Page 65 - Schaum's Outline of Theory and Problems of Advanced Calculus

56 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3

lim ðx þ 3Þ lim ð2x 1Þ 3
x! 1 x! 1 2 ð 3Þ
lim
ðx þ 3Þð2x 1Þ
ðbÞ 2 ¼ 2 ¼ ¼
x! 1 x þ 3x 2 4 2
lim ðx þ 3x 2Þ
x! 1
3 1
2
4
2x 3x þ 1 2 x 2 þ x 4
lim ¼ lim
ðcÞ 4 3
x!1 6x þ x 3x x!1 1 3
x x
6 þ 3
3 1
lim 2 þ lim þ lim
x!1 x 2 x!1 x 4 2 1
x!1
1 3 6 3
¼ ¼ ¼
lim 6 þ lim þ lim
x!1 x x!1 x 3
x!1
by Problem 3.19.
p ffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffi
4 þ h 2 4 þ h 2 4 þ h þ 2
lim ¼ lim
ðdÞ p ffiffiffiffiffiffiffiffiffiffiffi
h!0 h h!0 h 4 þ h þ 2
4 þ h 4 1 1 1
¼ lim p ffiffiffiffiffiffiffiffiffiffiffi ¼ lim p ffiffiffiffiffiffiffiffiffiffiffi ¼ ¼
h!0 4 þ h þ 2 2 þ 2 4
h!0 hð 4 þ h þ 2Þ
sin x sin x p ffiffiffi sin x p ffiffiffi
ffiffiffi ¼ lim x ¼ lim lim x ¼ 1 0 ¼ 0:
x x!0þ x x!0þ x
ðeÞ lim p
x!0þ x!0þ
Note that in (c), (d), and (e)ifweuse the theorems on limits indiscriminately we obtain the so
called indeterminate forms 1=1 and 0/0. To avoid such predicaments, note that in each case the form
of the limit is suitably modified. For other methods of evaluating limits, see Chapter 4.
CONTINUITY
(Assume that values at which continuity is to be demonstrated, are interior domain values unless
otherwise stated.)
2
3.21. Prove that f ðxÞ¼ x is continuous at x ¼ 2.
Method 1: By Problem 3.10, lim f ðxÞ¼ f ð2Þ¼ 4 and so f ðxÞ is continuous at x ¼ 2.
x!2
Method 2: We must show that given any > 0, we can find > 0(depending on )suchthat
2
j f ðxÞ f ð2Þj ¼ jx 4j < when jx 2j < . The proof patterns that are given in Problem 3.10.
x sin 1=x; x 6¼ 0

5; x ¼ 0
3.22. (a)Prove that f ðxÞ¼ is not continuous at x ¼ 0. (b) Can one redefine f ð0Þ
so that f ðxÞ is continuous at x ¼ 0?
(a)From Problem 3.13, lim f ðxÞ¼ 0. But this limit is not equal to f ð0Þ¼ 5, so that f ðxÞ is discontinuous
x!0
at x ¼ 0.
(b)By redefining f ðxÞ so that f ð0Þ¼ 0, the function becomes continuous. Because the function can be
made continuous at a point simply by redefining the function at the point, we call the point a removable
discontinuity.
3
2
4
2x 6x þ x þ 3
continuous at x ¼ 1?
3.23. Is the function f ðxÞ¼
x 1
f ð1Þ does not exist, so that f ðxÞ is not continuous at x ¼ 1. By redefining f ðxÞ so that f ð1Þ¼ lim
x!1
f ðxÞ¼ 8 (see Problem 3.11), it becomes continuous at x ¼ 1, i.e., x ¼ 1isa removable discontinuity.
3.24. Prove that if f ðxÞ and gðxÞ are continuous at x ¼ x 0 ,so also are (a) f ðxÞþ gðxÞ,(b) f ðxÞgðxÞ,
(c) f ðxÞ if f ðx 0 Þ 6¼ 0.
gðxÞ