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§10.1 Introduction to Limits
Dr .Hayk Melikyan
Departmen of Mathematics and CS
[email protected]
H.Melikian
1
Learning Objectives for Section 10.1
Introduction to Limits
The student will learn about:
■ Functions and graphs
■ Limits: a graphical approach
■ Limits: an algebraic approach
■ Limits of difference quotients
H.Melikian
Barnett/Ziegler/Byleen College
2
Functions and Graphs
A Brief Review
The graph of a function is the graph of the set of all ordered
pairs that satisfy the function. As an example, the following
graph and table represent the function f (x) = 2x – 1.
x
-2
-1
0
1
2
3
H.Melikian
Barnett/Ziegler/Byleen College
f (x)
-5
-3
-1
1
?
?
We will use this
point on the
next slide.
3
Analyzing a Limit
We can examine what occurs at a particular point by the limit
ideas presented in the previous chapter. 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
2.8
2.98 2.998 ? 3.002 3.02 3.2
4
We see that as x approaches 2, f (x) approaches 3.
H.Melikian
Barnett/Ziegler/Byleen College
4
Limits
In limit notation we have
lim 2 x  1  3.
x2
Definition: We write
lim f ( x)  L
3
2
xc
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).
H.Melikian
Barnett/Ziegler/Byleen College
5
One-Sided Limits
We write
lim f ( x)  K
xc
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
lim number
f ( x)  Lline.
of c on the real
xc
We write
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.
H.Melikian
Barnett/Ziegler/Byleen College
6
Example 1
4
2
On the other hand:
2
lim f ( x)  4
x4
4
lim f ( x)  4
x4
lim f ( x)  4
x2
lim f ( x)  2
x2
Since these two are not the
same, the limit does not exist
at 2.
H.Melikian
Since the limit from the left and
the limit from the right both
exist and are equal, the limit
exists at 4:
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lim f ( x)  4
x4
7
Limit Properties
Let f and g be two functions, and assume that the
following two limits exist and are finite:
lim f ( x)  L and
xc
lim g ( x)  M
x c
Then
 the limit of a constant is the constant.
 the limit of x as x approaches c is c.
 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.
H.Melikian
Barnett/Ziegler/Byleen College
8
Limit Properties
(continued)




H.Melikian
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.
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9
Examples 2, 3
lim x  3x  lim x  lim 3x  4  6  2
2
x 2
2
x 2
x 2
lim 2 x
2x
8
x 4
lim


x4 3x  1
lim 3x  1 13
x 4
From these examples we conclude that
1.lim f ( x)  f (c)
f any polynomial function
2.lim r ( x)  r (c)
r any rational function with a
nonzero denominator at x = c
x c
x c
H.Melikian
Barnett/Ziegler/Byleen College
10
Indeterminate Forms
It is important to note that there are restrictions on some of
the limit properties. In particular if lim r ( x)  0
x c
then finding lim
xc
denominator is 0.
If
f ( x ) may present difficulties, since the
r ( x)
lim f ( x )  0 and lim g ( x )  0 , then lim
x c
xc
x c
f ( x)
g ( x)
is said to be indeterminate.
The term “indeterminate” is used because the limit may or
may not exist.
H.Melikian
Barnett/Ziegler/Byleen College
11
Example 4
This example illustrates some techniques that can be useful for
indeterminate forms.
x2  4
( x  2)( x  2)
lim
 lim
 lim( x  2)  4
x 2 x  2
x 2
x 2
x2
Algebraic simplification is often useful when the numerator and
denominator are both approaching 0.
H.Melikian
Barnett/Ziegler/Byleen College
12
Difference Quotients
f ( a  h)  f ( a )
lim
.
Let f (x) = 3x - 1. Find h0
h
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Barnett/Ziegler/Byleen College
13
Difference Quotients
f ( a  h)  f ( a )
lim
.
Let f (x) = 3x - 1. Find h0
h
Solution:
f (a  h)  3(a  h)  1  3a  3h 1
f (a)  3a  1
f (a  h)  f (a)  3h
f ( a  h)  f ( a )
3h
lim
 lim  3.
h 0
h 0 h
h
H.Melikian
Barnett/Ziegler/Byleen College
14
Example 2
lim x2 – 3x =
x2
lim x2 - lim 3x = 4 – 6 = -2.
x2
x2
From this example we can see that for a
polynomial function
lim f (x) = f (c)
x c
H.Melikian
15
Indeterminate Form
f ( x)
lim
If lim
and
,
then
f
(
x
)

0
lim
g
(
x
)

0
xc
xc
xc
g ( x)
is said to be indeterminate.
The term indeterminate is used because the limit may
or may not exist.
H.Melikian
16
The Limit of a Difference Quotient.
f (a  h )  f (a )
Let f (x) = 3x –1. Find lim
h
h0
f (a + h) = 3 (a + h) - 1 = 3a + 3h - 1
f ( a ) = 3 ( a ) – 1 = 3a - 1
f (a + h) – f (a) = 3h
f (a  h )  f (a )
3 h lim 3 
3
lim
 lim

h0
h0
h0 h
h
H.Melikian
17
Practice Problems ( section 10.1)
# 1, 5, 11, 13, 15, 17, 21, 39, 43, 46.
49, 51, 55, 69.
Practice Problems ( section 10.2)
# 1, 5, 7, 11, 15, 19, 21, 25, 27, 31, 33, 41,
53, 63.
H.Melikian
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Summary
■ We started by using a table to investigate the idea of a limit.
This was an intuitive way to approach limits.
■ We saw that if the left and right limits at a point were the
same, we had a limit at that point.
■ We saw that we could add, subtract, multiply, and divide
limits.
■ We now have some very powerful tools for dealing with
limits and can go on to our study of calculus.
H.Melikian
Barnett/Ziegler/Byleen College
19
WARM UP EXERCISE
3x  2x  1
f ( x) 
x 2  3x  2
2
Let
Find
( A) xlim
f ( x)
3
(B) xlim
f ( x)
1
(C) lim
f ( x)
x2
H.Melikian
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§ 9.2 Continuity
The student will learn about:
continuity,
continuity properties,
and solving inequalities using continuity
properties.
H.Melikian
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H.Melikian
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Continuity
As we have seen some graphs have holes in them, some have
breaks and some have other irregularities. We wish to study
each of these oddities.
Then through a study of limits
we will examine the
instantaneous rate of change.
H.Melikian
23
Definition
A function f is continuous at a point x = c if
lim f (x) exists
1.
x c
2.
f (c) exists
3.
lim f (x)  f (c)
x c
THIS IS THE DEFINITION OF
CONTINUITY
H.Melikian
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Example 1
f (x) = x – 1 at x = 2.
a. lim x  1  1 The limit exist!
x2
b. f (2) = 1
AND !
c. lim x  1  1  f ( 2)
x2
1
2
Therefore the function is continuous at x = 2.
H.Melikian
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Definition
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.
H.Melikian
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Example 2
f (x) = (x2 – 9)/(x + 3) at x = -3
x2  9
 -6
a. lim
x  3 x  3
b. f (-3) = 0/0
The limit exist!
Is undefined! So
x 9
c. lim
 f ( 3)
x  3 x  3
2
-3
-6
Therefore the function is not
continuous at
H.Melikian
x = -3.
27
Example 3
x
a. lim
x0 x
f (x) = |x|/x at x = 0 and at x = 1
Does not exist!
b. f (0) = 0/0 Undefined!
0
x
c. lim
 f (0)
x0 x
Therefore the function is not continuous at
Notice that the function is continuous at
H.Melikian
x = 0.
x = 1.
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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.
H.Melikian
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Some Special Functions and Continuity
• A constant function is continuous for all x.
• For n > 0, f (x) = xn is continuous for all x.
• A polynomial function is continuous for all x.
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Some Special Functions and Continuity
• A rational function is continuous for all x except
those values that make the denominator 0.
• For n an odd positive integer, n f ( x) is continuous
wherever f (x) is continuous.
• For n an even positive integer,
n
f ( x) is
continuous wherever f (x) is continuous and non
negative.
H.Melikian
31
Solving Inequalities Using Continuity Properties.
A tool for analyzing graphs of functions is called a sign
chart. We find where the function is zero or undefined to
partition the number line into intervals for testing.
We then test each interval to determine if the function is
positive (above the x-axis) or negative (below the x-axis) in
those intervals.
H.Melikian
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Constructing Sign Charts.
1. Find all partition numbers.
a. Points of discontinuity – where the denominator is 0.
b. Points where the function is zero – where the numerator
is zero but the denominator is not.
2. Plot these partition numbers on the number line dividing
the line into intervals.
3. Select a test number in each interval and determine if f (x)
is positive (+) or negative (-) there.
4. Complete your sign chart showing the sign of f (x) on
each open interval.
H.Melikian
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Sign Chart - Example.
x 2  3 x x (x  3)
f (x) 

0
x2
x2
1. a. Points of discontinuity. Where the denominator
is zero. x = 2
1. b. Points where f (x) = 0. Where the numerator is
zero. x = 0, x = - 3
2. Place these partition values on a number line.
-3
0
2
continued
H.Melikian
34
Sign Chart - Example.
x 2  3 x x (x  3)
f (x) 

0
x2
x2
-3
0
2
3. Select test numbers and determine if f (x) is
positive or negative.
x
- 10
-1
1
10
f (x)
-
+
-
+
4. Complete the sign chart.
- - - - - - - + + + + - - - - + + + + + + +
-3
H.Melikian
0
2
continued
35
Sign Chart - Example.
x 2  3 x x (x  3)
f (x) 

0
x2
x2
- - - - - - - + + + + - - - - + + + + + + +
-3
0
2
Remember the plus signs mean the function is above the x-axis
while the minus signs mean the function is below the x – axis.
We can check this with a
graphing calculator.
H.Melikian
36
Summary.
We have developed a definition for determining if a function
is continuous. That is the function has no holes or oddities.
We have developed a set of properties for limits. AND
We have used sign charts to solve inequalities.
H.Melikian
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Sign Chart - Example.
x 2  3 x x (x  3)
f (x) 

0
x2
x2
- - - - - - - + + + + - - - - + + + + + + +
-3
0
2
The > 0 in the original problem means we want greater than
0, the positive signs (where the function is positive).
The answer to the problem is then – 3 < x < 0 and x > 2.
Or in interval notation (-3,0) and (2,).
H.Melikian
38
x2  4
lim
x 2 x  2
MATH 2000
Name_________________
H.Melikian
Quiz #1
Date___________
39
H.Melikian
40
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