The Real And Complex Number Systems

then we have
lim fx, y L if x 0andy 0,
x,ya,b L if x 0ory 0.
lim fx, y L.
x,ya,b
4.12 If x 0, 1 prove that the following limit exists,
lim
m
lim n cos2n m!x ,
and that its value is 0 or 1, according to whether x is irrational or rational.
Hence,
Proof: If x is rational, say x q/p, whereg. c. d. q, p 1, then p!x N. So,
lim n
cos 2n m!x
lim
m
1ifm p,
0ifm p.
lim n cos2n m!x 1.
If x is irrational, then m!x N for all m N. So, cos 2n m!x 1 for all irrational x.
Hence,
lim n
cos 2n m!x 0 lim m
Continuity of real-valued functions
lim n
cos 2n m!x 0.
4.13 Let f be continuous on a, b and let fx 0 when x is rational. Prove that
fx 0 for every x a, b.
Proof: Given any irrational number x in a, b, and thus choose a sequence x n Q
such that x n x as n . Note that fx n 0 for all n. Hence,
0 lim n
0
lim n
fx n
f lim n
x n by continuity of f at x
fx.
Since x is arbitrary, we have shown fx 0 for all x a, b. Thatis,f is constant 0.
Remark: Here is another good exercise, we write it as a reference. Let f be continuous
on R, andiffx fx 2 , then f is constant.
Proof: Since fx f x 2 fx 2 fx, we know that f is an even function. So,
in order to show f is constant on R, it suffices to show that f is constant on 0, . Given
any x 0, , sincefx 2 fx for all x R, wehavefx 1/2n fx for all n. Hence,
fx lim n
fx
lim n
fx 1/2n
f lim n
x 1/2n by continuity of f at 1
f1 since x 0.
So, we have fx f1 : c for all x 0, . In addition, given a sequence
x n 0, such that x n 0, then we have