MAA 4200: Homework Solutions
June 19, 2009
Sequential Limit Theorem = Theorem 2
.
1
Algebra of Limits Theorem = Theorem 2
.
4
[2.12] Prove that if lim
x
→
x
0
f
(
x
) =
L
then lim
x
→
x
0

f
(
x
)

=

L

.
Preliminary Work:
 
f
(
x
)
  
L
  ≤ 
f
(
x
)

L

for all
x
∈
D
by the triangle inequality.
Proof:
Let
>
0.
By definition of limit, there exists
δ
1
>
0 such that

f
(
x
)

L

<
for 0
<

x

x
0

< δ
1
.
Choose
δ
=
δ
1
and let 0
<

x

x
0

< δ
.
Then
 
f
(
x
)
  
L
  ≤ 
f
(
x
)

L

<
.
[2.13] In Example 2.6 on p. 71 it is proved that lim
x
→
x
0
[
x
] exists iff
x
0
is not an integer.
Thus, since lim
x
→
x
0
exists for all
x
0
∈
R
, the limit lim
x
→
x
0
(
x

[
x
]) exists for all
x
0
/
∈
Z
by
the Algebra of Limits Theorem.
We can also apply the Algebra of Limits Theorem to show that lim
x
→
x
0
(
x

[
x
0
]) does not
exist for
x
0
∈
Z
. Let
f
(
x
) =
x

[
x
]. Proof by contradiction: suppose lim
x
→
x
0
f
(
x
) exists
for some
x
0
∈
Z
. Then lim
x
→
x
0
[
x
] = lim
x
→
x
0
(
x

f
(
x
)) exists by the Algebra of Limits
Theorem, which is a contradiction.
[2.14] Show that lim
x
→
2
f
(
x
) = 16 and the limit does not exist at any other point.
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 Spring '07
 thieme
 Algebra, Limits, LG, Sequential Limit Theorem

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