Page 61 - Schaum's Outline of Theory and Problems of Advanced Calculus

52 FUNCTIONS, LIMITS, AND CONTINUITY [CHAP. 3

x
1
x
3.9. If f ðxÞ¼ cosh x ¼ ðe þ e Þ, prove that we can choose as the principal value of the inverse
2
p
ffiffiffiffiffiffiffiffiffiffiffiffiffi
2
function, cosh 1 x 1Þ, x A 1.
x ¼ lnðx þ
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
4y 4
1
x
x
x
x
If y ¼ ðe þ e Þ, e 2x 2ye þ 1 ¼ 0. Then using the quadratic formula, e ¼ 2y ¼
2
p ffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffiffiffi 2
2
2
y 1. y 1Þ.
y Thus x ¼ lnðy
!
p ffiffiffiffiffiffiffiffiffiffiffiffiffi
2
y 1 1
p ffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffiffiffi
2 2 y þ ffiffiffiffiffiffiffiffiffiffiffiffiffi,wecan also write
2
Since y y 1 ¼ðy y 1Þ p ffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ p 2
y 1 y 1
y þ y þ
q ffiffiffiffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffiffiffiffiffiffiffiffiffi
2 or cosh 1 2
x ¼ lnðy þ y 1 Þ y ¼ lnðy þ y 1 Þ
Choosing the þ sign as defining the principal value and replacing y by x,we have
2
x 1 Þ.
cosh 1 x ¼ lnðx þ p ffiffiffiffiffiffiffiffiffiffiffiffiffi The choice x A 1is made so that the inverse function is real.
LIMITS
2
2 x ; x 6¼ 2
3.10. If (a) f ðxÞ¼ x ,(b) f ðxÞ¼ , prove that lim f ðxÞ¼ 4.
0; x ¼ 2 x!2
2
(a)We must show that given any > 0wecan find > 0(depending on in general) such that jx 4j <
when 0 < jx 2j < .
2
Choose @ 1so that 0 < jx 2j < 1or 1 < x < 3, x 6¼ 2. Then jx 4j¼ jðx 2Þðx þ 2Þj ¼
jx 2jjx þ 2j < jx þ 2j < 5 .
2
Take as 1 or =5, whichever is smaller. Then we have jx 4j < whenever 0 < jx 2j < and
the required result is proved.
2
It is of interest to consider some numerical values. If for example we wish to make jx 4j <:05,
we can choose ¼ =5 ¼ :05=5 ¼ :01. To see that this is actually the case, note that if 0 < jx 2j <:01
2
2
then 1:99 < x < 2:01 ðx 6¼ 2Þ and so 3:9601 < x < 4:0401, :0399 < x 4 <:0401 and certainly
2
2
jx 4j <:05 ðx 6¼ 4Þ. The fact that these inequalities also happen to hold at x ¼ 2ismerely coin-
cidental.
2
If we wish to make jx 4j < 6, we can choose ¼ 1 and this will be satisfied.
(b) There is no difference between the proof for this case and the proof in (a), since in both cases we exclude
x ¼ 2.
2
4
3
2x 6x þ x þ 3
3.11. Prove that lim ¼ 8.
x!1 x 1

2x 6x þ x þ 3
4 3 2

ð 8Þ < when
x 1
We must show that for any > 0wecan find > 0suchthat

2
4
3
2x 6x þ x þ 3 3 2 3 2
0 < jx 1j < .Since x 6¼ 1, we can write ¼ ð2x 4x 3x 3Þðx 1Þ ¼ 2x 4x
x 1 x 1
3x 3oncancelling the common factor x 1 6¼ 0.
3
2
Then we must show that for any > 0, we can find > 0suchthat j2x 4x 3x þ 5j < when
0 < jx 1j < . Choosing @ 1, we have 0 < x < 2, x 6¼ 1.
3 2 2 2 2
Now j2x 4x 3x þ 5j¼jx 1jj2x 2x 5j < j2x 2x 5j < ðj2x jþj2xjþ 5Þ < ð8 þ 4 þ 5Þ
¼ 17 . Taking as the smaller of 1 and =17, the required result follows.
8
; x 6¼ 3
< jx 3j
x 3 , (a) Graph the function. (b) Find lim f ðxÞ. (c)Find
3.12. Let f ðxÞ¼
0; x ¼ 3
: x!3þ
lim f ðxÞ.(d)Find lim f ðxÞ.
x!3
x!3
x 3
(a) For x > 3, jx 3j ¼ ¼ 1.
x 3 x 3
For x < 3, jx 3j ¼ ðx 3Þ ¼ 1.
x 3 x 3