Math 117, F 09, HW#7
(sample solutions)
P.80, #27.
Set
X
=
D
∩
[
x
0

, x
0
+ ] =
E
∩
[
x
0

, x
0
+ ]. (These two sets are equal
by hypothesis.) We are also assuming that
f
(
x
) =
g
(
x
) for all
x
∈
X
.
Suppose that
f
has a limit at
x
0
, say
L
=
lim
x
→
x
0
f
(
x
). We claim that then
L
is also a
limit for
g
at
x
0
. So, let
>
0. Since
L
is the limit of
f
at
x
0
, there exists
δ >
0 such
that

f
(
x
)

L

<
for all
x
∈
D
with 0
<

x

x
0

< δ
. Set
δ
= min
{
δ,
}
, which is also
positive. If
x
∈
E
with 0
<

x

x
0

< δ
, then because
δ
≤
, we have
x
∈
E
∩
(
x
0

, x
0
+ )
⊆
X
=
D
∩
[
x
0

, x
0
+ ]
⊆
D.
Consequently,
f
(
x
) =
g
(
x
) and
x
∈
D
with 0
<

x

x
0

< δ
, so

f
(
x
)

L

<
and thus

g
(
x
)

L

<
. This proves that
g
has limit
L
at
x
0
, as claimed.
This proves one of the claimed implications: If
f
has a limit at
x
0
, then
g
has a limit
at
x
0
. By switching the roles of
f
and
g
, we have the other implication: If
g
has a limit at
x
0
, then
f
has a limit at
x
0
.
P.104, #1.
Since
f
(2) = 9, we need to show that
f
has limit 9 at 2. Then, Theorem 3.1
will tell us that
f
is continuous at 2.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Akhmedov,A
 Calculus, Sets, Limit, Continuous function

Click to edit the document details