Ma1a Final Solutions
Fall 2007
Problem 1.
Let
f
(
x
) be a realvalued function defined on
R
. Assume that
f
(
x
) is
differentiable at
x
= 1, that is,
f
(1) exists. Prove that
f
(
x
) is continuous at
x
= 1.
Solution 1.
Analytic Solution
Let
ε >
0. We want to show that
∃
δ >
0 such that

x

1

< δ
⇒ 
f
(
x
)

f
(1)

< ε
.
Set
c
=
f
(1). Then
∃
δ >
0 such that for 0
<

x

1

< δ
,

f
(
x
)

f
(1)
x

1

c

<
1
2
ε.
Now
c
is a fixed number, so we can choose
δ
>
0 such that
δ

c

< ε/
2. Now let
δ
= min
{
1
, δ , δ
}
, then if

x

1

< δ
,

f
(
x
)

f
(1)

c
(
x

1)

<

x

1

1
2
ε <
1
2
εδ
≤
1
2
ε.
Triangle inequality gives

f
(
x
)

f
(1)
 ≤ 
f
(
x
)

f
(1)

c
(
x

1)

+

c
(
x

1)

<
1
2
ε
+

c
(
x

1)
 ≤
1
2
ε
+
δ

c
 ≤
1
2
ε
+
δ

c

<
1
2
ε
+
1
2
ε
=
ε.
Alternate Limit Solution
To show that
f
is continuous at 1, it suffices to show that
lim
h
→
0
f
(1 +
h
) =
f
(1)
.
Note that for all
h
= 0, we may write this term as
f
(1 +
h
) =
f
(1) +
h
·
f
(1 +
h
)

f
(1)
h
.
Since
f
is differentiable at 1, the difference quotient will have a (finite) limit:
f
(1) = lim
h
→
0
f
(1 +
h
)

f
(1)
h
.
Thus, taking a limit of the entire expression, we obtain
lim
h
→
0
f
(1 +
h
) =
f
(1) + 0
·
f
(1) =
f
(1)
.
Thus,
f
is continuous at 1.
Problem 2.
Is the series
∑
∞
n
=1
(
√
n
4
+ 5

n
2
) absolutely convergent, conditionally
but not absolutely convergent, or divergent? Explain the tests you use and their
requirements.
Solution 2.
We have
(
n
4
+ 5

n
2
) =
(
√
n
4
+ 5

n
2
)(
√
n
4
+ 5 +
n
2
)
(
√
n
4
+ 5 +
n
2
)
=
5
√
n
4
+ 5 +
n
2
(
*
)
1
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2
Now,
n
4
+ 5
>
n
2
n
4
+ 5 +
n
2
>
2
n
2
5
√
n
4
+ 5 +
n
2
<
5
2
n
2
Since 0
≤
√
n
4

5 +
n
2
≤
5
2
n
2
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 Fall '08
 Borodin,A
 Calculus, Linear Algebra, Algebra, The Land, Cos, Limit, lim

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