Page 66 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 57

These results follow at once from the proofs given in Problem 3.19 by taking A ¼ f ðx 0 Þ and B ¼ gðx 0 Þ
and rewriting 0 < jx x 0 j < as jx x 0 j < , i.e., including x ¼ x 0 .

3.25. Prove that f ðxÞ¼ x is continuous at any point x ¼ x 0 .
We must show that, given any > 0, we can find > 0suchthat j f ðxÞ f ðx 0 Þj ¼ jx x 0 j < when
jx x 0 j < . By choosing ¼ ,the result follows at once.

3
3.26. Prove that f ðxÞ¼ 2x þ x is continuous at any point x ¼ x 0 .
3
3
2
2
Since x is continuous at any point x ¼ x 0 (Problem 3.25) so also is x x ¼ x , x x ¼ x ,2x , and
3
finally 2x þ x,using the theorem (Problem 3.24) that sums and products of continuous functions are
continuous.
ffiffiffiffiffiffiffiffiffiffiffi
p
x 5 for 5 @ x @ 9, then f ðxÞ is continuous in this interval.
3.27. Prove that if f ðxÞ¼
p ffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffiffiffi
If x 0 is any point such that 5 < x 0 < 9, then lim f ðxÞ¼ lim x 5 ¼ x 0 5 ¼ f ðx 0 Þ. Also,
x!x 0 x!x 0
p ffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffi
lim x 5 ¼ 0 ¼ f ð5Þ and lim x 5 ¼ 2 ¼ f ð9Þ. Thus the result follows.
x!5þ x!9 p ffiffiffiffiffiffiffiffiffi q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffi
Here we have used the result that lim f ðxÞ ¼ lim f ðxÞ ¼ f ðx 0 Þ if f ðxÞ is continuous at x 0 .An ,
x!x 0 x!x 0
proof, directly from the definition, can also be employed.
3.28. For what values of x in the domain of definition is each of the following functions continuous?
x
Ans. all x except x ¼ 1(where the denominator is zero)
x 1
(a) f ðxÞ¼ 2
1 þ cos x
Ans. all x
3 þ sin x
(b) f ðxÞ¼
1
(c) ffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ans. All x > 10
4 10 þ 4
f ðxÞ¼ p
2
(d) f ðxÞ¼ 10 1=ðx 3Þ Ans. all x 6¼ 3 (see Problem 3.55)
2
(e) f ðxÞ¼ 10 1=ðx 3Þ ; x 6¼ 3 Ans. all x, since lim f ðxÞ¼ f ð3Þ
0; x ¼ 3 x!3
x jxj
( f ) f ðxÞ¼
x
x x x þ x
¼ 2. At x ¼ 0, f ðxÞ is undefined. Then f ðxÞ is
If x > 0, f ðxÞ¼ ¼ 0. If x < 0, f ðxÞ¼
x x
continuous for all x except x ¼ 0.
8
; x < 0
x jxj
<
x
ðgÞ f ðxÞ¼
2; x ¼ 0
:
As in ð f Þ, f ðxÞ is continuous for x < 0. Then since
x þ x
lim x jxj ¼ lim ¼ lim 2 ¼ 2 ¼ f ð0Þ
x x
x!0 x!0 x!0
if follows that f ðxÞ is continuous (from the left) at x ¼ 0.
Thus, f ðxÞ is continuous for all x @ 0, i.e., everywhere in its domain of definition.
x
: Ans: all x except 0; ; 2 ; 3 ; ... :
ðhÞ f ðxÞ¼ x csc x ¼
sin x
x
(i) f ðxÞ¼ x csc x, f ð0Þ¼ 1. Since lim x csc x ¼ lim ¼ 1 ¼ f ð0Þ,we see that f ðxÞ is continuous for all x
x!0 x!0 sin x
except ; 2 ; 3 ; .. . [compare (h)].