INTRODUCTORY MATHEMATICAL ANALYSIS

1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL
ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
©2007 Pearson Education Asia
Chapter 10Chapter 10
Limits and ContinuityLimits and Continuity

2. ©2007 Pearson Education Asia
INTRODUCTORY MATHEMATICAL
ANALYSIS
0. Review of Algebra
1. Applications and More Algebra
2. Functions and Graphs
3. Lines, Parabolas, and Systems
4. Exponential and Logarithmic Functions
5. Mathematics of Finance
6. Matrix Algebra
7. Linear Programming
8. Introduction to Probability and Statistics

3. ©2007 Pearson Education Asia
9. Additional Topics in Probability
10. Limits and Continuity
11. Differentiation
12. Additional Differentiation Topics
13. Curve Sketching
14. Integration
15. Methods and Applications of Integration
16. Continuous Random Variables
17. Multivariable Calculus
INTRODUCTORY MATHEMATICAL
ANALYSIS

4. ©2007 Pearson Education Asia
• To study limits and their basic properties.
• To study one-sided limits, infinite limits, and
limits at infinity.
• To study continuity and to find points of
discontinuity for a function.
• To develop techniques for solving nonlinear
inequalities.
Chapter 10: Limits and Continuity
Chapter ObjectivesChapter Objectives

5. ©2007 Pearson Education Asia
Limits
Limits (Continued)
Continuity
Continuity Applied to Inequalities
10.1)
10.2)
10.3)
Chapter 10: Limits and Continuity
Chapter OutlineChapter Outline
10.4)

6. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.1 Limits10.1 Limits
Example 1 – Estimating a Limit from a Graph
• The limit of f(x) as x approaches a is the number L,
written as
a. Estimate limx→1 f (x) from the graph.
Solution:
b. Estimate limx→1 f (x) from the graph.
Solution:
( ) Lxf
ax
=
→
lim
( ) 2lim
1
=
→
xf
x
( ) 2lim
1
=
→
xf
x

7. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.1 Limits
Properties of Limits
1.
2. for any positive integer n
3.
4.
5.
( ) constantaiswherelimlim cccxf
axax
==
→→
nn
ax
ax =
→
lim
( ) ( )[ ] ( ) ( )xgxfxgxf
axaxax →→→
±=± limlimlim
( ) ( )[ ] ( ) ( )xgxfxgxf
axaxax →→→
⋅=⋅ limlimlim
( )[ ] ( )xfcxcf
axax →→
⋅= limlim

8. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.1 Limits
Example 3 – Applying Limit Properties 1 and 2
Properties of Limits
( ) 162limc.
366limb.
77lim;77lima.
44
2
22
6
52
=−=
==
==
→
→
−→→
t
x
t
x
xx
( )
( )
( )
( )
( ) 0limif
lim
lim
lim6. ≠=
→
→
→
→
xg
xg
xf
xg
xf
ax
ax
ax
ax
( ) ( )n
ax
n
ax
xfxf
→→
= limlim7.

9. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.1 Limits
Example 5 – Limit of a Polynomial Function
Find an expression for the polynomial function,
Solution:
where
( ) 01
1
1 ... cxcxcxcxf n
n
n
n ++++= −
−
( ) ( )
( )af
cacacac
ccxcxc
cxcxcxcxf
n
n
n
n
axax
n
ax
n
n
ax
n
n
n
n
n
axax
=
++++=
++++=
++++=
−
−
→→
−
→
−
→
−
−
→→
01
1
1
01
1
1
01
1
1
...
limlim...limlim
...limlim
( ) ( )afxf
ax
=
→
lim

10. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.1 Limits
Example 7 – Finding a Limit
Example 9 – Finding a Limit
Find .
Solution:
If ,find .
Solution:
1
1
lim
2
1 +
−
→ x
x
x
( ) 2111lim
1
1
lim
1
2
1
−=−−=−=
+
−
−→−→
x
x
x
xx
( ) 12
+= xxf
( ) ( )
h
xfhxf
h
−+
→0
lim
( ) ( ) [ ]
( ) xhx
h
xhxhx
h
xfhxf
h
hh
22lim
112
limlim
0
222
00
=+=
−−+++
=
−+
→
→→
Limits and Algebraic Manipulation
• If f (x) = g(x) for all x ≠ a, then
( ) ( )xgxf
axax →→
= limlim

11. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.2 Limits (Continued)10.2 Limits (Continued)
Example 1 – Infinite Limits
Infinite Limits
• Infinite limits are written as and .
Find the limit (if it exists).
Solution:
a. The results are becoming arbitrarily large. The limit
does not exist.
b. The results are becoming arbitrarily large. The limit
does not exist.
∞=+
−→ xx
1
lim
0
−∞=−
−→ xx
1
lim
0
1
2
lima.
1 ++
−→ xx 4
2
limb. 22 −
+
→ x
x
x

12. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.2 Limits (Continued)
Example 3 – Limits at Infinity
Find the limit (if it exists).
Solution:
a. b.
( )3
5
4
lima.
−∞→ xx
( )
0
5
4
lim 3
=
−∞→ xx
( )x
x
−
∞→
4limb.
( ) ∞=−
∞→
x
x
4lim
Limits at Infinity for Rational Functions
• If f (x) is a rational function,
and( ) m
m
n
n
xx xb
xa
xf
∞→∞→
= limlim ( ) m
m
n
n
xx xb
xa
xf
−∞→−∞→
= limlim

13. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.2 Limits (Continued)
Example 5 – Limits at Infinity for Polynomial Functions
Find the limit (if it exists).
Solution:
Solution: ( ) ∞=−=+−
−∞→−∞→
33
2lim92lim xxx
xx
( ) −∞==−+−
−∞→−∞→
323
lim2lim xxxx
xx
( ) 33
2lim92limb. xxx
xx
−=+−
−∞→−∞→
( ) 323
lim2lima. xxxx
xx −∞→−∞→
=−+−

14. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.3 Continuity10.3 Continuity
Example 1 – Applying the Definition of Continuity
Definition
• f(x) is continuous if three conditions are met:
a. Show that f(x) = 5 is continuous at 7.
Solution: Since , .
b. Show that g(x) = x2
− 3 is continuous at −4.
Solution:
( )
( )
( ) ( )afxf
xf
xf
=
→
→
ax
ax
lim3.
existslim2.
exists1.
( ) 55limlim
77
==
→→ xx
xf ( ) ( )75lim
7
fxf
x
==
→
( ) ( ) ( )43limlim 2
44
−=−=
−→−→
gxxg
xx

15. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.3 Continuity
Example 3 – Discontinuities
a. When does a function have infinite
discontinuity?
Solution:
A function has infinite discontinuity at a when at least
one of the one-sided limits is either ∞ or −∞ as x →a.
b. Find discontinuity for
Solution:
f is defined at x = 0 but limx→0 f (x) does not exist. f is
discontinuous at 0.
( )





<−
=
>
=
0if1
0if0
0if1
x
x
x
xf

16. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.3 Continuity
Example 5 – Locating Discontinuities in Case-Defined Functions
For each of the following functions, find all points of
discontinuity.
( )



<
≥+
=
3if
3if6
a. 2
xx
xx
xf
( )



<
>+
=
2if
2if2
b. 2
xx
xx
xf

17. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.3 Continuity
Example 5 – Locating Discontinuities in Case-Defined Functions
Solution:
a. We know that f(3) = 3 + 6 = 9. Because
and ,
the function has no points of discontinuity.
( ) ( ) 96limlim
33
=+= ++
→→
xxf
xx
( ) 9limlim 2
33
==
−→→ −
xxf
xx

18. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.3 Continuity
Example 5 – Locating Discontinuities in Case-Defined Functions
Solution:
b. It is discontinuous at 2,
limx→2 f (x) exists.
( ) ( )xfxxxf
xxxx +−−−
→→→→
=+===
22
2
22
lim2lim4limlim

19. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.4 Continuity Applied to Inequalities10.4 Continuity Applied to Inequalities
Example 1 – Solving a Quadratic Inequality
Solve .
Solution: Let .
To find the real zeros of f,
Therefore, x2
− 3x − 10 > 0 on (−∞,−2) ∪ (5,∞).
01032
>−− xx
( ) 1032
−−= xxxf
( )( )
5,2
052
01032
−=
=−+
=−−
x
xx
xx

20. ©2007 Pearson Education Asia
Chapter 10: Limits and Continuity
10.4 Continuity Applied to Inequalities
Example 3 – Solving a Rational Function Inequality
Solve .
Solution: Let .
The zeros are 1 and 5.
Consider the intervals: (−∞, 0) (0, 1) (1, 5) (5,∞)
Thus, f(x) ≥ 0 on (0, 1] and [5,∞).
0
562
≥
+−
x
xx
( ) ( )( )
x
xx
x
xx
xf
51562
−−
=
+−
=