Solução_Calculo_Stewart_6e

F.
e E
75. lim P (E) = lim + e −E
E→0 + E→0 + e E − e − 1
−E E
TX.10SECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE ¤ 171
E e E + e −E − 1 e E − e −E Ee E + Ee −E − e E + e −E
= lim
= lim
form is
0
E→0 + (e E − e −E ) E
E→0 + Ee E − Ee −E 0
H
= lim
E→0 + Ee E + e E · 1+E −e −E + e −E · 1 − e E + −e −E
Ee E + e E · 1 − [E(−e −E )+e −E · 1]
= lim
E→0 +
Ee E − Ee −E
= lim
Ee E + e E + Ee −E − e−E E→0 +
= 0
e E − e −E
, whereL = lim
2+L E→0 + E
Thus, lim P (E) = 0
E→0 + 2+2 =0.
e E − e −E
e E + eE E + e−E − e−E
E
form is
0
0
H = lim
E→0 + e E + e −E
1
77. We see that both numerator and denominator approach 0, so we can use l’Hospital’s Rule:
√
2a3 x − x
lim
4 − a 3√ aax
x→a a − 4√ ax 3
[divide by E]
= 1+1
1
=lim
H 1
2 (2a3 x − x 4 ) −1/2 (2a 3 − 4x 3 ) − a
1
3 (aax) −2/3 a 2
x→a − 1 4 (ax3 ) −3/4 (3ax 2 )
=
1
2 (2a3 a − a 4 ) −1/2 (2a 3 − 4a 3 ) − 1 3 a3 (a 2 a) −2/3
− 1 4 (aa3 ) −3/4 (3aa 2 )
=2
= (a4 ) −1/2 (−a 3 ) − 1 3 a3 (a 3 ) −2/3
− 3 4 a3(a4 )−3/4
= −a − 1 3 a
− 3 4
= 4 3
4
3 a = 16 9 a
79. Since f(2) = 0, the given limit has the form 0 0 .
f(2 + 3x)+f(2 + 5x) H f 0 (2 + 3x) · 3+f 0 (2 + 5x) · 5
lim
=lim
= f 0 (2) · 3+f 0 (2) · 5=8f 0 (2) = 8 · 7=56
x→0 x
x→0 1
81. Since lim
h→0
[f(x + h) − f(x − h)] = f(x) − f(x) =0 (f is differentiable and hence continuous) and lim
h→0
2h =0,weuse
l’Hospital’s Rule:
f(x + h) − f(x − h) H f 0 (x + h)(1) − f 0 (x − h)(−1)
lim
=lim
= f 0 (x)+f 0 (x)
= 2f 0 (x)
= f 0 (x)
h→0 2h
h→0 2
2
2
f(x + h) − f(x − h)
2h
is the slope of the secant line between
(x − h, f(x − h)) and (x + h, f(x + h)). Ash → 0,thislinegetscloser
to the tangent line and its slope approaches f 0 (x).
f(x)
83. (a) We show that lim
x→0 x =0for every integer n ≥ 0. Lety = 1 n
x .Then 2
f(x)
lim
x→0 x =lim e −1/x2 y n
2n x→0 (x 2 ) n = lim
y→∞ e y
= H ny n−1
lim
y→∞ e y
= H ··· H= n!
lim
y→∞ e =0 ⇒
y
f(x)
lim
x→0 x = lim f(x)
n x→0 xn x =lim f(x)
2n x→0 xn lim
x→0 x =0.Thus,f 0 f(x) − f(0) f(x)
(0) = lim
= lim
2n
x→0 x − 0 x→0 x =0.