CALCULUS For Business, Economics, and the Social and life by blc16452

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									CALCULUS
For Business, Economics, and the Social and life Sciences

                          Hoffmann, L.D. & Bradley, G.L.



    Instructor: Dr. Guangqing Lu
    TA: Niu Qiang
    Office: E409-R4
    Tel: 3620635(O) 156-028-63738 (Cell)
     Email: guangqinglu@uic.edu.hk

                        Uinted International College        1
They invented Calculus!




Sir Isaac Newton   Gottfriend Wilhelm von Leibniz
   (1642-1727)
                            (1646-1716)             2
Calculus has practical applications, such as understanding the true
meaning of the infinitesimals. (Image concept by Dr. Lachowska.)
                                                                      3
    What is Calculus all about?
 Calculus is the study of change, or to be more
  precise, changing quantities.


 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.
                                                   4
Why do I have to take Calculus?
             If you choose a career path that
              requires any level of mathematical
              sophistication (such as Engineering,
              Economics or any branch of Science),
              you will need Calculus.

             To succeed in Calculus (and
              Mathematics in general) you must be
              able to: solve problems, reason
              logically and understand abstract
              concepts. The ability to master these
              skills is a key indicators of success in
              future academic endeavors.
                                                    5
What can I do to maximize my chances
             for success?
Top five
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.
                                                6
          Some Suggestions
It is recommended that you spend at least 2
 hours study and review, for each hour of class
 time.

You must pay more attention to weekly
 tutorials where you can work on problems and
 get help from your demonstrator.


                                                  7
      20%


                  Home Works

                  Little Quizzes
50%         10%
                  Mid-term
                  Examination
                  Final Examination



      20%



                                   8
               Chapter 1
      Functions, Graphs and Limits

  In this Chapter, we will encounter some
  important concepts.

 Functions

 Limits

 One-sided Limits and Continuity

                                            9
        Section 1.1 Functions
 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 of the function, and
  the set of assigned objects in B is called the range.




                                                          10
Function, or not?
                        f
          A                          B


                     YES
     f
A
                                           f
              B                      A         B



                                          NO
     NO
                  Section 1.1 Functions            11
To be convenient, we represent a functional
 relationship by an equation y  f (x)

In this context, x and y are called variable,
 furthermore, we refer to y as the dependent
 variable and to x as the independent variable.
 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 2  4
                         Section 1.1 Functions    12
Function that describes tabular data
Table 1.1 Average Tuition and Fees for 4-Year Private Colleges

Academic Year                                           Tuition and
Ending in                     Period n                   Fees
1973                          1                         $1,898
1978                          2                         $2,700
1983                          3                         $4,639
1988                          4                         $7,048
1993                          5                         $10,448
1998                          6                         $13,785
2003                          7 Section 1.1 Functions   $18,273       13
 So how can we use functional notation to describe this
  kind of data that is listed in Table 1.1?

 We can describe this data as a function f defined by
  the rule
                average tuition and fees at the      
        f (n)  
                 beginning of the nth 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

                         Section 1.1 Functions               14
Piecewise-defined function
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
                    
                   Section 1.1 Functions          15
Example 1
Find f(-1/2), f(1), and f(2), where the piecewise-defined
function f(x) is given at the foregoing slide.
Solution:

           1
Since x   satisfies x<1, use the top part of the formula
           2
to find
               1     1       1        2
            f                   
               2  1/ 2 1  3 / 2    3
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:
 f (1)  3(1) 2  1  4 and f (2)  3(2) 2  1  13
                         Section 1.1 Functions           16
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

                    Section 1.1 Functions             17
Example 2
Find the domain and range of each of these functions
                1
a.   f ( x)                        b.   g (u )  4 u  2
              1 x 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.

                          Section 1.1 Functions                  18
 Functions Used in Economics
 A demand function 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 is given by the product
    R(x)=(number of items sold)(price per item)
         =xp=xD(x)

 If C(x) is the total cost of producing the x units, then the
  profile is given by the function P(x)=R(x)-C(x)=xD(x)-
  C(x)

                         Section 1.1 Functions              19
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  0.27 x  51 dollars. The cost of
producing the x thousand units is
              C ( x)  2.23 x 2  3.5 x  85
thousand dollars

 a. What are the revenue and profile functions, R(x) and
    P(x), for this production process?
 b. For what values of x is production of the coffee
    makers profitable? Section 1.1 Functions               20
Solution:

a. The demand function is         D( x)  0.27 x  51,   so the revenue is
               R( x)  xD( x)  0.27 x 2  51 x
  thousand dollars, and the profit is (thousand dollars)
              P( x)  R( x)  C ( x)
                     0.27 x 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  47.5 x  85
                    2.5( x 2  19 x  34)
                    2.5( x  2)( x  17)  0
  Thus, production is profitable for 2<x<17.
                       Section 1.1 Functions                           21
           Composition of Functions
 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 )  u  1 and
                                                        3

  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  x 3  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.
                                  Section 1.1 Functions                22
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
population of the community will be p(t )  10  0.1t 2
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?

                        Section 1.1 Functions               23
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 natural solution.
That is, 4 years from now the level of carbon monoxide will be
6.8 parts per million.                                                  24
        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, Horizontal axis, vertical axis.
 The below example shows that the function can be sketched by
  plotting a few points.


                                  f ( x)   x  x  2   2


                              x      -3    -2   -1   0       1   2   3     4

                              f(x)   -10   -4   0    2       2   0   -4    -10

                                                                          25
                    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).

                     Section 1.2 The Graph of a         26
                              Function
                    Parabolas
 Parabolas: The graph of y  Ax 2  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, and it always occurs where x   B
                                            2A




                                                         27
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 2  60 x
which is a parabola that opens downward (Since A=-1<0) and has its
high point (vertex) at x   B   60  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.                Section 1.2 The Graph of a           28
                               Function
       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.

                   Section 1.2 The Graph of a       29
                            Function
Power Functions, Polynomials, and
       Rational Functions
 A Power Function: A function of the form f ( x)  x , where n is
                                                        n



  a real number.
 A Polynomial Function: 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 of the polynomial.
                                      p( x)
 A Rational Function: A quotient     q( x)
                                              of two polynomials p(x)
  and q(x).




                                                                   30
         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.




                                                     31
    Section 1.3 Linear Functions
A linear function 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  mx  b
  where m and b are constants.
                                                           32
              The Slope of a Line
 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
                     change in y y y2  y1
             Slope               
                     change in x x x2  x1

 Sign and magnitude of slope




                                                                  33
 Forms of the equation of a line
 The Slope-Intercept Form: The equation y  mx  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.
                                     (0  0.5)      1
                             m                  
                                    ( 1.5  0)     3
                             The slope-Intercept form is
                                       1    1
                              y        x
                                       3    2
                             The point-slope form that passes
                             through the point (-1.5,0) is
                                                 1
                               y 0              ( x  1.5)
                    Section 1.3 Linear Functions 3              34
  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

         Number of Years   Percentage of
 Year    from 1991         Unemployed
 1991      0                  6.8
 1992      1                  7.5
 1993      2                  6.9
 1994      3                  6.1
 1995      4                  5.6
 1996      5                  5.4
 1997      6                  4.9
 1998      7                  4.5
 1999      8                  4.2
 2000      9                  4.0     Section 1.4 Functional Models   35
 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 “least-
squares 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
                                                                  36
in this example.
  Parallel and Perpendicular Lines
 Let m1 and m2 be the slope of the non-vertical lines L1
  and L2 . Then
     L1and L2 are parallel if and only if m1  m2
                                                      1
     L1and L2 are perpendicular if and only if m2  
                                                     m1




                    Section 1.3 Linear Functions       37
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
     4                                          4
y   x  1,   we see that L has slope mL  
     3                                          3
a. Any line parallel to L must also have slope -4/3. The required line
   L1 contains P(-1,4), we have y  4   4 ( x  1)  y   4 x  8
                                           3               3    3
b. A line perpendicular to L must have slope m=3/4.Since the required
   line L2 contains Q(2,-3), we have y  3  3 ( x  2)
                                                             4
                                                             3    9
                           Section 1.4 Functional Models y    x     38
                                                             4    2
  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
         Testing                           model

                                           Analysis
             Prediction
                          Interpretation                  39
         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.
                     Section 1.4 Functional Models      40
 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




                          Section 1.4 Functional Models             41
                 Proportionality
 The quantity Q is said to be:
  directly proportional to x if Q=kx for some constant k
  inversely proportional to x if Q=k/x for some constant k
  jointly proportional to x and y if Q=kxy for some
  constant k
 For example, When environmental factors impose an
  upper bound on its size, population grows at a rate that is
  jointly proportional to its current size and the difference
  between its current size and the upper bound.
  Let p denote the size of the population, R(p) the rate of
  population growth, and b the upper bound. So we have
                              where Models
             R(p)=kp(b-p)1.4 Functionalk is a constant.
                      Section                               42
           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?

                          Section 1.4 Functional Models              43
Solution:
a. As we know, Profit=(number of cassettes sold)(profit per
   cassette)
  Let p denotes 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  400 p)( p  2)
                                400 p 2  6800 p  12000
                        Section 1.4 Functional Models        44
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.
                    B     6800
   We know     p            8. 5
                     2A      2(400 )
  Thus, profit is maximized when the manufacturer
  charges $8.50 for each cassette, and the maximum
  monthly profit is
          Pm ax  P(8.5)  400 (8.5) 2  6800 (8.5)  12000  $16900




                          Section 1.4 Functional Models                 45
              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)
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
            S ( x)  x 2  14 D( x)  174  6 x
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.
                      Section 1.4 Functional Models       47
 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




                         Section 1.4 Functional Models            48
         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.




                    Section 1.4 Functional Models        49
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         Section 1.4 Functional Models       50
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.
                       Section 1.4 Functional Models      51
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  0.60 xdollars and the second will charge
C2 ( x)  30  0.50 x . So that x=50.
For shorter distances, the first agency offers the better deal,
and for longer distances, the second agency is better.
                         Section 1.4 Functional Models          52
              Section 1.5 Limits
 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  636 x  320
              C ( x) 
                         x 2  68 x  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? Functional Models
                      Section 1.4                       53
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 approaches 80 from the right
  x        79.8     79.99    79.999       80 80.0001    80.001   80.04

 C(x)      6.99782 6.99989 6.99999           7.000001   7.0000   7.00043
                                                        1

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 xlim C( x)  7Limits
                     80 Section 1.5                     54
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




                           Section 1.5 Limits              55
 Example 14
                                                             x 1
Use a table to estimate the limit                     lim
                                                       x 1 x  1

 Solution:
                    x 1
Let     f ( x) 
                   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
 x       0.99        0.999       0.9999           1      1.00001    1.0001    1.001
 f(x)    0.50126     0.50013     0.50001                 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
                                            x 1
                                 lim                    
                                          Section 1.5 Limits0.5
                                 x 1      x 1                                       56
It is important to remember that limits describe the behavior
of a function near a particular point, not necessarily at the
point itself.




                                  x)  4
 Three functions for which lim f (Section 1.5 Limits     57
                               x 3
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 so-
                                 discussed later.
called infinite limits will be 1.5 Limits
                         Section                         58
              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



                       f ( x)    lim f ( x)
                                 xc
                 lim [        ]            if lim g ( x)  0
                 x  c g ( x)    lim g ( x)    xc
                                 xc


               lim [ f ( x)] p  [ lim f ( x)] p if [ lim f ( x)] p exists
               xc               xc                   xc


                                  Section 1.5 Limits                         59
For any constant k,
                lim k  k       and        lim x  c
                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.




                            Section 1.5 Limits                 60
             Computation of Limits
 Example 15
                                                              3x 3  8
Find (a)      lim (3x 3  4 x  8)             (b)       lim
              x 1                                      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)3  4(1)  8  9
 x 1
         3
                       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
              3x  8
                3          3 lim x 3  lim 8
                                        0 8
         lim           x 0     x 0
                                            4
         x 0  x2     lim x  lim 2    02
                               x 0        x 0

                                Section 1.5 Limits                       61
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                       if q(c)  0
               x c q ( x )   q (c )

Example 16
             x 1
Find   lim
       x2   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.
                               Section 1.5 Limits        62
                      Indeterminate Form
                                                                    f ( x)
     lim f ( x)  0
If xc      and xc     lim g ( x)  0
                            , then             is said to be lim
                                                             x c   g ( x)
indeterminate. The term indeterminate is used since the
limit may or may not exist.
Example 17
                       x2 1                               x 1
(a) Find       lim 2                       (b) Find lim
                x 1 x  3 x  2                     x 1 x  1

 Solution:
              x2 1             ( x  1)( x  1)         x 1    2
a.    lim 2              lim                     lim              2
       x 1 x  3 x  2    x 1 ( x  1)( x  2)    x 1 x  2   1

            x 1
                  lim
                                  
                            x 1 x  1
                                        lim
                                                     x 1
                                                               lim
                                                                                  1
                                                                                     
                                                                                       1
b.   lim
      x 1 x  1    x 1 ( x  1)  
                                  x 1         
                                          x 1 ( x  1)  x  1 x1             x 1 2
                                        Section 1.5 Limits                          63
           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.




                                Section 1.5 Limits                   64
Reciprocal Power Rules: For constants A and k, with k>0
                            A                           A
                      lim  0           and       lim  0
                     x   x k                  x   x k

 Example 18
                      x2
 Find       lim
            x 1  x  2 x 2


  Solution:
          x2                        x2 / x2                        lim 1                 1
lim                 lim                                         x 
                                                                                               0.5
x 1  x  2 x 2   x  1 / x  x / x  2 x / x
                                2       2     2    2
                                                       lim 1 / x  lim 1 / x  lim 2 0  0  2
                                                                2
                                                    x        x       x




                                         Section 1.5 Limits                                   65
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).
Step 2. Compute xlim f ( x) or lim f ( x) using algebraic
                               x 
properties of limits and the reciprocal rules.

Exercise            3x 4  8 x 2  2 x              lim
                                                         sin( x)
               lim                                                 (Optional Question!)
               x      5x 4  1                  x     x
 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 Michaelis-
Menten 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?
                              Section 1.5 Limits                                66
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.
                          Section 1.5 Limits              67
Infinite Limits: If f(x) increases or decreases without bound
as x→c, we have

        lim f ( x)    or              lim f ( x)   
        xc                              xc
                        x
For example lim
            x2    ( x  2) 2
                                              From the figure, we can
                                              guest that
                                                          x
                                               lim              
                                               x  2 ( x  2) 2




                         Section 1.5 Limits                       68
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 left-
hand limit)
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.)                                           69
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:
                      Since f ( x)  1  x 2 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



                     Section 1.6 One-sided Limits and                   70
                                Continuity
Existence of a Limit: The two-sided limit lim2 f ( x)
                                             x
exists if and only if the two one-sided limits lim f ( x)               
                                                x 2
and lim f ( x) exist and are equal, and then
    x 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                   lim f ( x)  lim f ( x)
 x2                                                x 2        x 2



                      Section 1.6 One-sided Limits and                      71
                                 Continuity
                                             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
          x 1

          lim f  x   1
             
                                        Right-hand limit
          x 1

            f 1  1              value of the function
                  Section 1.6 One-sided Limits and                  72
                             Continuity
                                               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
          x 2

          lim f  x   1
             
                                        Right-hand limit
          x 2

           f  2  2              value of the function
                  Section 1.6 One-sided Limits and                73
                             Continuity
                                                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.




At x=3: x3 f  x   2
        lim                              Left-hand limit

           lim f  x   2
              
                                         Right-hand limit
           x 3

           f  3  2               value of the function
                   Section 1.6 One-sided Limits and                 74
                              Continuity
Nonexistent One-sided Limits
A simple example is provide 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.
                      Section 1.6 One-sided Limits and           75
                                 Continuity

								
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