5.7 Suppose f 0 (x), g 0 (x) exist, g 0 (x) 6= 0, and f (x) = g(x) = 0. Prove that
Rudin’s Ex. 7
f (t)
f 0 (x)
= 0
.
t→x g(t)
g (x)
lim
Show that it also holds for complex functions.
Proof By the definition of derivative,
lim
t→x
f (t) − f (x)
= f 0 (x),
t−x
lim
t→x
g(t) − g(x)
= g 0 (x) 6= 0.
t−x
By Theorem 4.4, we have
f (t) − f (x)
f (t) − f (x)
lim
f (t)
f 0 (x)
t→x
t−x
t−x
= lim
.
lim
=
= 0
t→x g(t) − g(x)
t→x g(t)
g(t) − g(x)
g (x)
lim
t→x
t−x
t−x
Since Theorem 4.4 is true for complex functions, we know that the limit also holds
for complex functions.
5.8 Suppose f 0 is continuous on [a, b] and > 0. Prove that there exists δ > 0 such that
f (t) − f (x)
0
<
−
f
(x)
t−x
whenever 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b. Does this hold for vector-valued
functions too?
0
0
Proof Since f is continuous on [a, b], by Theorem 4.19, f is uniformly continuous
on [a, b]. There exists δ > 0 such that
|f 0 (y) − f 0 (x)| < .
whenever |y − x| < δ, 0 ≤ x ≤ b, a ≤ y ≤ b.
For t and x satisfying 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b, by the Mean Value
Theorem, there is c between t and x, such that
f (t) − f (x)
= f 0 (c).
t−x
It is clear that c satisfies |c−x| < δ and c ∈ [a, b]. Hence, if 0 < |t−x| < δ, a ≤ x ≤ b,
a ≤ t ≤ b, we have
f (t) − f (x)
0
− f (x) < .
t−x
For vector-valued functions, the statement still holds. In fact, if we put f (x) =
(f1 (x), . . . , fk (x)), and if f is continuous on [a, b], then each fi , 1 ≤ i ≤ k, is uniformly
continuous on [a, b]. Hence, for fixed i, there exists δi > 0 such that
√
|fi0 (y) − fi0 (x)| < / k.
1
Rudin’s Ex. 8
The inequality seems
the same as in the
definition of derivative.
But the current
problem is different,
since it does not
require x to be a fixed
point in [a, b].
whenever |y − x| < δi , 0 ≤ x ≤ b, a ≤ y ≤ b. Put δ = min δi . Then δ > 0, and
1≤i≤k
√
|fi0 (y) − fi0 (x)| < / k,
i = 1, . . . , k,
whenever |y − x| < δ, 0 ≤ x ≤ b, a ≤ y ≤ b.
For t and x satisfying 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b, for each i, by the Mean
Value Theorem, there is ci between t and x, such that
fi (t) − fi (x)
= fi0 (ci ).
t−x
It is clear that ci satisfies |ci − x| < δ and c ∈ [a, b]. Hence, if 0 < |t − x| < δ,
a ≤ x ≤ b, a ≤ t ≤ b, we have
2
2
k X
f (t) − f (x)
fi (t) − fi (x)
0
0
=
−
f
(x)
−
f
(x)
i
t−x
t−x
i=1
=
k
X
i=1
or
2
(fi0 (ci ) − fi0 (x)) <
k
X
2 /k = 2 ,
i=1
f (t) − f (x)
0
− f (x) < .
t−x
5.9 Let f be a continuous real function on R, of which it is known that f 0 (x) exists for
all x 6= 0 and that f 0 (x) → 3 as x → 0. Does it follow that f 0 (0) exists?
Proof Put F (x) = f (x) − f (0) and G(x) = x. It is easy to verify the hypotheses of
Theorem 5.13 to have
F (x)
F 0 (x)
= lim
= lim f 0 (x) = 3,
x→0− G(x)
x→0− G0 (x)
x→0−
lim
and
F (x)
F 0 (x)
= lim
= lim f 0 (x) = 3,
x→0+ G(x)
x→0+ G0 (x)
x→0+
lim
These give
lim
x→0−
f (x) − f (0)
f (x) − f (0)
= lim
= 3.
x→0+
x
x
By the definition of limit, we have
lim
x→0
f (x) − f (0)
= 3,
x
thats is, f 0 (0) = 3 by the definition of derivative.
2
Rudin’s Ex. 9
5.10 Suppose f and g are complex differentiable functions on (0, 1), f (x) → 0, g(x) → 0,
f 0 (x) → A, g 0 (x) → B as x → 0, where A and B are complex numbers, B 6= 0.
Prove that
A
f (x)
= .
lim
x→0 g(x)
B
Proof Put f (x) = f1 (x) + if2 (x), where f1 , f2 are real functions on (0, 1). We know
that f 0 = f10 + if20 by the definition. Since f1 , f2 are real, it is easy to verify the
hypotheses of l’Hospital’s Rule to find the limits:
lim
x→0
f1 (x)
= lim f10 (x),
x→0
x
lim
x→0
f2 (x)
= lim f20 (x).
x→0
x
Hence,
f (x)
= lim
x→0
x→0 x
lim
f1 (x)
f2 (x)
+i
x
x
lim (f10 (x) + if20 (x)) = A.
x→0
g(x)
= B. Hence, for B 6= 0, we have
x
f (x)
f (x)
x
x
lim
= lim
−A ·
+A·
x→0 g(x)
x→0
x
g(x)
g(x)
1
A
1
= .
= (A − A) · + A ·
B
B
B
A similar argument gives lim
x→0
3
Rudin’s Ex. 10