Solutions to Test 1, Version A
Math 1501, Fall 02.
WG
*
September 10, 2002
Exercise 1
lim
x
→
2
(
x
2

x

6)
2
x
+ 2
= lim
x
→
2
[(
x
+ 2)(
x

3)]
2
x
+ 2
= lim
x
→
2
(
x
+ 2)(
x

3)
2
= 0
.
Exercise 2
Assume that
f
(
x
) =
A
2
x
2
if
x
≤
2
(1

A
)
x
if
x >
2
.
Then
f
is continuous at 2 iff
4
A
2
= lim
x
→
2
+
f
(
x
) = lim
x
→
2
+
(1

A
)
x
= 2(1

A
)
.
This means that 4
A
2
+ 2
A

2 = 0
,
and so,
A
= 1
/
2 or
A
=

1
.
Exercise 3
For all
c
∈
R
,
we have that lim
x
→
c
f
(
x
) = lim
x
→
c
x
2
=
c
2
.
Case 1:
c
∈
N
.
The function
f
is continuous a
c
if and only if
c
2
= lim
x
→
c
f
(
x
) =
f
(
c
) = 2
c,
i.e.
c
2

2
c
= 0
.
Hence
c
= 0 or
c
= 2
.
Case 2:
c
6∈
N
.
We have that lim
x
→
c
f
(
x
) =
c
2
=
f
(
c
)
.
Consequently,
f
is continuous
a
c.
Summary.
f
is continuous a
c
if and only if
c
= 0
,
or
c
= 2
,
or
c
6∈
N
.
Exercise 4
Recall that
f
(
x
) =
1
(
x

1)
2
+
1
x

10
√
2
,
and so,
f
is continuous on the interval
(1
,
10
√
2
)
.
Set
a
= 1
.
01
, b
=
10
√
2

0
.
001
.
We first use that
10
√
2

1
.
01
>
1 to deduce that
f
(
a
) = 10000 +
1
1
.
01

10
√
2
>
10000

1
>
0
.
*
School of Mathematics, Georgia Institute of technology.
1
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Then we use that
10
√
2

1
.
001
>
1 to deduce that
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 Fall '08
 N/A
 Math, Calculus, Order theory, School of Mathematics, Georgia Institute of Technology, greatest lower bound, Version A Math

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