Solutions for Test C1
I:
a)
5

h

2
< ε
⇒
δ
= (
ε
5
)
1
/
2
b) First statement is false. Take
a
=

2 and
b
= 1. Second statement is true.
c) The expression is positive for all
x <

2 and all
x >
1 since
x
2
+
x

2 = (
x
+ 2)(
x

1)
>
0
.
d)
‘
(
x
) = 2(
x

2) + 1 = 2
x

3
.
e)
‘
(
x
) = (
x

2) + 1 =
x

1
.
II:
a) Limit is 4
/
3
b) Limit is 3
/
2
c) Limit is 1
/
2
d) Limit is 0
e) Limit does not exist, since the limit from the right yields +1 and the limit from
the left yields

1.
III:
a) For all values of
x
.
b)
g
(
h
) =

8 +
h

(5 +
p
9 + (4 +
h
)
2
)
p
9 + (4 +
h
)
2
c) One reasonable answer to this problem is as follows. For

h

<
1
g
(
h
)
<
9
(5 +
√
9 + 3
2
)
√
9 + 3
2
=
3
(5 + 3
√
2)
√
2
Hence for

h

<
min(1
,
(5 + 3
√
2)
√
2
3
ε
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 Fall '08
 N/A
 Calculus, Limit, Expression, Leftwing politics, Test C1

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