Notes on Homework #7
Exercise 4.2.1 (c)
Most of you had the right idea, writing

x
3

8

=

x

2
 
x
2
+ 2
x
+ 4

.
Now you need to produce an upper bound for the second factor. If we commit ourselves to choosing
δ
≤
1
then

x

2

< δ
implies that

x

2

<
1, so that

1
< x

2
<
1
1
< x <
3
7 = 1
2
+ 2
·
1 + 4
< x
2
+ 2
x
+ 4
<
3
2
+ 2
·
3 + 4 = 19

x
2
+ 2
x
+ 4

<
19
.
(It is important that we produce these bounds
outside
the absolute value. That is because
a < b
does
not
imply

a

<

b

without a lower bound an
a
.) Now continuing in the usual way, you will come to the choice
of
δ
= min(1
, /
19).
Exercise 4.2.8
The most efficient way to do this is using Theorem 4.2.3 to relate the functional limits to
sequence limits, and then appealing to the Order Limit Theorem (Theorem 2.3.4) for sequences.
Some of you tried to work directly from the

δ
definition of limit. That is possible, but more difficult
than using sequences. Here is how it would go. Let
L
f
= lim
x
→
c
f
(
x
) and
L
g
= lim
x
→
c
g
(
x
). For any
>
0
there exist
δ
f
, δ
g
>
0 so that

f
(
x
)

L
f

<
whenever 0
<

x

c

< δ
f
and
x
∈
A,

g
(
x
)

L
g

<
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 '08
 KBHANNSGEN
 Calculus, Topology, Continuous function, LG, Limit of a sequence, Partially ordered set, Lf

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