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Solutions for Test E1
I:
a)
7

h

3
< ε
⇒
δ
= (
ε
7
)
1
/
3
b) First statement is false.
Take
a
= 1
/
100 then
√
a
= 1
/
10.
Second statement is
true.
c) The expression is positive for all
x <

2 and all
x >

1 since
x
2
+ 3
x
+ 2 = (
x
+ 2)(
x
+ 1)
>
0
.
d)
‘
(
x
) =

2(
x

1) + 2 =

2
x
+ 4
.
e)
‘
(
x
) = (
x

1) + 2 =
x
+ 1
.
II:
a) Limit is 3
b) Limit is

13
/
2
c) Limit is 1
/
2
d) Limit is 1
/
2
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
) =

2 +
h

(
√
2 +
p
1 + (1 +
h
)
2
)
p
1 + (1 +
h
)
2
c) One reasonable answer to this problem is as follows. For

h

<
1
g
(
h
)
<
3
(
√
2 + 1)
1
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View Full DocumentHence for

h

<
min(1
,
(1 +
√
2)
3
ε
)
we have that

f
(1 +
h
)

f
(1)

< ε .
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 Fall '08
 N/A
 Calculus

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