The Real And Complex Number Systems

5.28 Prove the following theorem:
Let f and g be two functions having finite nth derivatives in a, b. For some interior
point c in a, b, assume that fc f c ... f n1 c 0, and that
gc g c ... g n1 c 0, but that g n x is never zero in a, b. Show that
lim xc
fx
gx
fn c
g n c .
NOTE. f n and g n are not assumed to be continuous at c.
Hint. Let
Fx fx x cn2 f n c
,
n 2!
define G similarly, and apply Theorem 5.20 to the functions F and G.
Proof: Let
Fx fx
fn c
x cn2
n 2!
and
Gx gx
gn c
x cn2
n 2!
then inductively,
F k x f k f
x
n c
x cn2k
n 2 k!
and note that
F k c 0 for all k 0, 1, . . , n 3, and F n2 c f n c.
Similarly for G. Hence, by Theorem 5.20, wehave
n2
Fx
k0
F
k
k! x ck G n1 x 1 F n1 x 1 Gx
k0
n2
G
k
k!
x c k
where x 1 between x and c, which implies that
fxg n1 x 1 f n1 x 1 gx. *
Note that since g n is never zero on a, b; it implies that there exists a 0 such that
every g k is never zero in c , c c, wherek 0, 1, 2. . . , n. Hence, we have, by
(*),
lim xc
fx
gx lim xc
lim
x1 c
lim
x1 c
f n1 x 1
g n1 x 1
f n1 x 1 f n1 c
g n1 x 1 g n1 c since x c x 1 c
f n1 x 1 f n1 c/x 1 c
g n1 x 1 g n1 c/x 1 c
fn c
g n c since fn exists and g n exists 0 on a, b.
Remark: (1) The hint is not correct from text book. The reader should find the
difference between them.
(2) Here ia another proof by L-Hospital Rule and Remark in Exercise 5.27.
Proof: Since g n is never zero on a, b, it implies that there exists a 0 such that