Suppose first that
lim
x
→
a
+
0
f
(
x
)
=
lim
x
→
a
+
0
g
(
x
)
=
0
.
lim
x
→
a
+
0
f
(
x
)
=
lim
x
→
a
+
0
g
(
x
)
=
0
.
(6)Observe first that, because g'(x)≠0g'(x)≠0 for all xx in some interval (a,a+α),(a,a+α),g'(x)g'(x) is either always positive or always negative on that interval.
(This follows from part (d) of (Reference).)
Therefore
the function gg must be strictly monotonic on the interval (a,a+α).(a,a+α).
Hence, since limx→a+0g(x)=0,limx→a+0g(x)=0,
we must have that g(x)≠0g(x)≠0 on the interval (a,a+α).(a,a+α).
Now, given an ϵ>0,ϵ>0, choose
a positive δ<αδ<α such that
if a<c<a+δa<c<a+δ then |f'(c)g'(c)-L|<ϵ.|f'(c)g'(c)-L|<ϵ.
Then, for every natural number nn for which 1/n<δ,1/n<δ, and every a<x<a+δ,a<x<a+δ,
we have by the Cauchy Mean Value Theorem that
there exists a point cc between a+1/na+1/n and xx such that
|
f
(
x
)
-
f
(
a
+
1
/
n
)
g
(
x
)
-
g
(
a
+
1
/
n
)
-
L
|
=
|
f
'
(
c
)
g
'
(
c
)
-
L
|
<
ϵ
.
|
f
(
x
)
-
f
(
a
+
1
/
n
)
g
(
x
)
-
g
(
a
+
1
/
n
)
-
L
|
=
|
f
'
(
c
)
g
'
(
c
)
-
L
|
<
ϵ
.
(7)Therefore, taking the limit as nn approaches ∞,∞, we obtain
|
f
(
x
)
g
(
x
)
-
L
|
=
lim
n
→
∞
|
f
(
x
)
-
f
(
a
+
1
/
n
)
g
(
x
)
-
g
(
a
+
1
/
n
)
-
L
|
≤
ϵ
|
f
(
x
)
g
(
x
)
-
L
|
=
lim
n
→
∞
|
f
(
x
)
-
f
(
a
+
1
/
n
)
g
(
x
)
-
g
(
a
+
1
/
n
)
-
L
|
≤
ϵ
(8)for all xx for which a<x<a+δ.a<x<a+δ. This proves the theorem in this first case.
Next, suppose that
lim
x
→
a
+
0
f
(
x
)
=
lim
x
→
a
+
0
g
(
x
)
=
∞
.
lim
x
→
a
+
0
f
(
x
)
=
lim
x
→
a
+
0
g
(
x
)
=
∞
.
(9)This part of the theorem is a bit more complicated to prove.
First, choose a positive αα so that f(x)f(x) and g(x)g(x) are both
positive on the interval (a,a+α).(a,a+α).
This is possible because both functions are tending to infinity as xx approaches a.a.
Now, given an ϵ>0,ϵ>0, choose a positive number β<αβ<α such that
|
f
'
(
c
)
g
'
(
c
)
-
L
|
<
ϵ
2
|
f
'
(
c
)
g
'
(
c
)
-
L
|
<
ϵ
2
(10)for all a<c<a+β.a<c<a+β.
We express this absolute value inequality as the following pair of ordinary inequalities:
L
-
ϵ
2
<
f
'
(
c
)
g
'
(
c
)
<
L
+
ϵ
2
.
L
-
ϵ
2
<
f
'
(
c
)
g
'
(
c
)
<
L
+
ϵ
2
.
(11)Set y=a+β.y=a+β. Using the Cauchy Mean Value Theorem, and the preceding inequalities, we
have that for all a<x<ya<x<y
L
-
ϵ
2
<
f
(
x
)
-
f
(
y
)
g
(
x
)
-
g
(
y
)
<
L
+
ϵ
2
,
L
-
ϵ
2
<
f
(
x
)
-
f
(
y
)
g
(
x
)
-
g
(
y
)
<
L
+
ϵ
2
,
(12)implying that
(
L
-
ϵ
2
)
(
g
(
x
)
-
g
(
y
)
)
+
f
(
y
)
<
f
(
x
)
<
(
L
+
ϵ
2
)
(
g
(
x
)
-
g
(
y
)
)
+
f
(
y
)
.
(
L
-
ϵ
2
)
(
g
(
x
)
-
g
(
y
)
)
+
f
(
y
)
<
f
(
x
)
<
(
L
+
ϵ
2
)
(
g
(
x
)
-
g
(
y
)
)
+
f
(
y
)
.
(13)Dividing through by g(x)g(x) and simplifying we obtain
L
-
ϵ
2
-
(
L
-
ϵ
2
)
g
(
y
)
g
(
x
)
+
f
(
y
)
g
(
x
)
<
f
(
x
)
g
(
x
)
<
L
+
ϵ
2
-
(
L
+
ϵ
2
)
g
(
y
)
g
(
x
)
+
f
(
y
)
g
(
x
)
.
L
-
ϵ
2
-
(
L
-
ϵ
2
)
g
(
y
)
g
(
x
)
+
f
(
y
)
g
(
x
)
<
f
(
x
)
g
(
x
)
<
L
+
ϵ
2
-
(
L
+
ϵ
2
)
g
(
y
)
g
(
x
)
+
f
(
y
)
g
(
x
)
.
(14)Finally, using the hypothesis that limx→a+0g(x)=∞,limx→a+0g(x)=∞,
and the fact that L,ϵ,g(y),L,ϵ,g(y), and f(y)f(y) are all constants,
choose a δ>0,δ>0, with δ<β,δ<β, such that if a<x<a+δ,a<x<a+δ,
then
|
-
(
L
-
ϵ
2
)
g
(
y
)
g
(
x
)
+
f
(
y
)
g
(
x
)
|
<
ϵ
2
|
-
(
L
-
ϵ
2
)
g
(
y
)
g
(
x
)
+
f
(
y
)
g
(
x
)
|
<
ϵ
2
(15)and
|
-
(
L
+
ϵ
2
)
g
(
y
)
g
(
x
)
+
f
(
y
)
g
(
x
)
|
<
ϵ
2
.
|
-
(
L
+
ϵ
2
)
g
(
y
)
g
(
x
)
+
f
(
y
)
g
(
x
)
|
<
ϵ
2
.
(16)Then, for all a<x<a+δ,a<x<a+δ, we would have
L
-
ϵ
<
f
(
x
)
g
(
x
)
<
L
+
ϵ
,
L
-
ϵ
<
f
(
x
)
g
(
x
)
<
L
+
ϵ
,
(17)implying that
|
f
(
x
)
g
(
x
)
-
L
|
<
ϵ
,
|
f
(
x
)
g
(
x
)
-
L
|
<
ϵ
,
(18)and the theorem is proved.
