Page 91 - Schaum's Outline of Theory and Problems of Advanced Calculus

82 DERIVATIVES [CHAP. 4

tan 1 b tan 1 a 1
a < < b
b a ¼ 1 þ 2
2
2
2
2
Since > a,1=ð1 þ Þ < 1=ð1 þ a Þ. Since < b,1=ð1 þ Þ > 1=ð1 þ b Þ. Then
1 < tan 1 b tan 1 a < 1
1 þ b 2 b a 1 þ a 2
and the required result follows on multiplying by b a.
(b)Let b ¼ 4=3 and a ¼ 1inthe result of part (a). Then since tan 1 1 ¼ =4, we have
3 1 4 1 1 3 1 4 1
< tan tan 1 < or < tan <
25 3 6 4 þ 25 3 4 þ 6

4.25. Prove Cauchy’s generalized mean value theorem.
Consider GðxÞ¼ f ðxÞ f ðaÞ fgðxÞ gðaÞg, where is a constant. Then GðxÞ satisfies the conditions
of Rolle’s theorem, provided f ðxÞ and gðxÞ satisfy the continuity and differentiability conditions of Rolle’s
theorem and if GðaÞ¼ GðbÞ¼ 0. Both latter conditions are satisfied if the constant ¼ f ðbÞ f ðaÞ .
Applying Rolle’s theorem, G ð Þ¼ 0for a < < b,we have gðbÞ gðaÞ
0
0
f ð Þ g ð Þ¼ 0 or f ð Þ ¼ f ðbÞ f ðaÞ ; a < < b
0
0
0
g ð Þ gðbÞ gðaÞ
as required.

L’HOSPITAL’S RULE

4.26. Prove L’Hospital’s rule for the case of the ‘‘indeterminate forms’’ (a) 0/0, (b) 1=1.
(a)Weshall suppose that f ðxÞ and gðxÞ are differentiable in a < x < b and f ðx 0 Þ¼ 0, gðx 0 Þ¼ 0, where
a < x 0 < b.
By Cauchy’s generalized mean value theorem (Problem 25),

0
x 0 < < x
f ðxÞ f ðxÞ f ðx 0 Þ f ð Þ
¼ ¼
0
gðxÞ gðxÞ gðx 0 Þ g ð Þ
Then
0 0
lim f ðxÞ ¼ lim f ð Þ ¼ lim f ðxÞ ¼ L
0 0
x!x 0 þ gðxÞ x!x 0 þ g ð Þ x!x 0 þ g ðxÞ
since as x ! x 0 þ, ! x 0 þ.
Modification of the above procedure can be used to establish the result if x ! x 0 , x ! x 0 ,
x !1, x ! 1.
(b)Wesuppose that f ðxÞ and gðxÞ are differentiable in a < x < b, and lim f ðxÞ¼ 1, lim gðxÞ¼1
where a < x 0 < b. x!x 0 þ x!x 0 þ
Assume x 1 is such that a < x 0 < x < x 1 < b. By Cauchy’s generalized mean value theorem,
0
f ðxÞ f ðx 1 Þ f ð Þ
x < < x 1
¼
0
gðxÞ gðx 1 Þ g ð Þ
Hence
0
f ðxÞ f ðx 1 Þ f ðxÞ 1 f ðx 1 Þ=f ðxÞ f ð Þ
¼ ¼
0
gðxÞ gðx 1 Þ gðxÞ 1 gðx 1 Þ=gðxÞ g ð Þ