Since
M
0, there exists a positive number
d
1
such that for all
x
0
x
c
d
1
⇒
g x
M
M
2
.
(3)
continued

574
Appendices
For any numbers
A
and
B
it can be shown that
A
B
A
B
and
B
A
A
B
,
from which it follows that
A
B
A
B
. With
A
g x
and
B
M
, this
becomes
g x
M
g x
M
,
which can be combined with the inequality on the right in (3) to get, in turn,
g x
M
M
2
M
2
g x
M
M
2
M
2
g x
3
2
M
g
1
x
M
2
g
3
x
.
Multiply by 2
M
g x
.
(4)
Therefore, 0
x
c
d
implies that
g
1
x
M
1
M
Mg
g
x
x
M
1
•
g
1
x
•
M
g x
M
1
•
M
2
•
M
g x
.
Inequality (4)
(5)
Since 1 2
M
2
e
0, there exists a number
d
2
0 such that for all
x
in
D
0
x
c
d
2
⇒
M
g x
2
e
M
2
.
(6)
If we take
d
to be the smaller of
d
1
and
d
2
, the conclusions in (5) and (6) both hold for all
x
such that 0
x
c
d
. Combining these conclusions gives
0
x
c
d
2
⇒
g
1
x
M
1
e
.
This completes the proof of the Limit Quotient Rule.
■
The last proof we give is of the Sandwich Theorem (Theorem 4) of Section 2.1.
THEOREM 4
The Sandwich Theorem
If
g x
f x
h x
for all
x
c
in some interval about
c
, and
lim
x
→
c
g x
lim
x
→
c
h x
L
,
then
lim
x
→
c
f x
L.
Proof for Right-hand Limits
Suppose that lim
x
→
c
g x
lim
x
→
c
h x
L.
Then for any
e
0 there exists a
d
0 such that for all
x
the inequality
c
x
c
d
implies
L
e
g x
L
e
and
L
e
h x
L
e
.
continued