Page 67 - Schaum's Outline of Theory and Problems of Advanced Calculus

58 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3

UNIFORM CONTINUITY
2
3.29. Prove that f ðxÞ¼ x is uniformly continuous in 0 < x < 1.
Method 1: Using definition.
2
2
We must show that given any > 0wecan find > 0suchthat jx x 0 j < when jx x 0 j < , where
depends only on and not on x 0 where 0 < x 0 < 1.
If x and x 0 are any points in 0 < x < 1, then
2 2
jx x 0 j¼jx þ x 0 jjx x 0 j < j1 þ 1jjx x 0 j¼ 2jx x 0 j
2
2
2
2
Thus if jx x 0 j < it follows that jx x 0 j < 2 . Choosing ¼ =2, we see that jx x 0 j < when
2
jx x 0 j < , where depends only on and not on x 0 . Hence, f ðxÞ¼ x is uniformly continuous in
0 < x < 1.
2
The above can be used to prove that f ðxÞ¼ x is uniformly continuous in 0 @ x @1.
2
Method 2: The function f ðxÞ¼ x is continuous in the closed interval 0 @ x @ 1. Hence, by the theorem
on Page 48 is uniformly continuous in 0 @ x @ 1 and thus in 0 < x < 1.
3.30. Prove that f ðxÞ¼ 1=x is not uniformly continuous in 0 < x < 1.
Method 1: Suppose f ðxÞ is uniformly continuous in the given interval. Then for any > 0weshould be
able to find ,say, between 0 and 1, such that j f ðxÞ f ðx 0 Þj < when jx x 0 j < for all x and x 0 in the
interval.

: < :
1 þ 1 þ 1 þ
Let x ¼ and x 0 ¼ Then jx x 0 j¼ ¼

1 þ
1
1
1
However, ¼ ¼ > (since 0 < < 1Þ:
x x 0
Thus, we have a contradiction and it follows that f ðxÞ¼ 1=x cannot be uniformly continuous in
0 < x < 1.
Method 2: Let x 0 and x 0 þ be any two points in ð0; 1Þ. Then

1 1

j f ðx 0 Þ f ðx 0 þ Þj ¼ ¼
x 0 x 0 ðx 0 þ Þ
x 0 þ
can be made larger than any positive number by choosing x 0 sufficiently close to 0. Hence, the function
cannot be uniformly continuous.
MISCELLANEOUS PROBLEMS
3.31. If y ¼ f ðxÞ is continuous at x ¼ x 0 ,and z ¼ gðyÞ is continuous at y ¼ y 0 where y 0 ¼ f ðx 0 Þ, prove
that z ¼ gf f ðxÞg is continuous at x ¼ x 0 .
Let hðxÞ¼ gf f ðxÞg. Since by hypothesis f ðxÞ and gð yÞ are continuous at x 0 and y 0 , respectively, we
have

lim f ðxÞ¼ f ð lim xÞ¼ f ðx 0 Þ
x!x 0 x!x 0
lim gðyÞ¼ gð lim yÞ¼ gðy 0 Þ¼ gf f ðx 0 Þg
y!y 0 y!y 0
Then

lim hðxÞ¼ lim gf f ðxÞg ¼ gf lim f ðxÞg ¼ gf f ðx 0 Þg ¼ hðx 0 Þ
x!x 0 x!x 0 x!x 0
which proves that hðxÞ¼ gf f ðxÞg is continuous at x ¼ x 0 .

3.32. Prove Theorem 8, Page 48.