The Real And Complex Number Systems

f fy
c lim n fx n
n y n x n
.
(ii) There is a good exercise; we write it as follows. Let f C 2 a, b, andc a, b.
For small |h| such that c h a, b, write
fc h fc f c hhh
where 0 1. Show that if f c 0, then lim h0 h 1/2.
Proof: Since f C 2 a, b, byTaylor Theorem with Remainder Term, we have
fc h fc f ch f
h
2!
2 ,where c, h or h, c
f c hhh by hypothesis.
So,
f c hh f c
h
h f
2!
and let h 0, we have c by continuity of f at c. Hence,
limh 1/2 since f c 0.
h0
Note: We can modify our statement as follows. Let f be defined on a, b, and
c a, b. For small |h| such that c h a, b, write
fc h fc f c hhh
where 0 1. Show that if f c 0, and x x for x a h, a h, then
lim h0 h 1/2.
Proof: Use the exercise (c), we have
f fc h 2fc fc h
c lim
h0 h 2
f
lim
c hh f c hh
by hypothesis
h0 h
f
lim
c hh f c hh
2h since x x for x a h, a h.
h0 2hh
Since f c 0, we finally have lim h0 h 1/2.
5.25 Let f have a finite derivatiive in a, b and assume that c a, b. Consider the
following condition: For every 0, there exists a 1 ball Bc; , whose radius
depends only on and not on c, such that if x Bc; , andx c, then
fx fc
x c f c .
Show that f is continuous on a, b if this condition holds throughout a, b.
Proof: Given 0, we want to find a 0 such that as dx, y , x, y a, b, we
have
|f x f y| .
Choose any point y a, b, and thus by hypothesis, given /2, there is a 1 ball
By; , whose radius depends only on and not on y, such that if x By; , and
x y, then,
fx fy
x y f y /2 . *
Note that y Bx, , so, we also have
,