Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 3.11 HYPERBOLIC FUNCTIONS ¤ 127
e x − e −x
(c) lim sinh x = lim = ∞
x→∞ x→∞ 2
(d) lim sinh x =
x→−∞
lim
x→−∞
e x − e −x
2
= −∞
2
(e) lim sech x = lim =0
x→∞ x→∞ e x + e−x e x + e −x
(f ) lim coth x = lim
x→∞ x→∞ e x − e
(g)
(h)
−x ·
e−x
= lim
e−x x→∞
1+e −2x 1+0
= =1 [Or: Use part (a)]
1 − e−2x 1 − 0
cosh x
lim coth x = lim = ∞,sincesinh x → 0 through positive values and cosh x → 1.
x→0 + x→0 + sinh x
cosh x
lim coth x = lim = −∞,sincesinh x → 0 through negative values and cosh x → 1.
x→0− x→0 − sinh x
(i) lim csch x =
x→−∞
lim
x→−∞
2
=0
e x − e−x 25. Let y =sinh −1 x.Thensinh y = x and, by Example 1(a), cosh 2 y − sinh 2 y =1 ⇒ [with cosh y>0]
cosh y = 1+sinh 2 y = √ 1+x 2 .SobyExercise9,e y =sinhy +coshy = x + √ 1+x 2 ⇒ y =ln x + √ 1+x 2 .
27. (a) Let y =tanh −1 x.Thenx =tanhy = sinh y
cosh y = (ey − e −y )/2
(e y + e −y )/2 · ey
e = e2y − 1
y e 2y +1
1+x = e 2y − xe 2y ⇒ 1+x = e 2y (1 − x) ⇒ e 2y = 1+x
1+x
1 − x ⇒ 2y =ln 1 − x
(b) Let y =tanh −1 x.Thenx =tanhy, so from Exercise 18 we have
e 2y = 1+tanhy
1 − tanh y = 1+x
1+x
1 − x ⇒ 2y =ln 1 − x
⇒ y = 1 2 ln 1+x
1 − x
29. (a) Let y =cosh −1 x.Thencosh y = x and y ≥ 0 ⇒ sinh y dy
dx =1
dy
dx = 1
sinh y = 1
cosh 2 y − 1 = 1
√
x2 − 1
⇒
.
[since sinh y ≥ 0 for y ≥ 0]. Or: Use Formula 4.
⇒ xe 2y + x = e 2y − 1 ⇒
1+x
⇒ y = 1 ln .
2
1 − x
(b) Let y =tanh −1 x.Thentanh y = x ⇒ sech 2 y dy
dy
=1 ⇒
dx dx = 1
sech 2 y = 1
1 − tanh 2 y = 1
1 − x . 2
Or: Use Formula 5.
(c) Let y =csch −1 x.Thencsch y = x ⇒ −csch y coth y dy
dy
=1 ⇒
dx dx = − 1
. By Exercise 13,
csch y coth y
coth y = ± csch 2 y +1=± √ x 2 +1.Ifx>0,thencoth y>0,socoth y = √ x 2 +1.Ifx0.]
1 − x2 ⇒