Page 63 - Schaum's Outline of Theory and Problems of Advanced Calculus

54 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3

Area of triangle OAC < Area of sector OAD < Area of triangle OBD B

1
1
i.e., 1 sin cos < < tan A
2 2 2
1
Dividing by sin , tan
2
1 sin
cos < <
sin cos
sin 1 O C D
or cos < < cos
cos
sin
As ! 0, cos ! 1 and it follows that lim ¼ 1.
!0 Fig. 3-13
THEOREMS ON LIMITS
3.17. If lim f ðxÞ exists, prove that it must be unique.
x!x 0
We must show that if lim f ðxÞ¼ l 1 and lim f ðxÞ¼ l 2 ,then l 1 ¼ l 2 .
x!x 0 x!x 0
By hypothesis, given any > 0wecan find > 0suchthat
j f ðxÞ l 1 j < =2 when 0 < jx x 0 j <
j f ðxÞ l 2 j < =2 when 0 < jx x 0 j <
Then by the absolute value property 2 on Page 3,
jl 1 l 2 j¼jl 1 f ðxÞþ f ðxÞ l 2 j @ jl 1 f ðxÞj þ j f ðxÞ l 2 j < =2 þ =2 ¼
i.e., jl 1 l 2 j is less than any positive number (however small) and so must be zero. Thus l 1 ¼ l 2 .

3.18. If lim gðxÞ¼ B 6¼ 0, prove that there exists > 0 such that
x!x 0
1
2
jgðxÞj > jBj for 0 < jx x 0 j <
1
Since lim gðxÞ¼ B,we can find > 0suchthat jgðxÞ Bj < jBj for 0 < jx x 0 j < .
2
x!x 0
Writing B ¼ B gðxÞþ gðxÞ,we have
1
2
jBj @ jB gðxÞj þ jgðxÞj < jBjþjgðxÞj
1
1
i.e., jBj < jBjþjgðxÞj, from which jgðxÞj > jBj.
2 2
3.19. Given lim f ðxÞ¼ A and lim gðxÞ¼ B, prove (a) lim ½ f ðxÞþ gðxފ ¼ A þ B, (b) lim
x!x 0 x!x 0 x!x 0 x!x 0
1 1 A
f ðxÞgðxÞ¼ AB, (c) lim ¼ if B 6¼ 0, (d) lim f ðxÞ ¼ if B 6¼ 0.
B B
x!x 0 gðxÞ x!x 0 gðxÞ
(a)We must show that for any > 0wecan find > 0suchthat
j½ f ðxÞþ gðxފ ðA þ BÞj < when 0 < jx x 0 j <
Using absolute value property 2, Page 3, we have
j½ f ðxÞþ gðxފ ðA þ BÞj ¼ j½ f ðxÞ AŠþ½gðxÞ BŠj @ j f ðxÞ AjþjgðxÞ Bj ð1Þ
By hypothesis, given > 0wecan find 1 > 0 and 2 > 0suchthat
j f ðxÞ Aj < =2 when 0 < jx x 0 j < 1 ð2Þ
jgðxÞ Bj < =2 when 0 < jx x 0 j < 2 ð3Þ
Then from (1), (2), and (3),
j½ f ðxÞþ gðxފ ðA þ BÞj < =2 þ =2 ¼ when 0 < jx x 0 j <
where is chosen as the smaller of 1 and 2 .