
s t 4
2
32
Function that describes tabular data
Table 1.1 Average Tuition and Fees for 4-Year Private Colleges
Academic Year
Ending in
1973
1978
1983
1988
1993
1998
2003
Period n
1
2
3
4
5
6
7
Tuition and
Fees
$1,898
$2,700
$4,639
$7,048
$10,448
$13,785
$18,273
33
Solution:

We can describe this data as a function f defined
by the rule
averagetuitionand fees at the

f (n)  

beginning
of
the
n
th
5
year
period


Thus, f (1)  1,898, f (2)  2,700,......., f (7)  18,273

Noted that the domain of f is the set of integers
A  {1,2,....,7}
34
Piecewise-defined function
（分段函数）


A piecewise-defined function is such a function that
is often defined using more than one formula,
where each individual formula describes the
function on a subset of the domain.
Here is an example of such a function
 1
if x  1

f ( x)   x  1
 3x 2  1 if x  1
35
Example 1
Find f(-1/2), f(1), and f(2), where the piecewise-defined
function f(x) is given at the above slide.
Solution:
1
2
Since x   satisfies x<1, use the top part of the formula to find
1
1
2
 1
f   


3
 2  1/ 2 1  3 / 2
However, x=1 and x=2 satisfy x≥1, so f(1) and f(2) are both
found by using the bottom part of the formula:
2
2
f
(
2
)

3
(
2
)
 1  13
and
f (1)  3(1)  1  4
36
Domain Convention


We assume the domain of f to be the set of all
numbers for which f(x) is defined (as a real number).
We refer to this as the natural domain of f.
In general, there are two situations where a number
is not in the domain of a function:
1) division by 0
2) The even number root of a negative number
37
Example 2
Find the domain and range of each of these functions
a.
1
f ( x) 
1 x2
b.
g (u)  4 u  2
Solution:
a. Since division by any number other than 0 is possible, the
domain of f is the set of all numbers except -1 and 1. The range
of f is the set of all numbers y except 0.
b. Since negative numbers do not have real fourth roots, so the
domain of g is the set of all numbers u such as u≥-2. The range
of g is the set of all nonnegative numbers.
38
Functions Used in Economics

A demand function (需求函数) p=D(x) is a function that
relates the unit price p for a particular commodity to the
number of units x demanded by consumers at that price.

The total revenue （总收入）is given by the product
R(x)=(number of items sold)(price per item)
=xp=xD(x)

If C(x) is the total cost （总成本）of producing the x units,
then the profit （利润）is given by the function
P(x)=R(x)-C(x)=xD(x)-C(x)
39
Example 3
Market research indicates that consumers will buy x
thousand units of a particular kind of coffee maker when
the unit price is p  0.27x  51 dollars. The cost of
producing the x thousand units is
C( x)  2.23x  3.5x  85
2
thousand dollars
a. What are the revenue and profit functions, R(x) and
P(x), for this production process?
b. For what values of x is production of the coffee
makers profitable?
40
Solution:
a. The demand function is D( x)  0.27x  51 , so the revenue is
R( x)  xD( x)  0.27x 2  51x
thousand dollars, and the profit is (thousand dollars)
P( x)  R( x)  C ( x)
 0.27x 2  51x  (2.23x 2  3.5 x  85)
 2.5 x 2  47.5 x  85
b. Production is profitable when P(x)>0. We find that
P ( x )  2.5 x 2  4 7.5 x  8 5
 2.5( x 2  1 9x  3 4)
 2.5( x  2)( x  1 7)  0
Thus, production is profitable for 2<x<17.
41
Composition of Functions
（复合函数）

Composition of Functions: Given functions f(u) and g(x), the
composition f(g(x)) is the function of x formed by substituting
u=g(x) for u in the formula for f(u).
Example 4
Find the composition function f(g(x)), where f (u)  u3  1 and
g ( x)  x  1
Solution:
Replace u by x+1 in the formula for f(u) to get
f ( g ( x))  ( x  1)3  1  x3  3x 2  3x  2
Question: How about g(f(x))?
Note: In general, f(g(x)) and g(f(x)) will not be the same.
42
Example 5
An environmental study of a certain community suggests
that the average daily level of carbon monoxide in the air
will be c( p)  0.5 p  1 parts per million when the population
is p thousand. It is estimated that t years from now the
2
p
(
t
)

10

0
.
1
t
population of the community will be
thousand.
a. Express the level of carbon monoxide in the air as a
function of time.
b. When will the carbon monoxide level reach 6.8 parts
per million?
43
Solution:
a. Since the level of carbon monoxide is related to the variable p
by the equation c( p)  0.5 p  1 , and the variable p is related to
the variable t by the equation p(t )  10  0.1t 2
It follows that the composite function
c( p(t ))  c(10  0.1t 2 )  0.5(10  0.1t 2 )  1  6  0.05t 2
expresses the level of carbon monoxide in the air as a function of
the variable t.
b. Set c(p(t)) equal to 6.8 and solve for t to get
6  0.05t 2  6.8
0.05t 2  0.8
t 2  16
t4
t  4 is not a naturalsolution.
That is, 4 years from now the level of carbon monoxide will be
6.8 parts per million.
44
Section 1.2 The Graph of a Function



The graph of a function f consists of all points (x,y) where x
is in the domain of f and y=f(x), that is, all points of the form
(x,f(x)).
Rectangular coordinate system （平面直角坐标系）,
Horizontal axis （横坐标）, vertical axis （纵坐标）.
The below example shows that the function can be sketched
by plotting a few points.
2
f ( x)   x  x  2
x
-3
-2
-1
0
1
2
3
4
f(x)
-10
-4
0
2
2
0
-4
-10
45
Intercepts





x intercepts: The points where a graph crosses the x axis.
A y intercept: A point where the graph crosses the y axis.
How to find the x and y intercepts: The only possible y
intercept for a function is y0  f (0) , to find any x
intercept of y=f(x), set y=0 and solve for x.
Note: Sometimes finding x intercepts may be difficult.
Following above example, the y intercept is f(0)=2. To
find the x intercepts, solve the equation f(x)=0, we have
x=-1 and 2. Thus, the x intercepts are (-1,0) and (2,0).
46
Parabolas （抛物线）



Parabolas: The graph of y  Ax2  Bx  C as long as A≠0.
All parabolas have a “U shape” and the parabola opens up
if A>0 and down if A<0.
The “peak” or “valley” of the parabola is called its vertex
（顶点）, and it always occurs where x   B
2A
47
Example 6
A manufacturer determines that when x hundred units of a particular
commodity are produced, they can all be sold for a unit price given by
the demand function p=60-x dollars. At what level of production is
revenue maximized? What is the maximum revenue?
Solution:
The revenue function R(x)=x(60-x) hundred dollars. Note that R(x) ≥0
only for 0≤x≤60. The revenue function can be rewritten as
R( x)   x  60x
2
which is a parabola that opens downward (Since A=-1<0) and has its
B
 60
high point (vertex) at x 

 30
2A
2( 1)
Thus, revenue is maximized when x=30 hundred units are produced,
and the corresponding maximum revenue is R(30)=900 hundred
dollars.
48
Intersections of Graphs

Sometimes it is necessary to determine when two
functions are equal.
For example, an
economist may wish to
compute the market
price at which the
consumer demand for
a commodity will be
equal to supply.

49
Power Functions, Polynomials, and Rational
Functions


n
A Power Function （幂函数）: A function of the form f ( x)  x ,
where n is a real number.
A Polynomial Function（多项式）: A function of the form
p( x)  an x n  an1 x n1   a1 x  a0
where n is a nonnegative integer and a0 , a1 ,, an are constants.
If an  0 , the integer n is called the degree （阶）of the polynomial.

p( x)
A Rational Function （有理函数）: A quotient
of two
q( x)
polynomials p(x) and q(x).
50
The Vertical Line Test
The Vertical Line Test: A curve is the graph of a
function if and only if no vertical line intersects the
curve more than once.
51
Section 1.3 Linear Functions
A linear function （线性函数）is a function that
changes at a constant rate with respect to its
independent variable.
The graph of a linear function is a straight line.
The equation of a linear function can be written in the
form
y  m x b
where m and b are constants.
52
The Slope of a Line （斜率）

The Slope of a Line: The slope of the non-vertical line
passing through the points ( x1 , y1 ) and( x2 , y2 ) is given by the
formula
changein y y y2  y1
Slope 


changein x x x2  x1
 Sign and magnitude of slope
53
Forms of the equation of a line


The Slope-Intercept Form: The equation y  m x  b is the
equation of a line whose slope is m and whose y intercept is
(0,b).
The Point-Slope Form: The equation y  y0  m( x  x0 ) is an
equation of the line that passes through the point
( x0 , y0 )and that has slope equal to m.
m
(0  0.5)
1

( 1.5  0)
3
The slope-Intercept form is
y 
1
1
x
3
2
The point-slope form that passes
through the
point (-1.5,0) is
1
y0  
3
( x  1.5)
54
Example 7
Table 1.2 lists the percentage of the labour force that was unemployed
during the decade 1991-2000. Plot a graph with the time (years after
1991) on the x axis and percentage of unemployment on the y axis.
Do the points follow a clear pattern? Based on these data, what would
you expect the percentage of unemployment to be in the year 2005?
Table 1.2 Percentage of Civilian Unemployment
Year
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
Number of Years
Percentage of
from 1991
Unemployed
0
1
2
3
4
5
6
7
8
9
6.8
7.5
6.9
6.1
5.6
5.4
4.9
4.5
4.2
4.0
55
Solution:
The pattern does suggest that we may be able to get useful
information by finding a line that “best fits” the data in
some meaningful way. One such procedure, called “leastsquares approximation”, require the approximating line to
be positioned so that the sum of squares of vertical distances
from the data points to the line is minimized.
It produces the “best-fitting line” y  0.389x  7.338 .
Based on this formula, we can attempt a prediction of the
unemployment rate in the year 2005:
y(14)  0.389(14)  7.338  1.892
Note: Care must be taken when making predictions by extrapolating
from known data, especially when the data set is as small as the one
in this example.
56
Parallel （平行）and Perpendicular（垂直）
Lines

Let m1 and m2 be the slope of the non-vertical lines L1
and L2 . Then
L1 and L2 are parallel if and only if m1  m2
1
L1 and L2 are perpendicular if and only if m2  
m1
57
Example 8
Let L be the line 4x+3y=3
a. Find the equation of a line L1 parallel to L through P(-1,4).
b. Find the equation of a line L2 perpendicular to L through Q(2,-3).
Solution:
By rewriting the equation 4x+3y=3 in the slope-intercept form
y
4
x  1,
3
4
we see that L has slope mL  
3
a. Any line parallel to L must also have slope -4/3. The required line
L1 contains P(-1,4), we have
4
4
8
y  4   ( x  1)  y   x 
3
3
3
b. A line perpendicular to L must have slope m=3/4.Since the required
3
y

3

( x  2)
line L2 contains Q(2,-3), we have
4
y 
3
9
x 
4
2
58
Section 1.4 Functional Models

To analyze a real world problem, a common procedure is to
make assumptions about the problem that simplify it enough to
allow a mathematical description. This process is called
mathematical modelling and the modified problem based on
the simplifying assumptions is called a mathematical model.
Real-world
problem
Formulation
adjustments
Mathematical
model
Testing
Analysis
Prediction
Interpretation
59
Elimination of Variables

In next example, the quantity you are seeking is expressed
most naturally in term of two variables. We will have to
eliminate one of these variables before you can write the
quantity as a function of a single variable.
Example 9
The highway department is planning to build a picnic area
for motorists along a major highway. It is to be rectangular
with an area of 5,000 square yards and is to be fenced off
on the three sides not adjacent to the highway. Express the
number of yards of fencing required as a function of the
length of the unfenced side.
60
Solution:
We denote x and y as the lengths of the sides of the picnic area.
Expressing the number of yards F of required fencing in terms of
these two variables, we get F  x  2 y . Using the fact that the area
is to be 5,000 square yards that is xy  5,000  y  5000
x
and substitute the resulting expression for y into the formula for F to
get F ( x)  x  2 5000  x  10000
 x 
x
61
Modelling in Business and Economics
Example 10
A manufacturer can produce blank videotapes at a cost of $2 per
cassette. The cassettes have been selling for $5 a piece. Consumers
have been buying 4000 cassettes a month. The manufacturer is
planning to raise the price of the cassettes and estimates that for each
$1 increase in the price, 400 fewer cassettes will be sold each month.
a: Express the manufacturer’s monthly profit as a function of the
price at which the cassettes are sold.
b: Sketch the graph of the profit function. What price corresponds to
maximum profit? What is the maximum profit?
62
Solution:
a. As we know, Profit=(number of cassettes sold)(profit per
cassette)
Let p denote the price at which each cassette will be sold
and let P(p) be the corresponding monthly profit.
Number of cassettes sold
=4000-400(number of $1 increases)
=4000-400(p-5)=6000-400p
Profit per cassette=p-2
The total profit is P( p)  (6000 400p)( p  2)
 400p 2  6800p  12000
63
b. The graph of P(p) is the downward opening parabola
shown in the bottom figure. Profit is maximized at the
value of p that corresponds to the vertex of the parabola.
We know p   B   6800  8.5
2A
2(400)
Thus, profit is maximized when the manufacturer charges
$8.50 for each cassette, and the maximum monthly profit
is Pmax  P(8.5)  400(8.5)2  6800(8.5) 12000 $16900
64
Market Equilibrium
The law of supply and demand: In a competitive market
environment, supply tends to equal demand, and when this
occurs, the market is said to be in equilibrium.
The demand function: p=D(x)
The supply function: p=S(x)
The equilibrium price:
pe  D( xe )  S ( xe )
Shortage: D(x)>S(x)
Surplus: S(x)>D(x)
65
Example 11
Market research indicates that manufacturers will supply x
units of a particular commodity to the marketplace when the
price is p=S(x) dollars per unit and that the same number of
units will be demanded by consumers when the price is
p=D(x) dollars per unit, where the supply and demand
functions are given by
2
S ( x)  x  14 D( x)  174 6x
a. At what level of production x and unit price p is market
equilibrium achieved?
b. Sketch the supply and demand curves, p=S(x) and
p=D(x), on the same graph and interpret.
66
Solution:
a. Market equilibrium occurs when S(x)=D(x), we have
x 2  14  174 6 x
( x  10)(x  16)  0
x  10 or  16
Only positive values are meaningful, pe  D(10)  174 6(10)  114
67
Break-Even Analysis
At low levels of production, the manufacturer suffers
a loss. At higher levels of production, however, the total
revenue curve is the higher one and the manufacturer
realizes a profit.
Break-even point : The total revenue equals total cost.
68
Example 12
A manufacturer can sell a certain product for $110 per unit.
Total cost consists of a fixed overhead of $7500 plus
production costs of $60 per unit.
a. How many units must the manufacturer sell to break even?
b.What is the manufacturer’s profit or loss if 100 units are
sold?
c.How many units must be sold for the manufacturer to
realize a profit of $1250?
Solution:
If x is the number of units manufactured and sold, the total
revenue is given by R(x)=110x and the total cost by
C(x)=7500+60x
69
a. To find the break-even point, set R(x) equal to C(x) and solve
110x=7500+60x, so that x=150.
It follows that the manufacturer will have to sell 150 units to break
even.
b. The profit P(x) is revenue minus cost. Hence,
P(x)=R(x)-C(x)=110x-(7500+60x)=50x-7500
The profit from the sale of 100 units is P(100)=-2500
It follows that the manufacturer will lose $2500 if 100 units are
sold.
c. We set the formula for profit P(x) equal to 1250 and solve for x,
we have P(x)=1250, x=175. That is 175 units must be sold to
generate the desired profit.
70
Example 13
A certain car rental agency charges $25 plus 60 cents per
mile. A second agency charge $30 plus 50 cents per mile.
Which agency offers the better deal?
Solution:
Suppose a car is to be driven x miles, then the first agency
will charge C1 ( x)  25  0.60xdollars and the second will charge
C2 ( x)  30  0.50x . So that x=50.
For shorter distances, the first agency offers the better deal,
and for longer distances, the second agency is better.
71
Section 1.5 Limits （极限）


Roughly speaking, the limit process involves examining the
behavior of a function f(x) as x approaches a number c that
may or may not be in the domain of f.
To illustrate the limit process, consider a manager who
determines that when x percent of her company’s plant
capacity is being used, the total cost is
8 x 2  636x  320
C ( x) 
x 2  68x  960
hundred thousand dollars. The company has a policy of
rotating maintenance in such a way that no more than
80% of capacity is ever in use at any one time. What cost
should the manager expect when the plant is operating at
full permissible capacity?
72
It may seem that we can answer this question by simply
evaluating C(80), but attempting this evaluation results in
the meaningless fraction 0/0. However, it is still possible
to evaluate C(x) for values of x that approach 80 from the
left (x<80) and the right (x>80), as indicated in this table:
x approaches 80 from the left →
x
C(x)
79.8
79.99
79.999
6.99782
6.99989
6.99999
←x approaches 80 from the right
80
80.0001
80.001
80.04
7.000001
7.00001
7.00043
The values of C(x) displayed on the lower line of this table
suggest that C(x) approaches the number 7 as x gets closer
and closer to 80. The functional behavior in this example
can be describe by lim C ( x)  7
x 80
73
Limit: If f(x) gets closer and closer to a number L as x gets
closer and closer to c from both sides, then L is the limit of
f(x) as x approaches c. The behavior is expressed by writing
lim f ( x)  L
x c
74
Example 14
Use a table to estimate the limit
x 1
lim
x 1 x  1
Solution:
Let
f ( x) 
x 1
x 1
and compute f(x) for a succession of values
of x approaching 1 from the left and from the right.
x→ 1
x
0.99
0.999
0.9999
f(x)
0.50126
0.50013
0.50001
←x
1
1.00001
1.0001
1.001
0.499999
0.49999
0.49988
The numbers on the bottom line of the table suggest that
f(x) approaches 0.5 as x approaches 1. That is
lim
x 1
x 1
 0.5
x 1
75
It is important to remember that limits describe the behavior
of a function near a particular point, not necessarily at the
point itself.
f ( x)  4
Three functions for which lim
x 3
76
The figure below shows that the graph of two functions
that do not have a limit as x approaches 2.
Figure (a): The limit does not exist; Figure (b): The
function has no finite limit as x approaches 2. Such socalled infinite limits will be discussed later.
77
Properties of Limits
If lim
f ( x) and lim
g ( x) exist, then
x c
x c
lim [ f ( x)  g ( x)]  lim f ( x)  lim g ( x)
xc
xc
xc
lim [ f ( x)  g ( x)]  lim f ( x)  lim g ( x)
xc
xc
xc
lim kf ( x)  k lim f ( x) for any constant k
xc
xc
lim [ f ( x) g ( x)]  [ lim f ( x)][ lim g ( x)]
xc
xc
xc
lim f ( x)
f ( x)
xc
lim [
]
if lim g ( x)  0
x  c g ( x)
xc
lim g ( x)
xc
lim [ f ( x)] p  [ lim f ( x)] p if [ lim f ( x)] p exists
xc
xc
xc
78
For any constant k,
lim k  k
x c
and
lim x  c
x c
That is, the limit of a constant is the constant itself, and
the limit of f(x)=x as x approaches c is c.
79
Computation of Limits
Example 15
Find (a)
lim (3 x 3  4 x  8)
(b)
x  1
3x 3  8
lim
x 0
x2
Solution:
a. Apply the properties of limits to obtain
lim (3x  4 x  8)  3lim x   4lim x  lim 8  3(1)  4(1)  8  9
3
3
x 1
x 1
3
x 1
x 1
b. Since lim( x  2)  0, you can use the quotient rule for
x 0
limits to get
3 lim x 3  lim 8
3x  8
0 8
x 0
x 0
lim


4
x 0
x2
lim x  lim 2
02
3
x 0
x 0
80
Limits of Polynomials and Rational Functions: If p(x)
and q(x) are polynomials, then
lim p ( x )  p (c )
x c
and
p ( x)
p (c )
lim

x c q ( x )
q (c )
if q (c)  0
Example 16
Find
lim
x2
x 1
x2
Solution:
The quotient rule for limits does not apply in this case
since the limit of the denominator is 0 and the limit of the
numerator is 3. So the limit of the quotient does not exist.
81
Indeterminate Form （不定形）
g ( x )  0 , then lim f ( x ) is said to be
f ( x)  0 and lim
If lim
x c
xc
g ( x)
indeterminate. The term indeterminate is used since the
limit may or may not exist.
xc
Example 17
x2 1
lim 2
x 1 x  3 x  2
(a) Find
x 1
(b) Find lim
x 1 x  1
Solution:
a.
x2 1
( x  1)(x  1)
x 1
2
lim 2
 lim
 lim

 2
x 1 x  3x  2
x 1 ( x  1)(x  2)
x 1 x  2
1
b. lim x 1  lim  x 1 x  1  lim
x1
x 1
x1


( x 1) x  1
x 1
1
1
 lim

x1 ( x  1) x  1
x1
x 1 2


82
Limits Involving Infinity
Limits at Infinity: If the value of the function f(x) approach the
number L as x increases without bound, we write
lim f ( x )  L
x
Similarly, we write lim f ( x)  M
x  
when the functional values f(x) approach the number M as x
decreases without bound.
83
Reciprocal Power Rules: For constants A and k, with k>0
A
lim k  0
x  x
and
A
lim k  0
x   x
Example 18
Find
x2
lim
x  1  x  2 x 2
Solution:
lim 1
x2
x2 / x2
1
x 
lim
 lim 2


 0.5
2
2
2
2
2
x   1  x  2 x
x  1 / x  x / x  2 x / x
lim 1 / x  lim 1 / x  lim 2 0  0  2
x 
x  
x  
84
Procedure for Evaluating a Limit at Infinity of f(x)=p(x)/q(x)
Step 1. Divide each term in f(x) by the highest power xk that
appears in the denominator polynomial q(x).
f ( x ) or lim f ( x ) using algebraic
Step 2. Compute xlim

x  
properties of limits and the reciprocal rules.
Exercise
3x 4  8 x 2  2 x
lim
x 
5x 4  1
sin( x)
lim
x 
x
(Optional Question!)
Example 19
If a crop is planted in soil where the nitrogen level is N,
then the crop yield Y can be modeled by the MichaelisMenten function Y ( N )  AN
N 0
BN
where A and B are positive constants. What happens to
crop yield as the nitrogen level is increased indefinitely?
85
Solution:
We wish to compute
AN
lim Y ( N )  lim
N 
N  B  N
AN / N
 lim
N  B / N  N / N
A
A
 lim

N  B / N  1
0 1
A
Thus, the crop yield tends toward the constant value A as
the nitrogen level N increases indefinitely. For this reason,
A is called the maximum attainable yield.
86
Infinite Limits （无穷极限）: If f(x) increases or decreases
without bound as x→c, we have
lim f ( x)   
x c
or
lim f ( x)   
x c
x
For example lim
x  2 ( x  2) 2
From the figure, we can
guest that
x
lim

2
x 2 ( x  2)
87
Section 1.6 One-sided Limits and Continuity
If f(x) approaches L as x tends toward c from the
left (x<c), we write
lim f ( x)  L
x c
where L is called the limit from the left (or lefthand limit) （左极限）
Likewise if f(x) approaches M as x tends toward
c from the right (x>c), then lim f ( x )  M
xc 
M is called the limit from the right (or right-hand
limit.) （右极限）
88
Example 20
For the function
1  x 2 if x  2
f ( x)  
2 x  1 if x  2
evaluate the one-sided limits lim f ( x) and lim f ( x)
x2

x2
Solution:
2
Since f ( x)  1  x for x<2, we have
lim f ( x)  lim (1  x 2 )  3
x 2
x 2
Similarly, f(x)=2x+1 if x≥2, so
lim f ( x )  lim ( 2 x  1)  5
x2
x2
89
f ( x)
Existence of a Limit: The two-sided limit lim
x 2
exists if and only if the two one-sided limits lim f ( x)
x2
f ( x) exist and are equal, and then
and xlim
2


lim f ( x)  lim f ( x)  lim f ( x)
x2
x2
x2
Notice that the limit of the piecewise-define
function f(x) in example 20 does not exist, that is
lim f ( x ) does not exist, since xlim
2
x 2

f ( x )  lim f ( x )
x2
90
lim f ( x ) does not exist!
x1
Since the left and right
hand limits are not
equal.
At x=1:
lim f  x   0
Left-hand limit
lim f  x   1
Right-hand limit
f 1  1
value of the function
x 1
x 1
91
lim f ( x ) does exist!
x 2
Since the left and right
hand limits are equal,
However, the limit is
not equal to the value
of function.
At x=2:
lim f  x   1
Left-hand limit
lim f  x   1
Right-hand limit
x 2
x 2
f  2  2
value of the function
92
lim f ( x ) does exist!
x 3
Since the left and right
hand limits are equal,
and the limit is equal to
the value of function.
f  x  2
At x=3: xlim

3
lim f  x   2
x 3
f  3  2
Left-hand limit
Right-hand limit
value of the function
93
Nonexistent One-sided Limits
A simple example is provided by the function
f ( x)  sin(1 / x)
As x approaches 0 from either the left or the right, f(x) oscillates
between -1 and 1 infinitely often. Thus neither one-sided limit at 0
exists.
94
Continuity （连续性）
A continuous function is one whose graph can be
drawn without the “pen” leaving the paper. (no holes or
gaps )
95
A “hole “ at x=c
96
A “gap” at x=c
97
So what properties will guarantee that f(x) does not
have a “hole” or “gap” at x=c?
Continuity: A function f is continuous at c if all three of
these conditions are satisfied:
a. f (c) is defined
b. lim f ( x ) exists
x c
c. lim f ( x)  f (c)
x c
If f(x) is not continuous at c, it is said to have a
discontinuity there.
98
f(x) is continuous at
x=3 because the left
and right hand limits
exist and equal to f(3).
At x=1: lim f ( x )  lim f ( x )
x 1

x 1

At x=2: lim f ( x )  lim f ( x )  f ( 2)
x  2
At x=3:
x  2
lim f ( x )  lim f ( x )  f ( 3)
x  3
x3
Discontinuous
Discontinuous
Continuous
99
Continuity Polynomials and Rational Functions
 If p(x) and q(x) are polynomials, then
lim p( x)  p(c)
xc
p ( x ) p (c )
lim

if q(c)  0
x  c q ( x)
q (c )
A polynomial or a rational function is continuous
wherever it is defined
100
Example 21
x 1
Show that the rational function f ( x) 
is continuous
x2
at x=3.
Solution:
( x  2)  0 , you
Note that f(3)=(3+1)/(3-2)=4, since lim
x 3
will find that
lim( x  1)
x 1
4
lim f ( x)  lim
 x 3
  4  f (3)
x 3
x 3 x  2
lim( x  2) 1
x 3
as required for f(x) to be continuous at x=3, since the
three criteria for continuity are satisfied.
101
Example 22
Determine where the function below is not continuous.
Solution:
Rational functions are continuous everywhere except
where we have division by zero.
The function given will not be continuous at t=-3 and
t=5.
102
Exercise
Discuss the continuity of each of the following functions
1
a. f ( x) 
x
x2 1
b. g ( x) 
x 1
 x  1 if x  1
c. h( x)  
2  x if x  1
103
Example 23
For what value of the constant A is the following function
continuous for all real x?
Solution:
if x  1
 Ax  5
f ( x)   2
 x  3x  4 if x  1
Since Ax+5 and x 2  3x  4 are both polynomials, it follows
that f(x) will be continuous everywhere except possibly at
x=1 . According to the three criteria for continuity, we
have
lim f ( x)  lim f ( x)  f (1)  A  5  2  f (1)  A  3
x 1
x 1
This means that f is continuous for all x only when A=-3
104
Example 24
Find numbers a and b so that the following function is
continuous everywhere.
ax
if x  -1


f ( x )   x 2  a  b if  1  x  1

bx
if x  1

Solution:
Since the “parts” f are polynomials, we only need to
choose a and b so that f is continuous at x=-1 and 1.
f ( x)  lim f ( x)  f (1)  a  1  a  b  2a  b  1
At x=-1 xlim
 1
x  1
f ( x)  lim f ( x)  f (1)  1  a  b  b  a  2b  1
At x=1 xlim
1
x 1




We have a=-1/3 and b=1/3 for f is continuous everywhere
105
Continuity on an Interval


A function f(x) is said to be continuous on an open interval
a<x<b if it is continuous at each point x=c in that interval.
f is continuous on closed interval a≤x≤b, if it continuous
on the open interval a<x<b and
lim f ( x)  f (a )
xa

lim f ( x )  f (b)
x b
f ( x)  1  x 2 is continuous on [-1,1]
106
Example 25
f ( x) 
x2
x 3
Discuss the continuity of the function
on the
open interval -2<x<3 and on the closed interval -2≤x≤3
Solution:
The rational function f(x)is continuous for all x except
x=3. Therefore, it is continuous on the open interval
-2<x<3 but not on the closed interval -2≤x≤3,since it is
discontinuous at the endpoint 3 (where its denominator is
zero). The graph of f is shown in below Figure.
107
The Intermediate Value Property

Suppose that f(x) is continuous on the interval a≤x≤b and L
is a number between f(a) and f(b), then there exists a
number c between a and b, such that f(c)=L.
108
Corollary
If f is continuous on the closed interval [a,b], and f(a)
and f(b) have opposite signs, then there exists a number c
in (a,b) where f(c)=0.
109
Example 26
Show that the equation
1<x<2
x2  x 1 
1
x 1
has a solution for
Solution:
f ( x)  x 2  x  1 
1
.
x1
Let
Then
f(1)=-3/2 and f(2)=2/3. Since f(x)
is continuous for 1≤x≤2, it
follows from the intermediate
value property that the graph
must cross the x axis somewhere
between x=1 and x=2.
110
Summary
 Function:
domain and range of a function
composition of function f(g(x))
 Graph of a function:
x and y intercepts,
Piecewise-defined function, power function
Polynomial, Rational function, Vertical line test
 Linear function:
Slope, Slope-intercept formula, point-slope formula
Parallel and perpendicular lines.
111
Function Models:
Market equilibrium: law of supply and demand
Shortage and surplus, Break-even analysis
 Limits: lim f ( x )  L
xc
Calculation of limits: limits of polynomial and
rational function, limits at infinity: calculation of
limits at the infinity (Reciprocal power Rules),
Infinite limit, One sided limit, Existence of limit
 Continuity of f(x) at x=c:
Continuity on an interval, Continuity of polynomials
and rational function, Intermediate value property
112