The Real And Complex Number Systems

Remark: In this exercise, it tells us that the converse of L-Hospital Rule is NOT
necessary true. Here is a good exercise very like L-Hospital Rule, but it does not! We
write it as follows.
Suppose that f a and g a exist with g a 0, and fa ga 0. Prove that
Proof: Consider
lim xa
fx
gx lim xa
lim xa
fx
gx
fx fa
gx ga lim xa
f a
g a .
fx fa/x a
gx ga/x a
f a
g a since f a and g a exist with g a 0.
Note: (i) It should be noticed that we CANNOT use L-Hospital Rule since the
statement tells that f and g have a derivative at a, we do not make sure of the situation of
other points.
(ii) This holds also for complex functions. Let us recall the proof of L-Hospital Rule,
we need use the order field R; however, C is not an order field. Hence, L-Hospital Rule
does not hold for C. Infact,noordercanbedefinedinthecomplexfieldsincei 2 1.
Supplement on L-Hospital Rule
We do not give a proof about the following fact. The reader may see the book named A
First Course in Real Analysis written by Protter and Morrey, Charpter 4, pp 88-91.
Theorem ( 0 ) Let f and g be continuous and differentiable on a, b with 0 g 0ona, b.
If
then
lim
xa
lim
xa
fx 0 lim gx 0and
*
xa f x
g x L,
lim
xa
fx
gx L.
Remark: 1. The size of the interval a, b is of no importance; it suffices to
have g 0ona, a , forsome 0.
2. (*) is a sufficient condition, not a necessary condition. For example,
fx x 2 ,andgx sin 1/x both defined on 0, 1.
3. We have some similar results: x a ; x a; x 1/x 0 ;
x 1/x 0 .
Theorem ( ) Let f and g be continuous and differentiable on a, b with g 0on
a, b. If
then
lim
xa
lim
xa
fx lim gx and
*
xa f x
g x L,