MAT 319/320
Solutions for HW8
Exercise 1.
Section 4.3, #5a.
Proof.
We have that lim
x
→
1+
1
x

1
=+
∞
. Indeed, for any given
α >
0
if we take
x
∈
(1
,
1+
1
α
)
,
then we have that
1
x

1
> α
. Now since lim
x
→
1+
x
=1
, we know that by the comparison the
orem, lim
x
→
1+
x
x

1
=+
∞
if and only if lim
x
→
1+
1
x

1
=+
∞
, which is the case. In conclusion,
lim
x
→
1+
x
x

1
=+
∞
.
square
Exercise 2.
Section 4.3, #5c.
Proof.
lim
x
→
0+
x
+2=2
and lim
x
→
0+
1
x
√
=+
∞
: indeed for any given
α >
0
if we take
x
∈
(0
,
1
α
2
)
, then we have that
1
x
√
> α
.Thus again by the comparison theorem we know that
lim
x
→
0+
x
+2
x
√
=+
∞
.
square
Exercise 3.
Section 4.3, #8.
Proof.
lim
x
→
+
∞
f
(
x
)=
L
means the following:
for any given
ε >
0
there exists an
α >
0
such that
x > α
⇒
f
(
x
)
∈
(
L

ε, L
+
ε
)
.
If one writes
δ
=1/
α
the condition
x > α
is equivalent to
y
=1/
x < δ
, so the condition is now
equivalent to:
for any given
ε >
0
there exists a
δ >
0
such that 0<
y < δ
⇒
f
(1/
y
)
∈
(
L

ε, L
+
ε
)
. But this
exactly means that lim
x
→
0+
f
(1/
y
)=
L
.
square
Exercise 4.
Section 5.1, #7.
Proof.
See the solutions for the Practice midterm II, where this is proved...
square
Exercise 5.
Section 5.1, #12.
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 Fall '10
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