The Real And Complex Number Systems

x r y r rz r1 x y
ry r1 /2.
So, if we choose y 2 1
r1
, then we have
r
x r y r 1
which is absurb with (*). Hence, x r is not uniformly continuous on 0, .
Ps: The reader should try to realize why x r is not uniformly continuous on 0, , for
r 1. The ruin of non-uniform continuity comes from that x is large enough. At the same
time, compare it with theorem that a continuous function defined on a compact set K is
uniflormly continuous on K.
(sin x r )Asr 0, it means that x r is a constant function. So, it is obviuos. As r 0, 1,
given 0, there is a 1/r 0 such that as |x y| , x, y 0, , wehave
|sin x r sin y r x
| 2cos
r y r
sin
xr y r
2
2
|x r y r |
|x y| r by the note in the Remark 4.
r
.
So, sin x r is uniformly continuous on 0, , ifr 0, 1.
As r 1, assume that sin x r is uniformly continuous on 0, , thengiven 1, there
is a 0 such that as |x y| , x, y 0, , wehave
|sin x r sin y r | 1. **
Consider a sequence n 2 1/r n 1/r , it is easy to show that the sequence tends to
0asn . So, there exists a positive integer N such that |x y| , x n 2 1/r ,
y n 1/r . Then
sin x r sin y r 1
which contradicts (**). So, we know that sin x r is not uniformly continuous on 0, .
Ps: For n 2 1/r n 1/r : x n 0asn 0, here is a short proof by using
L-Hospital Rule.
Proof: Write
x n n 1/r
n
1/r
2
n 1/r 1 1 2n
1/r
1
and thus consider the following limit
1 1
2n 1/r 1
n 1/r