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sin Оё
lim = 1.
Оё
+
Оёв†’0

Since
sin(в€’Оё) sin Оё
= ,
в€’Оё Оё
sin Оё
lim = 1.
Оё
Оёв€’0в€’

Therefore,
sin Оё
lim = 1.
Оё
Оёв†’0

Part (iv)

1 в€’ cos Оё (1 в€’ cos Оё)(1 + cos Оё)
lim = lim
Оё Оё(1 + cos Оё)
Оёв†’0 Оёв†’0

1 в€’ cos2 Оё 1
В·
= lim
Оё (1 + cos Оё)
Оёв†’0

sin Оё sin Оё
В·
= lim
Оё 1 + cos Оё
Оёв†’0
0
=1В·
2
= 0.

Example 2.1.22 Show that

(i) sin Оё and cos Оё are continuous for all real Оё.
2.1. INTUITIVE TREATMENT AND DEFINITIONS 59

ПЂ
(ii) tan Оё and sec Оё are continuous for all Оё = 2nПЂ В± , n integer.
2
(iii) cot Оё and csc Оё are continuous for all Оё = nПЂ, n integer.

Part (i) First, we show that for all real c,

lim sin Оё = sin c or equivalently lim | sin Оё в€’ sin c| = 0.
Оёв†’c Оёв†’c

We observe that
Оёв€’c
Оё+c
0 в‰¤ | sin Оё в€’ sin c| = 2 cos sin
2 2
(Оё в€’ c)
в‰¤ 2 sin
2
sin (Оёв€’c)
2
= |(Оё в€’ c)| (Оёв€’c)
2

Therefore, by squeeze theorem,

0 в‰¤ lim | sin Оё в€’ sin c| в‰¤ 0 В· 1 = 0.
Оёв€’c

It follows that for all real c, sin Оё is continuous at c.
Next, we show that

lim cos x = cos c or equivalently lim | cos x в€’ cos c| = 0.
xв†’c xв†’c

We observe that
(x в€’ c)
x+c
0 в‰¤ | cos x в€’ cos c| = в€’2 sin sin
2 2
xв€’c
sin x+c
2
в‰¤ |Оё в€’ c| в‰¤1
sin
;
xв€’c
2
2

Therefore,
0 в‰¤ lim | cos x в€’ cos c| в‰¤ 0 В· 1 = 0
xв†’c

and cos x is continuous at c.
60 CHAPTER 2. LIMITS AND CONTINUITY

Part (ii) Since for all Оё = 2nПЂ В± ПЂ , n integer,
2

sin Оё 1
tan Оё = , sec Оё =
cos Оё cos Оё
it follows that tan Оё and sec Оё are continuous functions.

Part (iii) Both cot Оё and csc Оё are continuous as quotients of two continuous
functions where the denominators are not zero for n = nПЂ, n integer.

Exercises 2.1 Evaluate each of the following limits.
x2 в€’ 1 sin(2x) sin 5x
1. lim 3 2. lim 3. lim
xв†’1 x в€’ 1 x sin 7x
xв†’0 xв†’0

xв€’2
1 1
4. lim 5. lim 6. lim
x2 в€’ 4 x2 в€’ 4 x2 в€’ 4
в€’ xв†’2
+
xв†’2 xв†’2

xв€’2 xв€’2 xв€’2
7. lim 8. lim 9. lim
|x в€’ 2| |x в€’ 2| |x в€’ 2|
в€’ xв†’2
+
xв†’2 xв†’2

x2 в€’ 9 x2 в€’ 9
10. lim 11. lim 12. lim tan x
xв†’3 x в€’ 3 xв†’3 x + 3 ПЂ
xв†’ 2

13. lim+ tan x 14. lim csc x 15. lim csc x
в€’ +
ПЂ xв†’0 xв†’0
xв†’ 2

16. lim cot x 17. lim cot x 18. lim+ sec x
в€’
+ ПЂ
xв†’0 xв†’0 xв†’ 2

в€љ
xв€’2
sin 2x + sin 3x
19. lim sec x 20. lim 21. lim
xв€’4
x
ПЂ в€’
xв†’0
xв†’ 2 xв†’4

в€љ в€љ
x4 в€’ 81
xв€’2 xв€’2
22. lim 23 lim 24. lim 2
xв€’4 xв€’4 xв†’3 x в€’ 9
xв†’4
xв†’4+

Sketch the graph of each of the following functions. Determine all the
discontinuities of these functions and classify them as (a) removable type,
(b) п¬Ѓnite jump type, (c) essential type, (d) oscillation type, or other types.
2.2. LINEAR FUNCTION APPROXIMATIONS 61

xв€’1 xв€’2 x
в€’
25. f (x) = 2 26. f (x) =
x2 в€’ 9
|x в€’ 1| |x в€’ 2|

if x в‰¤ 0
for x в‰¤ 0 sin x
2x
27. f (x) = 28. f (x) = 2
x2 + 1 for x > 0 if x > 0
sin x

xв€’1 |x в€’ 1| if x в‰¤ 1
29. f (x) = 30. f (x) =
|x в€’ 2| if x > 1
(x в€’ 2)(x в€’ 3)

0 if x < 0
Recall the unit step function u(x) =
1 if x в‰Ґ 0.
Sketch the graph of each of the following functions and determine the left
hand limit and the right hand limit at each point of discontinuity of f and
g.

31. f (x) = 2u(x в€’ 3) в€’ u(x в€’ 4)

32. f (x) = в€’2u(x в€’ 1) + 4u(x в€’ 5)

33. f (x) = u(x в€’ 1) + 2u(x + 1) в€’ 3u(x в€’ 2)
ПЂ ПЂ
в€’u xв€’
34. f (x) = sin x u x +
2 2
ПЂ ПЂ
в€’u xв€’
35. g(x) = (tan x) u x +
2 2
36. f (x) = [u(x) в€’ u(x в€’ ПЂ)] cos x

2.2 Linear Function Approximations
One simple application of limits is to approximate a function f (x), in a small
neighborhood of a point c, by a line. The approximating line is called the
tangent line. We begin with a review of the equations of a line.
A vertical line has an equation of the form x = c. A vertical line has no
slope. A horizontal line has an equation of the form y = c. A horizontal
line has slope zero. A line that is neither horizontal nor vertical is called an
oblique line.
62 CHAPTER 2. LIMITS AND CONTINUITY

Suppose that an oblique line passes through two points, say (x1 , y1 ) and
(x2 , y2 ). Then the slope of this line is deп¬Ѓne as
y2 в€’ y1 y1 в€’ y2
m= = .
x2 в€’ x1 x1 в€’ x2
If (x, y) is any arbitrary point on the above oblique line, then
y в€’ y2
y в€’ y1
= .
m=
x в€’ x1 x в€’ x2
By equating the two forms of the slope m we get an equation of the line:
y в€’ y1 y2 в€’ y1 y в€’ y2 y2 в€’ y1
= or = .
x в€’ x1 x2 в€’ x1 x в€’ x2 x2 в€’ x1
On multiplying through, we get the вЂњtwo pointвЂќ form of the equation of the
line, namely,
y2 в€’ y1 y2 в€’ y1
y в€’ y1 = (x в€’ x1 ) or y в€’ y2 = (x в€’ x2 ).
x2 в€’ x1 x2 в€’ x1

Example 2.2.1 Find the equations of the lines passing through the follow-
ing pairs of points:

(i) (4, 2) and (6, 2) (ii) (1, 3) and (1, 5)
(iii) (3, 4) and (5, в€’2) (iv) (0, 2) and (4, 0).

Part (i) Since the y-coordinates of both points are the same, the line is
horizontal and has the equation y = 2. This line has slope 0.

Part (ii) Since the x-coordinates of both points are equal, the line is vertical
and has the equation x = 1.

Part (iii) The slope of the line is given by
в€’2 в€’ 4
= в€’3.
m=
5в€’3
The equation of this line is

y в€’ 4 = в€’3(x в€’ 3) or y + 2 = в€’3(x в€’ 5).
2.2. LINEAR FUNCTION APPROXIMATIONS 63

On solving for y, we get the equation of the line as

y = в€’3x + 13.

This line goes through the point (0, 13). The number 13 is called the y-
intercept. The above equation is called the slope-intercept form of the line.

Example 2.2.2 Determine the equations of the lines satisfying the given
conditions:

(i) slope = 3, passes through (2, 4)

(ii) slope = в€’2, passes through (1, в€’3)

(iii) slope = m, passes through (x1 , y1 )

(iv) passes through (3, 0) and (0, 4)

(v) passes through (a, 0) and (0, b)

Part (i) If (x, y) is on the line, then we equate the slopes and simplify:

yв€’4
or y в€’ 4 = 3(x в€’ 2).
3=
xв€’2

Part (ii) If (x, y) is on the line, then we equate slopes and simplify:

y+3
в€’2 = or y + 3 = в€’2(x в€’ 1).
xв€’1

Part (iii) On equating slopes and clearing fractions, we get
y в€’ y1
or y в€’ y1 = m(x в€’ x1 ).
m=
x в€’ x1
This form of the line is called the вЂњpoint-slopeвЂќ form of the line.
64 CHAPTER 2. LIMITS AND CONTINUITY

Part (iv) Using the two forms of the line we get

yв€’0 4в€’0 4
or y = в€’ (x в€’ 3).
=
xв€’3 0в€’3 3
If we divide by 4 we get
xy
+ = 1.
34
The number 3 is called the x-intercept and the number 4 is called the y-
intercept of the line. This form of the equation is called the вЂњtwo-interceptвЂќ
form of the line.

Part (v) As in Part (iv), the вЂњtwo-interceptвЂќ form of the line has the equation
xy
+ = 1.
ab
In order to approximate a function f at the point c, we п¬Ѓrst deп¬Ѓne the slope
m of the line that is tangent to the graph of f at the point (c, f (c)).

graph

f (x) в€’ f (c)
.
m = lim
xв€’c
xв†’c

Then the equation of the tangent line is

y в€’ f (c) = m(x в€’ c),

written in the point-slope form. The point (c, f (c)) is called the point of
tangency. This tangent line is called the linear approximation of f about
x = c.

Example 2.2.3 Find the equation of the line tangent to the graph of f (x) =
x2 at the point (2, 4).
2.2. LINEAR FUNCTION APPROXIMATIONS 65

The slope m of the tangent line at (3, 9) is

x2 в€’ 9
m = lim
xв†’3 x в€’ 3

= lim (x + 3)
xв†’3
= 6.

The equation of the tangent line at (3, 9) is

y в€’ 9 = 6(x в€’ 3).

Example 2.2.4 Obtain the equation of the line tangent to the graph of
в€љ
f (x) = x at the point (9, 3).
The slope m of the tangent line is given by

в€љ
xв€’3
m = lim
в€љв€’ 9
x
xв†’9
в€љ
( x в€’ 3)( x + 3)
в€љ
= lim
(x в€’ 9)( x + 3)
xв†’9
xв€’9
в€љ
= lim
(x в€’ 9)( x + 3)
xв†’9
1
в€љ
= lim
x+3
xв†’9
1
=.
6
The equation of the tangent line is
1
y в€’ 3 = (x в€’ 9).
6

Example 2.2.5 Derive the equation of the line tangent to the graph of
ПЂ1
, .
f (x) = sin x at
62
The slope m of the tangent line is given by
66 CHAPTER 2. LIMITS AND CONTINUITY

ПЂ
sin x в€’ sin 6
m = lim
xв€’ ПЂ
xв†’ ПЂ
6
6

x+ПЂ/6 xв€’ПЂ/6
2 cos sin
2 2
= lim
(x в€’ ПЂ/6)
ПЂ
xв†’ 6

xв€’ПЂ/6
sin 2
= cos(ПЂ/6) В· limПЂ xв€’ПЂ/6
xв†’ 6
2

= cos(ПЂ/6)
в€љ
3
= .
2
The equation of the tangent line is
в€љ
1 3 ПЂ
yв€’ = xв€’ .
2 2 6

Example 2.2.6 Derive the formulas for the slope and the equation of the
line tangent to the graph of f (x) = sin x at (c, sin c).
As in Example 27, replacing ПЂ/6 by c, we get

sin x в€’ sin c
m = lim
xв€’c
xв†’c

2 cos x+c sin xв€’c
2 2
= lim
xв€’c
xв†’c

sin xв€’c
x+c 2
В· lim
= lim cos xв€’c
2
xв†’c xв†’c
2
= cos c.

Therefore the slope of the line tangent to the graph of f (x) = sin x at (c, sin c)
is cos c.
The equation of the tangent line is

y в€’ sin c = (cos c)(x в€’ c).
2.2. LINEAR FUNCTION APPROXIMATIONS 67

Example 2.2.7 Derive the formulas for the slope, m, and the equation of
the line tangent to the graph of f (x) = cos x at (c, cos c). Then determine
ПЂ1
the slope and the equation of the tangent line at , .
32
As in Example 28, we replace the sine function with the cosine function,

cos x в€’ cos c
m = lim
xв€’c
xв†’c

в€’2 sin x+c sin xв€’c
2 2
= lim
xв€’c
xв†’c

sin xв€’c
x+c 2
= lim sin lim xв€’c
2
xв†’c xв†’c
2
= в€’ sin(c).

The equation of the tangent line is

y в€’ cos c = в€’ sin c(x в€’ c).
в€љ
ПЂ 3
ПЂ
For c = , slope = в€’ sin =в€’ and the equation of the tangent line
3 3 2
в€љ
1 3 ПЂ
yв€’ =в€’ xв€’ .
2 2 3

Example 2.2.8 Derive the formulas for the slope, m, and the equation of
the line tangent to the graph of f (x) = xn at the point (c, cn ), where n is a
natural number. Then get the slope and the equation of the tangent line for
c = 2, n = 4.
By deп¬Ѓnition, the slope m is given by

xn в€’ c n
.
m = lim
xв€’c
xв†’c

To compute this limit for the general natural number n, it is convenient to
68 CHAPTER 2. LIMITS AND CONTINUITY

let x = c + h. Then
(c + h)n в€’ cn
m = lim
h
hв†’0
n(n в€’ 1) nв€’2 2
1
cn + ncnв€’1 h + c h + В· В· В· + hn в€’ cn
= lim
hв†’0 h 2!
n(n в€’ 1) nв€’2 2
1
ncnв€’1 h + c h + В· В· В· + hn
= lim
hв†’0 h 2!
n(n в€’ 1) nв€’2
= lim ncnв€’1 + c h + В· В· В· + hnв€’1
2!
hв†’0

= ncnв€’1 .

Therefore, the equation of the tangent line through (c, cn ) is

y в€’ cn = ncnв€’1 (x в€’ c).

For n = 4 and c = 2, we п¬Ѓnd the slope, m, and equation for the tangent line
to the graph of f (x) = x4 at c = 2:

m = 4c3 = 32
y в€’ 24 = 32(x в€’ 2) or y в€’ 16 = 32(x в€’ 2).

Deп¬Ѓnition 2.2.1 Suppose that a function f is deп¬Ѓned on a closed interval
[a, b] and a < c < b. Then c is called a critical point of f if the slope of the
line tangent to the graph of f at (c, f (c)) is zero or undeп¬Ѓned. The slope
function of f at c is deп¬Ѓned by
f (c + h) в€’ f (c)
slope (f (x), c) = lim
h
hв†’0
f (x) в€’ f (c)
= lim .
xв€’c
xв†’c

Example 2.2.9 Determine the slope functions and critical points of the
following functions:

(i) f (x) = sin x, 0 в‰¤ x в‰¤ 2ПЂ (ii) f (x) = cos x, 0 в‰¤ x в‰¤ 2ПЂ
(iv) f (x) = x3 в€’ 4x, в€’2 в‰¤ x в‰¤ 2
(iii) f (x) = |x|, в€’1 в‰¤ x в‰¤ 1
2.2. LINEAR FUNCTION APPROXIMATIONS 69

Part (i) In Example 28, we derived the slope function formula for sin x,
namely
slope (sin x, c) = cos c.
Since cos c is deп¬Ѓned for all c, the non-end point critical points on [0, 2ПЂ]
are ПЂ/2 and 3ПЂ/2 where the cosine has a zero value. These critical points
correspond to the maximum and minimum values of sin x.
Part (ii) In Example 29, we derived the slope function formula for cos x,
namely
slope (cos x, c) = в€’ sin c.
The critical points are obtained by solving the following equation for c:
в€’ sin c = 0, 0 в‰¤ c в‰¤ 2ПЂ
c = 0, ПЂ, 2ПЂ.
These values of c correspond to the maximum value of cos x at c = 0 and 2ПЂ,
and the minimum value of cos x at c = ПЂ.

|x| в€’ |c|
Part (iii) slope (|x|, c) = lim
xв€’c
xв†’c

|x| в€’ |c| |x| + |c|
В·
= lim
x в€’ c |x| + |c|
xв†’c

x2 в€’ c 2
= lim
xв†’c (x в€’ c)(|x| + |c|)

x+c
= lim
|x| + |c|
xв†’c

2c
=
2|c|
c
=
|c|
пЈ±
пЈІ1 if c > 0
в€’1 if c < 0
=
undeп¬Ѓned if c = 0
пЈі
70 CHAPTER 2. LIMITS AND CONTINUITY

The only critical point is c = 0, where the slope function is undeп¬Ѓned. This
critical point corresponds to the minimum value of |x| at c = 0. The slope
function is undeп¬Ѓned because the tangent line does not exist at c = 0. There
is a sharp corner at c = 0.

Part (iv) The slope function for f (x) = x3 в€’ 4x is obtained as follows:

1
[((c + h)3 в€’ 4(c + h)) в€’ (c3 в€’ 4c)]
slope (f (x), c) = lim
hв†’0 h
13
[c + 3c2 h + 3ch2 + h3 в€’ 4c в€’ 4h в€’ c3 + 4c]
= lim
hв†’0 h
1
[3c2 h + 3ch2 + h3 в€’ 4h]
= lim
hв†’0 h
= lim [3c2 + 3ch + h2 в€’ 4]
hв†’0
2
= 3c в€’ 4

graph

The critical points are obtained by solving the following equation for c:

3c2 в€’ 4 = 0
2
c = В±в€љ
3
в€’2 16 2
At c = в€љ , f has a local maximum value of в€љ and at c = в€љ , f has a
3 33 3
в€’16
local minimum value of в€љ . The end point (в€’2, 0) has a local end-point
33
minimum and the end point (2, 0) has a local end-point maximum.

Remark 7 The zeros and the critical points of a function are helpful in
sketching the graph of a function.
2.2. LINEAR FUNCTION APPROXIMATIONS 71

Exercises 2.2

1. Express the equations of the lines satisfying the given information in the
form y = mx + b.

(a) Line passing through (2, 4) and (5, в€’2)
(b) Line passing through (1, 1) and (3, 4)
(c) Line with slope 3 which passes through (2, 1)
(d) Line with slope 3 and y-intercept 4
(e) Line with slope 2 and x-intercept 3
(f) Line with x-intercept 2 and y-intercept 4.

2. Two oblique lines are parallel if they have the same slope. Two oblique
lines are perpendicular if the product of their slopes is в€’1. Using this
information, solve the following problems:

(a) Find the equation of a line that is parallel to the line with equation
y = 3x в€’ 2 which passes through (1, 4).
(b) Solve problem (a) when вЂњparallelвЂќ is changed to вЂњperpendicular.вЂќ
(c) Find the equation of a line with y-intercept 4 which is parallel to
y = в€’3x + 1.
(d) Solve problem (c) when вЂњparallelвЂќ is changed to вЂњperpendicular.вЂќ
(e) Find the equation of a line that passes through (1, 1) and is
(i) parallel to the line with equation 2x в€’ 3y = 6.
(ii) perpendicular to the line with equation 3x + 2y = 6

3. For each of the following functions f (x) and values c,

(i) derive the slope function, slope (f (x), c) for arbitrary c;
(ii) determine the equations of the tangent line and normal line (perpen-
dicular to tangent line) at the point (c, f (c)) for the given c;
(iii) determine all of the critical points (c, f (c)).
(a) f (x) = x2 в€’ 2x, c = 3
(b) f (x) = x3 , c = 1
72 CHAPTER 2. LIMITS AND CONTINUITY

ПЂ
(c) f (x) = sin(2x), c=
12
ПЂ
(d) f (x) = cos(3x), c =
9 в€љв€љ
4 2
(e) f (x) = x в€’ 4x , c = в€’2, 0, 2, в€’ 2, 2.

2.3 Limits and Sequences
We begin with the deп¬Ѓnitions of sets, sequences, and the completeness prop-
erty, and state some important results. If x is an element of a set S, we write
x в€€ S, read вЂњx is in S.вЂќ If x is not an element of S, then we write x в€€ S,/
read вЂњx is not in S.вЂќ

Deп¬Ѓnition 2.3.1 If A and B are two sets of real numbers, then we deп¬Ѓne

A в€© B = {x : x в€€ A and x в€€ B}

and

A в€Є B = {x : x в€€ A or x в€€ B or both}.

We read вЂњA в€© BвЂќ as the вЂњintersection of A and B.вЂќ We read вЂњA в€Є BвЂќ as the
вЂњunion of A and B.вЂќ If A в€© B is the empty set, в€…, then we write A в€© B = в€….

Deп¬Ѓnition 2.3.2 Let A be a set of real numbers. Then a number m is said
to be an upper bound of A if x в‰¤ m for all x в€€ A. The number m is said to
be a least upper bound of A, written lub(A) if and only if,

(i) m is an upper bound of A, and,

(ii) if q < m, then there is some x в€€ A such that q < x в‰¤ m.

Deп¬Ѓnition 2.3.3 Let B be a set of real numbers. Then a number is said
to be a lower bound of B if в‰¤ y for each y в€€ B. This number is said to
be the greatest lower bound of B, written, glb(b), if and only if,

(i) is a lower bound of B, and,
2.3. LIMITS AND SEQUENCES 73

< p, then there is some element y в€€ B such that в‰¤ y < p.
(ii) if

Deп¬Ѓnition 2.3.4 A real number p is said to be a limit point of a set S if
and only if every open interval that contains p also contains an element q of
S such that q = p.

Example 2.3.1 Suppose A = [1, 10] and B = [5, 15].
Then Aв€©B = [5, 10], Aв€ЄB = [1, 15], glb(A) = 1, lub(A) = 10, glb(B) = 5
and lub(B) = 15. Each element of A is a limit point of A and each element
of B is a limit point of B.

1
: n is a natural number .
Example 2.3.2 Let S =
n
Then no element of S is a limit point of S. The number 0 is the only
limit point of S. Also, glb(S) = 0 and lub(S) = 1.
Completeness Property: The completeness property of the set R of all real
numbers states that if A is a non-empty set of real numbers and A has an
upper bound, then A has a least upper bound which is a real number.

Theorem 2.3.1 If B is a non-empty set of real numbers and B has a lower
bound, then B has a greatest lower bound which is a real number.

Proof. Let m denote a lower bound for B. Then m в‰¤ x for every x в€€ B.
Let A = {в€’x : x в€€ B}. then в€’x в‰¤ в€’m for every x в€€ B. Hence, A is a
non-empty set that has an upper bound в€’m. By the completeness property,
A has a least upper bound lub(A). Then, -lub(A) = glb(B) and the proof is
complete.

Theorem 2.3.2 If x1 and x2 are real numbers such that x1 < x2 , then
1
x1 < (x1 + x2 ) < x2 .
2
Proof. We observe that
1
x1 в‰¤ (x1 + x2 ) < x2 в†” 2x1 < x1 + x2 < 2x2
2
в†” x1 < x2 < x2 + (x2 в€’ x1 ).
This completes the proof.
74 CHAPTER 2. LIMITS AND CONTINUITY

Theorem 2.3.3 Suppose that A is a non-empty set of real numbers and
m = lub(A). If m в€€ A, then m is a limit point of A.
/

Proof. Let an open interval (a, b) contain m. That is, a < m < b. By the
deп¬Ѓnition of a least upper bound, a is not an upper bound for A. Therefore,
there exists some element q of A such that a < q < m < b. Thus, every open
interval (a, b) that contains m must contain a point of A other than m. It
follows that m is a limit point of A.

Theorem 2.3.4 (Dedekind-Cut Property). The set R of all real numbers is
not the union of two non-empty sets A and B such that
(i) if x в€€ A and y в€€ B, then x < y,
(ii) A contains no limit point of B, and,
(iii) B contains no limit point of A.

Proof. Suppose that R = A в€Є B where A and B are non-empty sets that
satisfy conditions (i), (ii) and (iii). Since A and B are non-empty, there exist
real numbers a and b such that a в€€ A and b в€€ B. By property (i), a is
a lower bound for B and b is an upper bound for A. By the completeness
property and theorem 2.3.1, A has a least upper bound, say m, and B has a
greatest lower bound, say M . If m в€€ A, then m is a limit point of A. Since
/
B contains no limit point of A, m в€€ A. Similarly, M в€€ B. It follows that
m < M by condition (i). However, by Theorem 2.3.2,
1
m < (m + M ) < M.
2
1
The number (m + M ) is neither in A nor in B. This is a contradiction,
2
because R = A в€Є B. This completes the proof.

Deп¬Ѓnition 2.3.5 An empty set is considered to be a п¬Ѓnite set. A non-empty
set S is said to be п¬Ѓnite if there exists a natural number n and a one-to-one
function that maps S onto the set {1, 2, 3, . . . , n}. Then we say that S has n
elements. If S is not a п¬Ѓnite set, then S is said to be an inп¬Ѓnite set. We say
that an inп¬Ѓnite set has an inп¬Ѓnite number of elements. Two sets are said to
have the same number of elements if there exists a one-to-one correspondence
between them.
2.3. LIMITS AND SEQUENCES 75

Example 2.3.3 Let A = {a, b, c}, B = {1, 2, 3}, C = {1, 2, 3, . . . }, and D =
{0, 1, в€’1, 2, в€’2, . . . }.
In this example, A and B are п¬Ѓnite sets and contain three elements each.
The sets C and D are inп¬Ѓnite sets and have the same number of elements. A
one-to-one correspondence f between n, C and D can be deп¬Ѓned as f : C в†’
D such that
f (1) = 0, f (2n) = n and f (2n + 1) = в€’n for n = 1, 2, 3, . . . .

Deп¬Ѓnition 2.3.6 A set that has the same number of elements as C =
{1, 2, 3, . . . } is said to be countable. An inп¬Ѓnite set that is not countable
is said to be uncountable.

Remark 8 The set of all rational numbers is countable but the set of all real
numbers is uncountable.

Deп¬Ѓnition 2.3.7 A sequence is a function, say f , whose domain is the set
of all natural numbers. It is customary to use the notation f (n) = an , n =
1, 2, 3, . . . . We express the sequence as a list without braces to avoid confusion
with the set notation:
{an }в€ћ .
a1 , a2 , a3 , . . . , an , . . . or, simply, n=1

The number an is called the nth term of the sequence. The sequence is said
to converge to the limit a if for every > 0, there exists some natural number,
say N , such that |am в€’ a| < for all m в‰Ґ N . We express this convergence
by writing
lim an = a.
nв†’в€ћ
If a sequence does not converge to a limit, it is said to diverge or be divergent.

Example 2.3.4 For each natural number n, let
(в€’1)n
в€’n
n n
.
an = (в€’1) , bn = 2 , cn = 2 , dn =
n
The sequence {an } does not converge because its terms oscillate between в€’1
and 1. The sequence {bn } converges to 0. The sequence {cn } diverges to в€ћ.
The sequence {dn } converges to 0.
76 CHAPTER 2. LIMITS AND CONTINUITY

Deп¬Ѓnition 2.3.8 A sequence {an }в€ћ diverges to в€ћ if, for every natural
n=1
number N , there exists some m such that

am+j в‰Ґ N for all j = 1, 2, 3, В· В· В· .

The sequence {an }в€ћ is said to diverge to в€’в€ћ if, for every natural number
n=1
N , there exists some m such that

am+j в‰¤ в€’N , for all j = 1, 2, 3, . . . .

Theorem 2.3.5 If p is a limit point of a non-empty set A, then every open
interval that contains p must contain an inп¬Ѓnite subset of A.
Proof. Let some open interval (a, b) contain p. Suppose that there are only
two п¬Ѓnite subsets {a1 , a2 , . . . , an } and {b1 , b2 , . . . , bm } of distinct elements of
A such that

a < a1 < a2 < В· В· В· < an < p < bm < bmв€’1 < В· В· В· < b1 < b.

Then the open interval (an , bm ) contains p but no other points of A distinct
from p. Hence p is not a limit point of A. The contradiction proves the
theorem.

Theorem 2.3.6 If p is a limit point of a non-empty set A, then there exists
a sequence {pn }в€ћ , of distinct points pn of A, that converges to p.
n=1

1 1
Proof. Let a1 = p в€’ , b1 = p + . Choose a point p1 of A such that p1 = p
2 2
and a1 < p1 < p < b1 or a1 < p < p1 < b1 . If a1 < p1 < p < b1 , then deп¬Ѓne
1 1
1
a2 = max p1 , p в€’ 2 and b2 = p + 2 . Otherwise, deп¬Ѓne a2 = p в€’ 2 and
2 2 2
1
b2 = min p1 , p + 2 . Then the open interval (a2 , b2 ) contains p but not p1
2
1
and b2 в€’ a2 в‰¤ . We repeat this process indeп¬Ѓnitely to select the sequence
2
{pn }, of distinct points pn of A, that converges to p. The fact that {pn } is an
inп¬Ѓnite sequence is guaranteed by Theorem 2.3.5. This completes the proof.

Theorem 2.3.7 Every bounded inп¬Ѓnite set A has at least one limit point p
and there exists a sequence {pn }в€ћ , of distinct points of A, that converges
n=1
to p.
2.3. LIMITS AND SEQUENCES 77

Proof. We will show that A has a limit point. Since A is bounded, there
exists an open interval (a, b) that contains all points of A. Then either
1 1
a, (a + b) contains an inп¬Ѓnite subset of A or (a + b), b contains an
2 2
inп¬Ѓnite subset of A. Pick one of the two intervals that contains an inп¬Ѓnite
subset of A. Let this interval be denoted (a1 , b1 ). We continue this process
repeatedly to get an open interval (an , bn ) that contains an inп¬Ѓnite subset of
|b в€’ a|
. Then the lub of the set {an , a2 , . . . } and glb of
A and |bn в€’ an | =
2n
the set {b1 , b2 , . . . } are equal to some real number p. It follows that p is a
limit point of A. By Theorem 2.3.6, there exists a sequence {pn }, of distinct
points of A, that converges to p. This completes the proof.

Deп¬Ѓnition 2.3.9 A set is said to be a closed set if it contains all of its limit
points. The complement of a closed set is said to be an open set. (Recall
that the complement of A is {x в€€ R : x в€€ A}.)
/

Theorem 2.3.8 The interval [a, b] is a closed and bounded set. Its comple-
ment (в€’в€ћ, a) в€Є (b, в€ћ) is an open set.
Proof. Let p в€€ (в€’в€ћ, a) в€Є (b, в€ћ). Then в€’в€ћ < p < a or b < p < в€ћ. The
11 1 1
intervals p в€’ , (a + p) or (b + p), p + contain no limit point of
22 2 2
[a, b]. Thus [a, b] must contain its limit points, because they are not in the
complement.

Theorem 2.3.9 If a non-empty set A has no upper bound, then there exists
a sequence {pn }в€ћ , of distinct points of A, that diverges to в€ћ. Furthermore,
n=1
every subsequence of {pn }в€ћ diverges to в€ћ
n=1

Proof. Since 1 is not an upper bound of A, there exists an element p1 of A
such that 1 < p1 . Let a1 = max{2, p1 }. Choose a point, say p2 , of A such
that a1 < p2 . By repeating this process indeп¬Ѓnitely, we get the sequence
{pn } such that pn > n and p1 < p2 < p3 < . . . . Clearly, the sequence
{pn }в€ћ diverges to в€ћ. It is easy to see that every subsequence of {pn }в€ћn=1
n=1
also diverges to в€ћ.

Theorem 2.3.10 If a non-empty set B has no lower bound, then there exists
a sequence {qn }в€ћ , of distinct points of B, that diverges to в€’в€ћ. Further-
n=1
more, every subsequence of {qn }в€ћ diverges to в€’в€ћ.
n=1
78 CHAPTER 2. LIMITS AND CONTINUITY

Proof. Let A = {в€’x : x в€€ B}. Then A has no upper bound. By Theorem
2.3.9, there exists a sequence {pn }в€ћ , of distinct points of A, that diverges to
n=1
в€ћ
в€ћ. Let qn = в€’pn . Then {qn }n=1 is a sequence that meets the requirements
of the Theorem 2.3.10. Also, every subsequence of {qn }в€ћ diverges to в€’в€ћ.
n=1

Theorem 2.3.11 Let {pn }в€ћ be a sequence of points of a closed set S that
n=1
converges to a point p of S. If f is a function that is continuous on S, then
the sequence {f (pn )}в€ћ converges to f (p). That is, continuous functions
n=1
preserve convergence of sequences on closed sets.
Proof. Let > 0 be given. Since f is continuous at p, there exists a Оґ > 0
such that

|f (x) в€’ f (p)| < whenever |x в€’ p| < Оґ, and x в€€ S.

The open interval (p в€’ Оґ, p + Оґ) contains the limit point p of S. The sequence
{pn }в€ћ converges to p. There exists some natural numbers N such that for
n=1
all n в‰Ґ N ,
p в€’ Оґ < pn < p + Оґ.
Then
|f (pn ) в€’ f (p)| < whenever n в‰Ґ N.
By deп¬Ѓnition, {f (pn )}в€ћ converges to f (p). We write this statement in the
n=1
following notation:
lim f (pn ) = f lim pn .
nв†’в€ћ nв†’в€ћ

That is, continuous functions allow the interchange of taking the limit and
applying the function. This completes the proof of the theorem.

Corollary 1 If S is a closed and bounded interval [a, b], then Theorem 2.3.11
is valid for [a, b].

Theorem 2.3.12 Let a function f be deп¬Ѓned and continuous on a closed
and bounded set S. Let Rf = {f (x) : x в€€ S}. Then Rf is bounded.
Proof. Suppose that Rf has no upper bound. Then there exists a sequence
{f (xn )}в€ћ , of distinct points of Rf , that diverges to в€ћ. The set A =
n=1
{x1 , x2 , . . . } is an inп¬Ѓnite subset of S. By Theorem 2.3.7, the set A has
some limit point, say p. Since S is closed, p в€€ S. There exists a sequence
2.3. LIMITS AND SEQUENCES 79

{pn }в€ћ , of distinct points of A that converges to p. By the continuity of
n=1
f, {f (pn )}в€ћ converges to f (p). Without loss of generality, we may assume
n=1
that {f (pn )}в€ћ is a subsequence of {f (xn )}в€ћ . Hence {f (pn )}в€ћ diverges
n=1
n=1
n=1
to в€ћ, and f (p) = в€ћ. This is a contradiction, because f (p) is a real number.
This completes the proof of the theorem.

Theorem 2.3.13 Let a function f be deп¬Ѓned and continuous on a closed
and bounded set S. Let Rf = {f (x) : x в€€ S}. Then Rf is a closed set.
Proof. Let q be a limit point of Rf . Then there exists a sequence {f (xn )}в€ћ ,
n=1
of distinct points of Rf , that converges to q. As in Theorem 2.3.12, the set
A = {x1 , x2 , . . . } has a limit point p, p в€€ S, and there exists a subsequence
{pn }в€ћ , of {xn }в€ћ that converges to p. Since f is deп¬Ѓned and continuous
n=1
n=1
on S,
q = lim f (pn ) = f lim pn = f (p).
nв†’в€ћ nв†’в€ћ

Therefore, q в€€ Rf and Rf is a closed set. This completes the proof of the
theorem.

Theorem 2.3.14 Let a function f be deп¬Ѓned and continuous on a closed
and bounded set S. Then there exist two numbers c1 and c2 in S such that
for all x в€€ S,
f (c1 ) в‰¤ f (x) в‰¤ f (c2 ).
Proof. By Theorems 2.3.12 and 2.3.13, the range, Rf , of f is a closed and
bounded set. Let
m = glb(Rf ) and M = lub(Rf ).
Since Rf is a closed set, m and M are in Rf . Hence, there exist two numbers,
say c1 and c2 , in S such that

m = f (c1 ) and M = f (c2 ).

This completes the proof of the theorem.

Deп¬Ѓnition 2.3.10 A set S of real numbers is said to be compact, if and
only if S is closed and bounded.

Theorem 2.3.15 A continuous function maps compact subsets of its domain
onto compact subsets of its range.
80 CHAPTER 2. LIMITS AND CONTINUITY

Proof. Theorems 2.3.13 and 2.3.14 together prove Theorem 2.1.15.

Deп¬Ѓnition 2.3.11 Suppose that a function f is deп¬Ѓned and continuous on
a compact set S. A number m is said to be an absolute minimum of f on S
if m в‰¤ f (x) for all x в€€ S and m = f (c) for some c in S.
A number M is said to be an absolute maximum of f on S if M в‰Ґ f (x)
for all x в€€ S and M = f (d) for some d in S.

Theorem 2.3.16 Suppose that a function f is continuous on a compact set
S. Then there exist two points c1 and c2 in S such that f (c1 ) is the absolute
minimum and f (c2 ) is the absolute maximum of f on S.
Proof. Theorem 2.3.14 proves Theorem 2.3.16.

Exercises 2.3

1. Find lub(A), glb(A) and determine all of the limit points of A.

(a) A = {x : 1 в‰¤ x2 в‰¤ 2}
(b) A = {x : x sin(1/x), x > 0}
(c) A = {x2/3 : в€’8 < x < 8}
(d) A = {x : 2 < x3 < 5}
(e) A = {x : x is a rational number and 2 < x3 < 5}

2. Determine whether or not the following sequences converge. Find the
limit of the convergent sequences.
в€ћ
n
(a)
n + 1 n=1
nв€ћ
(b)
n2 n=1
в€ћ
n
(в€’1)n
(c)
3n + 1 n=1
в€ћ
n2
(d)
n+1 n=1
(в€’1)n }в€ћ
(e) {1 + n=1
2.3. LIMITS AND SEQUENCES 81

3. Show that the Dedekind-Cut Property is equivalent to the completeness
property.

4. Show that a convergent sequence cannot have more than one limit point.

5. Show that the following principle of mathematical induction is valid: If
1 в€€ S, and k + 1 в€€ S whenever k в€€ S, then S contains the set of all
natural numbers. (Hint: Let A = {n : n в€€ S}. A is bounded from below
/
by 2. Let m = glb(A). Then k = m в€’ 1 в€€ S but k + 1 = m в€€ S. This is
/

6. Prove that every rational number is a limit point of the set of all rational
numbers.

7. Let {an }в€ћ be a sequence of real numbers. Then
n=1

(i) {an }в€ћ is said to be increasing if an < an+1 , for all n.
n=1

(ii) {an }в€ћ is said to be non-decreasing if an в‰¤ an+1 for all n.
n=1

(iii) {an }в€ћ is said to be non-increasing if an в‰Ґ an+1 for all n.
n=1

(iv) {an }в€ћ is said to be decreasing if an > an+1 for all n.
n=1

(v) {an }в€ћ is said to be monotone if it is increasing, non-decreasing,
n=1
non-increasing or decreasing.

(a) Determine which sequences in Exercise 2 are monotone.
(b) Show that every bounded monotone sequence converges to some
point.
(c) A sequence {bm }в€ћ is said to be a subsequence of the {an }в€ћ if and
n=1
m=1
only if every bm is equal to some an , and if

bm1 = an1 and bm2 = an2 and n1 < n2 , then m1 < m2 .

That is, a subsequence preserves the order of the parent sequence.
Show that if {an }в€ћ converges to p, then every subsequence of
n=1
в€ћ
{an }n=1 also converges to p
(d) Show that a divergent sequence may contain one or more convergent
sequences.
82 CHAPTER 2. LIMITS AND CONTINUITY

(e) In problems 2(c) and 2(e), п¬Ѓnd two convergent subsequences of each.
Do the parent sequences also converge?

8. (Cauchy Criterion) A sequence {an }в€ћ is said to satisfy a Cauchy Crite-
n=1
rion, or be a Cauchy sequence, if and only if for every > 0, there exists
some natural number N such that (an в€’ am ) < whenever n в‰Ґ N and
m в‰Ґ N . Show that a sequence {an }в€ћ converges if and only if it is a
n=1
Cauchy sequence. (Hint: (i) If {an } converges to p, then for every > 0
there exists some N such that if n в‰Ґ N , then |an в€’ p| < /2. If m в‰Ґ N
and n в‰Ґ N , then
|an в€’ am | = |(an в€’ p) + (p в€’ am )|
в‰¤ |an в€’ p| + |am в€’ p| (why?)
< + =.
22
So, if {an } converges, then it is Cauchy.
(ii) Suppose {an } is Cauchy. Let > 0. Then there exists N > 0 such
that
|an в€’ am | < whenever n в‰Ґ N and m в‰Ґ N.
In particular,
|an в€’ aN | < whenever n в‰Ґ N.
Argue that the sequence {an } is bounded. Unless an element is repeated
inп¬Ѓnitely many times, the set consisting of elements of the sequence has a
limit point. Either way, it has a convergent subsequence that converges,
say to p. Then show that the Cauchy Criterion forces the parent sequence
{an } to converge to p also.)
9. Show that the set of all rational numbers is countable. (Hint: First show
that the positive rationals are countable. List them in reduced form
without repeating according to denominators, as follows:

01234
, , , , ,В·В·В· .
11111
1352
, , , ,В·В·В· .
2222
1 2 4 5 7 8 10
, , , , , , ,В·В·В· .
333333 3
2.3. LIMITS AND SEQUENCES 83

Count them as shown, one-by-one. That is, list them as follows:
113 5211
, , , 2, 3, , , , , В· В· В·
0, 1, .
232 2345

Next, insert the negative rational right after its absolute value, as follows:
1 11 1
0, 1, в€’1, , в€’ , , в€’ , В· В· В· .
2 23 3
Now assign the even natural numbers to the positive rationals and the
odd natural numbers to the remaining rationals.)

10. A non-empty set S has the property that if x в€€ S, then there is some
open interval (a, b) such that x в€€ (a, b) вЉ‚ S. Show that the complement
of S is closed and hence S is open.
в€ћ
ПЂ (в€’1)n
11. Consider the sequence an = + . Determine the conver-
2 n n=1
gent or divergent properties of the following sequences:

(a) {sin(an )}в€ћ
n=1

(b) {cos(an )}в€ћ
n=1

(c) {tan(an )}в€ћ
n=1

(d) {cot(an )}в€ћ
n=1

(e) {sec(an )}в€ћ
n=1

(f) {csc(an )}в€ћ
n=1

12. Let

(a) f (x) = x2 , в€’2 в‰¤ x в‰¤ 2
(b) g(x) = x3 , в€’2 в‰¤ x в‰¤ 2
в€љ
(c) h(x) = x, 0 в‰¤ x в‰¤ 4
(d) p(x) = x1/3 , в€’8 в‰¤ x в‰¤ 8

Find the absolute maximum and absolute minimum of each of the func-
tions f, g, h, and p. Determine the points at which the absolute maximum
and absolute minimum are reached.
84 CHAPTER 2. LIMITS AND CONTINUITY

13. A function f is said to have a п¬Ѓxed point p if f (p) = p. Determine all of
the п¬Ѓxed points of the functions f, g, h, and p in Exercise 12.

14. Determine the range of each of the functions in Exercise 12, and show
that it is a closed and bounded set.

2.4 Properties of Continuous Functions
We recall that if two functions f and g are deп¬Ѓned and continuous on a
common domain D, then f + g, f в€’ g, af + bg, g В· f are all continuous on
D, for all real numbers a and b. Also, the quotient f /g is continuous for all
x in D where g(x) = 0. In section 2.3 we proved the following:

(i) Continuous functions preserve convergence of sequences.

(ii) Continuous functions map compact sets onto compact sets.

(iii) If a function f is continuous on a closed and bounded interval [a, b], then
{f (x) : x в€€ [a, b]} вЉ† [m, M ], where m and M are absolute minimum and
absolute maximum of f , on [a, b], respectively.

Theorem 2.4.1 Suppose that a function f is deп¬Ѓned and continuous on
some open interval (a, b) and a < c < b.

(i) If f (c) > 0, then there exists some Оґ > 0 such that f (x) > 0 whenever
c в€’ Оґ < x < c + Оґ.

(ii) If f (c) < 0, then there exists some Оґ > 0 such that f (x) < 0 whenever
c в€’ Оґ < x < c + Оґ.

1
|f (c)|. For both cases (i) and (ii), > 0. Since f is
Proof. Let =
2
continuous at c and > 0, there exists some Оґ > 0 such that a < (c в€’ Оґ) <
c < (c + Оґ) < b and

|f (x) в€’ f (c)| < whenever |x в€’ c| < Оґ.
2.4. PROPERTIES OF CONTINUOUS FUNCTIONS 85

We observe that
1
|f (x) в€’ f (c)| < в†” |f (x) в€’ f (c)| < |f (c)|
2
1 1
в†”в€’ |f (c)| < f (x) в€’ f (c) < |f (c)|
2 2
1 1
в†” f (c) в€’ |f (c)| < f (x) < f (c) + |f (c)|.
2 2
1 1
We note also that the numbers f (c) в€’ |f (c)| and f (c) + |f (c)| have the
2 2
same sign as f (c). Therefore, for all x such that |xв€’c| < Оґ, we have f (x) > 0
in part (i) and f (x) < 0 in part (ii) as required. This completes the proof.

Theorem 2.4.2 Suppose that a function f is deп¬Ѓned and continuous on
some closed and bounded interval [a, b] such that either
(i) f (a) < 0 < f (b) or (ii) f (b) < 0 < f (a).
Then there exists some c such that a < c < b and f (c) = 0.
Proof. Part (i) Let A {x : x в€€ [a, b] and f (x) < 0}. Then A is non-
empty because it contains a. Since A is a subset of [a, b], A is bounded. Let
c1 = lub(A). We claim that f (c1 ) = 0. Suppose f (c1 ) = 0. Then f (c1 ) > 0
or f (c1 ) < 0. By Theorem 2.4.1, there exists Оґ > 0 such that f (x) has the
same sign as f (c1 ) for all x such that c1 в€’ Оґ < x < c1 + Оґ.
If f (c1 ) < 0, then f (x) < 0 for all x such that c1 < x < c1 + Оґ and hence
c1 = lub(A). If f (c1 ) > 0, then f (x) > 0 for all x such that c1 в€’ Оґ < x < c1
and hence c1 = lub(A). This contradiction proves that f (c1 ) = 0.

Part (ii) is proved by a similar argument.

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