This preview shows pages 1–3. Sign up to view the full content.
PART III. FUNCTIONS: LIMITS AND CONTINUITY
III.1.
LIMITS OF FUNCTIONS
This chapter is concerned with functions
f
:
D
→
R
where
D
is a nonempty subset of
R
. That is, we will be considering realvalued functions of a real variable. The set
D
is
called the
domain
of
f
.
Defnition 1.
Let
f
:
D
→
R
and let
c
be an accumulation point of
D
. A number
L
is the
limit o±
f
at
c
if to each
±>
0
there exists a
δ>
0
such that

f
(
x
)

L

<±
whenever
x
∈
D
and
0
<

x

c

<δ
.
This deFnition can be stated equivalently as follows:
Defnition.
Let
f
:
D
→
R
and let
c
be an accumulation point of
D
. A number
L
is
the
limit o±
f
at
c
if to each neighborhood
V
of
L
there exists a deleted neighborhood
U
of
c
such that
f
(
U
∩
D
)
⊆
V
.
Notation
lim
x
→
c
f
(
x
)=
L
.
Examples:
(a) lim
x
→
2
(
x
2

2
x
+ 4) = 12.
(b) lim
x
→
2
x
2

4
x

2
=4
.
(c) lim
x
→
3
x
2
+3
x
+5
x

3
does not exist.
(d) lim
x
→
1

x

1

x

1
does not exist.
Example:
Let
f
(
x
)=4
x

5. Prove that lim
x
→
3
f
(
x
)=7
.
Proo±:
Let
0.

f
(
x
)

7

=

(4
x

5)

7

=

4
x

12


x

3

.
Choose
δ
=
±/
4. Then

f
(
x
)

7


x

3

<
4
±
4
=
±
whenever
0
<

x

3

.
Two Obvious Limits:
23
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document (a) For any constant
k
and any number
c,
lim
x
→
c
k
=
k
.
(b) For any number
c,
lim
x
→
c
x
=
c
.
THEOREM 1.
Let
f
:
D
→
R
and let
c
be an accumulation point of
D
. Then
lim
x
→
c
f
(
x
)=
L
if and only if for every sequence
{
s
n
}
in
D
such that
s
n
→
c, s
n
±
=
c
for all
n
,
f
(
s
n
)
→
L
.
Proof:
Suppose that lim
x
→
c
f
(
x
L
. Let
{
s
n
}
be a sequence in
D
which converges to
c, s
n
±
=
c
for all
n
. Let
±>
0. There exists
δ>
0 such that

f
(
x
)

L

<±
whenever
0
<

x

c

<δ
(
x
∈
D
)
.
Since
s
n
→
c
there exists a positive integer
N
such that

c

s
n

for all
n>N
.
Therefore

f
(
s
n
)

L

for all
and
f
(
s
n
)
→
L.
Now suppose that for every sequence
{
s
n
}
in
D
which converges to
c
,
f
(
s
n
)
→
L
.
Suppose that lim
x
→
c
f
(
x
)
±
=
L
. Then there exists an
0 such that for each
0 there is
an
x
∈
D
with 0
<

x

c

but
f
(
x
)

L
≥
±
. In particular, for each positive integer
n
there is an
s
n
∈
D
such that

c

s
n

<
1
/n
and

f
(
s
n
)

L
±
. Now,
s
n
→
c
but
{
f
(
s
n
)
}
does not converge to
L
, a contradiction.
Corollary
Let
f
:
D
→
R
and let
c
be an accumulation point of
D
. If lim
x
→
c
f
(
x
) exists,
then it is unique. That is,
f
can have only one limit at
c
.
THEOREM 2.
f
:
D
→
R
and let
c
be an accumulation point of
D
.I
f
lim
x
→
c
f
(
x
)
does not exist, then there exists a sequence
{
s
n
}
in
D
such that
s
n
→
c
, but
{
f
(
s
n
)
}
does not converge.
Proof:
Suppose that lim
x
→
c
f
(
x
) does not exist. Suppose that for every sequence
{
s
n
}
in
D
such that
s
n
→
c
(
s
n
±
=
c
),
{
f
(
s
n
)
}
converges. Let
{
s
n
}
and
{
t
n
}
be sequences in
D
which converge to
c
. Then
{
f
(
s
n
)
}
and
{
f
(
t
n
)
}
are convergent sequences. Let
{
u
n
}
be
the sequence
{
s
1
,t
1
,s
2
2
,...
}
. Then
{
u
n
}}
converges to
c
and
{
f
(
u
n
)
}
converges to
some number
L
. Since
{
f
(
s
n
)
}
and
{
f
(
t
n
)
}
are subsequences of
{
f
(
u
n
)
}
,
f
(
s
n
)
→
L
and
f
(
t
n
)
→
L
. Therefore, for every sequence
{
s
n
}
in
D
such that
s
n
→
c, s
n
±
=
c
for
all
n
,
f
(
s
n
)
→
L
and lim
x
→
c
f
(
x
L
.
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 01/06/2011 for the course MATH 3333 taught by Professor Staff during the Spring '08 term at University of Houston.
 Spring '08
 Staff
 Continuity, Limits

Click to edit the document details