Este documento apresenta a resolução de vários exercícios de cálculo de limites matemáticos do 12o ano em 3 frases: 1) São resolvidos exercícios de cálculo de limites unilaterais e bilaterais de funções racionais, irracionais e trigonométricas. 2) Incluem-se também exercícios sobre limites de funções compostas, limites de funções contínuas e limites de sequências. 3) São usados métodos como o critério de Heine, substituição de

1. EXERCÍCIOS DE MATEMÁTICA A 12.° ANO
PROPOSTAS DE RESOLUÇÃO
Capítulo 6
Pág. 136
1.
Atendendo à definição de limite de uma função (de Heine) e dado
que lim un = lim (1 + œn ) = +?, lim g(un) = lim g(x) = 2.
nS +?
1
-5
2
6
-6
1 n
=e
n
lim yn = lim [ 4 + ln (xn) ]
= 4 + lim ln (xn)
= 4 + lim ln x
1
-3
0
=4+1=5
1
5x(x - 2)
5x
(x - 2) = lim
lim 2
xS2 x - 5x + 6
xS2 (x - 3)(x - 2)
3
4
2
Resposta: (B)
5
5x(œx + x)
lim x *
= lim
xS0
xS0
x-x
x - x)(œx + x)
(œ
œ
1
2
= lim
xS0
=
5x(œx + x)
5x(œx + x)
= lim
xS0
x - x2
x(1 - x)
Como lim
5*0
=0
1
1.1 lim
xS4
(x - 1)2
x2 - 2x + 1
x-1
= lim
= lim
=0
2
xS1 (x - 1)(x + 1)
xS1 x + 1
x -1
1.3 lim
aS2
(a - 2)(a + 2)
a2 - 4
= lim
=4
a - 2 aS2
(a - 2)
1.4 lim
bS0
b(b2 - b)
b3 - b 2
0
= lim
=
=0
2
bS0 b(3b - 1)
3b - b
-1
1.5 tS -1
lim
4.1 D = {x å R: x 0 0 ‹ 1 - x > 0}
x-4
x-4
1
= lim
=
x2 - 16 xS 4 (x - 4)(x + 4) 8
1.2 lim
xS1
2
(t 2 - 1)(t 2 + 1)
(t - 1)(t + 1)(t 2 + 1)
t4 - 1
= lim
= lim
tS -1
t + 1 tS -1
t+1
t+1
= (-1 - 1)(1 + 1) = -4
1 - x2 = 0 § x = -1 › x = 1
D = ] -1, 0[ ∂ ]0, 1[
Como A ƒ D, A só pode ser ]-1, 1[ {0}.
Resposta: (A)
-
-
1
ln(1 - x2) = 1 * ln 0 + = -?
x
Resposta: (C)
5.
lim
xS0 +
ln x
-?
= + = -?
ex - 1
0
6.
(x + 2)(x3 - 2x2 + 4x - 8)
x4 - 16
= lim
xS -2
x3 + 8
(x + 2)(x2 - 2x + 4)
-8 - 8 - 8 - 8
32
8
=
==4+4+4
12
3
Cálculo auxiliar:
1
0
0
0 -16
0
1
-2
-2
4 -8
16
-2
-2
0
1 -2 4 -8
1 -2
2x x2
2 -x
3x 3x
lim
lim
x = xS + ?
xS + ? x * 3
x
2 x x2
- x
3
3
0-0
= lim
=
=0
xS + ?
x
+?
Resposta: (D)
Pág. 137
2
12
7.
e3x x
- x
e3x - x
ex
e
lim
= lim
x
xS + ?
xS + ?
2e
2
x
e2x - x
e
+? - 0
= lim
=
= +?
xS + ?
2
2
2.
5
f(x) - 2
h(x) = f(x)
f(x) + 2
ekx - 1
se x < 0
3x
f(x) =
3x + 2 se x ≥ 0
5
+
CEXMA12 © Porto Editora
xS0
-
ekx - 1 k
ekx - 1
= lim
xS0
3x
3 xS0 kx
k
k
= *1=
3
3
-
-
lim f(x) = lim (3x + 2) = 2
xS0 +
xS0
-
xS0
+
k
=2 § k=6
3
; lim f(x) = b
xSa
+
-
-
5
1
se x > -1
x2 + 1
-
1
2x
se x ≤ -1
1
lim h(x) = lim -
xS -1-
xS -1-
lim h(x) = lim
xS -1 +
xS -1 +
2
1
1
=
2x
2
1x
2
lim h(x) = lim h(x) =
xS -1-
xS -1+
2
1
1
=
+1
2
xS0 +
lim f(x) = lim f(x) §
8
-8
0
lim
lim
2.2 xSa f(x) = xSa (f(x) - 2) = b - 2
3.1 h(x) =
lim f(x) = lim
se x < a
se x = a
se x > a
0
4
4
2.1 xSa h(x) = xSa (f(x) + 2) = b + 2
lim
lim
Resposta: (A)
8.
Pág. 138
1.6 xS -2
lim
Resposta: (C)
x
1
1
= 0, se xn = 2 , lim h(xn) = 1.
n2
n
Resposta: (B)
1
ln(1 - x2)
x
4.2 xS1 f(x) = xS1
lim
lim
ex - 1
=1
x
11. lim h(x) = lim
xS0
xS0
Resposta: (B)
f(x) =
2
x"e
Resposta: (A)
4.
xS0
10. lim xn = lim 1 +
10
= -10
=
-1
3.
2x
x
= 2 lim
=2*1=2
xS0 ln(x + 1)
ln(x + 1)
lim
Resposta: (C)
xS +?
Resposta: (B)
2.
9.
1
1
§ lim h(x) =
xS -1
2
2
Resposta: (D)
43

2. EXERCÍCIOS DE MATEMÁTICA A 12.° ANO
PROPOSTAS DE RESOLUÇÃO
3.2 i(x) = (3x + 2) *
x
;
x|
CAPÍTULO 6
se x ≥ 0
se x < 0
5- x
x
lim
lim
8.1 xS - ?(-x + x3) = xS - ? x3 = -?
lim i(x) = lim (3x + 2) *
x
= (0 + 2) * (-1) = -2
-x
lim i(x) = lim (3x + 2) *
x
= (0 + 2) * 1 = 2
x
xS0 -
xS0 -
xS0+
xS0+
8.2 xS + ?(-x + x2) = xS + ? x2 = +?
lim
lim
8.3 xS + ?(œx + x4) = +? + ? = +?
lim
8.4 xS + ?(-x2 - œx ) = -? - ? = -?
lim
3
8.5 xS + ?(œ-x - x2) = -? - ? = -?
lim
Não existe lim i(x).
xS0
4.
h(x) =
x + 2a
2
- ax + 1
5x
se x < -1
se x ≥ -1
8.6 xS - ?(x2 + œ-x ) = +? + ? = +?
lim
lim h(x) = lim (x + 2a) = -1 + 2a
8.7 x " + ?[(x2 + x)(2x3 + 5x)] = 1? * (1?) = 1?
lim
lim h(x) = lim (x2 - ax + 1) = 1 + a + 1 = a + 2
8.8 xS + ?
lim
lim h(x) = lim h(x) § -1 + 2a = a + 2 § a = 3
lim
lim
lim
8.9 x " + ?œx3 - 3x + 1 = œx " + ? (x3 - 3x + 1) = œx " +? x3 = + ?
xS -1-
xS -1-
xS -1 +
xS -1+
xS -1-
5.
xS -1 +
8.10
Por exemplo:
3
3
=
=0
x3 + 5x + 1 +?
lim
x" -?
-1
œ3 - x
=
lim f(x) = lim
xS - ?
lim f(x) = lim
6.1 xS + ?[f(x) + 2] = xS + ? f(x) + 2 = 1 + 2 = 3
lim
lim
lim
6.2 xS0 [f(x) + g(x)] = -? + 0 = -?
Pag. 139
-2
=0
x
-2x2 + 5
-2x2
= lim
= -2
xS + ?
x2 + 3x
x2
9.2 xS + ? f(x) = xS + ?
lim
lim
xS - ?
xS - ?
9.3 xS + ? f(x) = xS + ?
lim
lim
-2x2
= -2
x2
x3 + 5x + 2
x3
= lim
= lim x2 = +?
xS + ? x
xS + ?
x-1
lim f(x) = lim x2 = +?
+
xS - ?
g(x)
g(1)
1
6.3 lim
=
=
= -1
xS1 h(x - 1)
h(0) -1
xS - ?
lim
lim
9.4 xS + ? f(x) = xS + ?
lim
6.4 xS0
h(x)
-1
=
=0
f(x) -?
lim
6.5 xS1
g(x)
1
=
= -?
f(x) 0-
lim
6.6 xS0
g(x)
0=
=0
f(x) +?
lim f(x) =
6.7 xS0
lim
f(x)
-?
= + = -?
h(x) + 1
0
105x + 2000
105x
105
= lim
=0
lim
2
2 = xS + ?
xS + ? 0,3x
0,3x + 0,1
0,3x
9.5 xS + ? f(x) = xS + ?
lim
lim
+
-
-
+
lim f(x) = lim
xS - ?
7.
xS - ?
105
=0
0,3x
x+1
3
1 lim
xS - ?
x
-x
3
2
(?- ?)
lim
lim
10.1 x " +? f(x) = x " + ?(œx + 1 - œx ) =
lim g(x) = lim
1x - 2 + x - 42 = 0
CEXMA12 © Porto Editora
xS2 -
1
x
1
x
+
1
-
- 1 = +?
44
x " 2+
2= lim
Não existe lim g(x) , já que lim g (x) 0 lim g (x) .
x"2
2x - œx ?
lim
lim
= lim
10.2 xS + ? f(x) = xS + ?
xS + ?
x+1
- 1 = -?
x " 2-
xS + ?
1
x
1+0
œx + 1 + œx
x+1-x
œx + 1 + œx
2-
œx
x
1
1+
x
Œ1 = 2 - 0 = 2
x
1+
(œx + 1 - œx )(œx + 1 + œx)
lim
x " +?
xS + ?
1x - 2 + x - 42 = 0
xS2 -
= (-1)3 = -1
xS - ?
= lim
lim g(x) = lim
1
3
lim f(x) = lim 2 = 2
xS - ?
1
x
+
x-2 x-4
xS2 +
2
= (-1)3 = -1
?
xS2 +
x
-x
xS + ?
lim
lim
9.6 xS + ? f(x) = xS + ? x2 = +?
f(x)
1
= lim =
=0
g(x) xS + ? +?
g(x) =
xS - ?
1 -x - 3 2 = 1 lim
g(x)
mx + b
mx
m
= lim
= lim
lim
= lim
=0
6.8 xS + ?
h(x) xS + ? ax2 + c xS +? ax2 xS +? ax
6.9 xS + ?
lim
Pag. 140
-2x + 1
-2x
-2
= lim
= lim
=0
xS + ? x
x2 + 3x xS + ? x2
9.1 xS + ? f(x) = xS + ?
lim
lim
xS - ?
-1
=0
+?
=
1
=0
+?

3. EXERCÍCIOS DE MATEMÁTICA A 12.° ANO
PROPOSTAS DE RESOLUÇÃO
CAPÍTULO 6
lim
lim
10.3 xS + ? f(x) = xS + ?(œx2 + 1 - œx )
(? - ?)
=
(œx2 + 1 - œx )(œx2 + 1 + œx )
lim
x " +?
xS + ?
lim
lim
x " +?
=
x - 2|
x-2
= lim
=1
xS2 x - 2
x-2
xS2 -
-
œx + 1 + œx
1
1
x2 x
1
1
1
+
+
x2 x4
x3
Œ
xS2 +
Œ
3
= lim
2 œx
x " +?
x " +?
1
2
6
xS1
2
12.8 x " -1
lim
3
-
1x + 1 - x
3
2
x-1
2
= lim
x " -1-
1
1
lim 6
2 x " +? œx
1
1
1
*
= *0=0
2 +? 2
13.
f(x) - f(2)
x2 + 1 - 5
x2 - 4
= lim
= lim
xS2
xS2 x - 2
x-2
x-2
lim
xS2
= lim
xS2
§ -m = 5 § m = -5
14.
3x
-3
=
= -?
(x + 1)2 0 +
f(x) = -
(x - 2)(x + 2)
=4
x-2
Pág. 141
1
x
f (œ3 + h) - f (œ3)
= lim
lim
hS0
hS0
h
(x - 3)(x - 5)
x2 - 8x + 15 (0)
12.2 lim
= lim
xS3
xS3 (x - 3)(x + 3)
x2 - 9
0
Cálculo auxiliar:
1 -8
15
3
3 -15
1 -5
0
-2
1
=6
3
= lim
h"0
-4
4
1 -2
12.4 xlim1
"
6
-8
4
-2
0
1
0
12
-4
6
-12
2 -3
6
0
(x + 1)(x2 - x + 1)
x3 + 1 ( )
12.5 xlim
= lim
" -1 x + 1
xS -1
(x + 1)
=1+1+1=3
Cálculo auxiliar:
1
0 0
-1
f(x) =
ex + x
h * œ3(œ3 + h)
œ3(œ3 + 0)
=
1
3
se x ≤ 0
5ln(x + 1) se x > 0
15.1 xS + ? f(x) = xS + ?ln(x + 1) = +?
lim
lim
lim
lim
15.2 xS0 f(x) = xS0 (ex + x) = 1
-
0
(x - 1)
0
œx - 1 (=) lim (œx - 1)(œx + 1) = lim
1
=
x"1
x-1
(œx + 1) x " 1 (x - 1)(œx + 1) 2
(x - 1)
0
0
CEXMA12 © Porto Editora
2
-2
œ3
h
h
1
=
15.
4
1
+
œ3(œ3 + h)
h"0
13
-26
= 20
10
-2
œ3 + h
h
= lim
0
-2
1
-
-œ 3 + œ3 + h
(x + 2)(x3 - 2x2 + 4x - 2)
x4 + 6x - 4 (0)
12.3 xS -2 3
lim
= lim
2
x " -2
2x + x + 12
(x + 2)(2x2 - 3x + 6)
Cálculo auxiliar:
0 0
1
3x - 7 -10
= + = -?
x2 - 1
0
f(x) = x2 + 1
(2 - mx)(x + 1)
-mx2
lim
= 5 § lim
=5
xS + ?
xS +?
x2 - 4
x2
=
1
4
-
6
=
=
(x - 1)(œx + 3 + 2)
4
3x - 3 - 4
(?-?)
= lim
x " -1
-1
x2 - 1
x " +?
=
12.1 xS -1
lim
x+3-4
(x - 1)(œx + 3 + 2)
(x - 1)
=
11.
= lim
Œx
x
1
lim Œ
x
1
2
= lim
=
xS1
3
1 œx
*
2 œx
= lim
se x ≥ 2
se x < 2
0
0
œx + 3 - 2 (=) lim (œx + 3 - 2)(œx + 3 + 2)
x"1
x-1
(x - 1)(œx + 3 + 2)
lim
lim
10.4 x " + ? f(x) = x " + ? œx
1 ?2
?
x-2
5-(x - 2)
x - 2|
.
x-2
xS2
12.7 xlim1
"
x - 2|=
+
Não existe lim
1
=+?
0+
Nota auxiliar:
x - 2|
-(x - 2)
= lim
= -1
xS2
x-2
(x - 2)
2
1+
= lim
12
Calculemos os limites laterais:
œx2 + 1 + œx
x2 + 1 - x
= lim
x - 2|
0
=
x-2
0
12.6 xlim2
"
1
-1
1
1
0
xS0 +
xS0 +
Não existe lim f(x).
xS0
31
24
= lim
ln x
+
x
1
?
ln x 1 +
x
f(x)
ln(x + 1) 1 ? 2
15.3 xS + ?
lim
= lim
= lim
xS + ?
xS + ?
x
x
x
1
ln x + ln 1 +
-1
1 -1
-
lim f(x) = lim ln(x + 1) = ln1 = 0
= lim
xS + ?
x
1
x
2
xS + ?
1
ln 1 +
1
x
2
x
=0+0=0
45

4. EXERCÍCIOS DE MATEMÁTICA A 12.° ANO
PROPOSTAS DE RESOLUÇÃO
15.4 xS - ?
lim
CAPÍTULO 6
f(x)
ex + x
ex
= lim
= lim
+1 =0+1=1
xS - ?
xS -? x
x
x
1
2
y
x-3
= lim
=1
yS0 ln(y + 1)
ln(x - 2)
16.15 lim
xS3
Mudança de variável:
x-2=y+1
f(x)
ln(x + 1)
15.5 xS0
lim
= lim
=1
xS0
x
x
+
lim
15.6 xS0
-
§ y=x-3
x"3± y"0
+
f(x) - 1
ex + x - 1
ex - 1 x
= lim
= lim
+
xS0
xS0
x
x
x
x
-
-
= lim
xS0 -
1
2
e -1
+1=1+1=2
x
x
16.1 xlim0
"
e(ex - 1)
ex+1 - e
ex - 1
= lim
= e * lim
=e*1=e
x"0
x"0
x
x
x
16.3 xlim0
"
e3 - e x+3
-e3
ex - 1
e3
-e3
=
* lim
*1= =
x"0
x
2x
2
2
2
ln(-1 - x)
ln(y + 1)
= lim
yS0 -2 - y + 2
x+2
e2x - 1
e2x - 1
= 2 * lim
=2*1=2
x"0
x
2x
16.2 xlim0
"
16.16 xlim
" -2
16.4 xlim0
"
yS0
16.6
16.7
17.1
1
= lim
xS0
17.2
xS0
f(x) - f(0)
xex - 0
= lim
= lim ex = 1
xS0
xS0
x
x
18.
f(x) = x ln x
lim
hS0
f(1 + h) - f(1)
(1 + h) ln(1 + h) - 1 ln 1
= lim
hS0
h
h
= lim
ax
a
bx
a
a
= * lim
= *1=
e - 1 b xS0 ebx - 1 b
b
hS0
bx
xS0
?
16.9
lim
x " +?
hS0
19.1
19.2
lim
20
20
=0
=
1 + 4e -0,1x 1 + ?
lim
20
20
= 20
=
1 + 4e -0,1x 1 + 0
xS - ?
xS + ?
lim
e3x - x2 - 1 (0)
e3x - 1
x2
= lim
- lim
x"0
x " 0 2x
2x
2x
x"0
=
3
e3x - 1
3
3
* lim
-0= *1=
2 x " 0 3x
2
2
2(ex-1 - 1)
2ex-1 - 2 (0)
= lim
2
x " 1 (x - 1)(x + 1)
x -1
0
19.4
lim
x"1
= lim
x"1
4
lim
x"0
2
ex-1 - 1
* lim
=1*1=1
x + 1 x"1 x - 1
2x - 1 (0)
eln 2 - 1
ex ln 2 - 1
= lim
= lim
xS0
xS0
x
x
x
0
19.5
ln(x + 1)
1 ln(x + 1)
=?
= lim
*
x"0 x
x
x2
3
x
= ln 2 * lim
xS0
Calculemos os limites laterais:
1 ln(x + 1)
*
= -? * 1 = -?
x
x
lim
3
lim
1 ln(x + 1)
*
= +? * 1 = +?
x
x
CEXMA12 © Porto Editora
xS0 +
4
Não existe lim
xS0
16.14 lim
xS0
46
19.6
ln(x + 1)
ln (x + 1)
ln (x + 1)
0 lim
, já que lim
.
x"0
x"0
x2
x2
x2
+
ln(x + 1)
ln(x + 1)
x
= lim
* lim x
=1*1=1
xS0
xS0 e - 1
x
ex - 1
-
2x = eln 2
ex ln 2 - 1
= ln 2 * 1 = ln 2
x ln 2
ln(x + a) - ln a (0)
lim
= lim
x"0
x"0
x
0
x " 0-
ln(1 + h)
h
0
19.3
ln(2x + 1) 2
ln(2x + 1) 2
2
= * lim
= *1=
3x
3 x"0
2x
3
3
16.13 xlim0
"
hS0
=1*1=1
ln(x + 1) 1
ln(x + 1) 1
1
= * lim
= *1=
x
3x
3 x"0
3
3
16.12 xlim0
"
(1 + h) * ln(h + 1) - 0
h
= lim(1 + h) * lim
e2x -1
e2x * e -1 ?
e -1
= lim
= lim
3x
x " +? 1 - e3x
x " +? 1
1-e
e3x
e2x e2x
e -1
e -1
e -1
=
= lim
=
=0
x " +? 1
0-? -?
- ex
e2x
eax - 1
eax - 1
* ax
lim
eax - 1
ax
a xS0 ax
16.10 lim bx
= lim
= *
xS0 e
- 1 xS0 ebx - 1
b
ebx - 1
* bx
lim
xS0
bx
bx
a 1 a
= * =
b 1 b
16.11 xlim0
"
x
e -1
+ lim
=1+1=2
x xS0 x
Pág. 142
x3
x2 * x
x2
x
= lim
lim x +3
= lim
* lim
xS0 e
- e3 xS0 e3(ex - 1) xS0 e3 xS0 ex - 1
lim
2
x
f(x) = xex
lim
=0*1=0
16.8
f(x) - f(0)
x + ex - 1
x ex - 1
= lim
= lim
+
xS0
xS0 x
x
x
x
x
ex-3 - 1 1
ex-3 - 1 1
1
= * lim
= *1=
2x - 6
2 x"3 x - 3
2
2
lim
§ x = -2 - y
f(x) = x + ex
xS0
2x
2
5x
2
2
=
* lim 5x
= *1=
x"0 e
e -1 5
-1 5
5
x"3
§ y = -2 - x
x " -2 § y " 0
lim
ex(ex - 1)
e -e
ex
ex - 1
= lim
16.5 lim
= lim
* lim
xS0 -5x
xS0
xS0 -5
xS0
x
-5x
1
1
=- *1=5
5
-1 - x = y + 1
ln(y + 1)
= -1
y
= - lim
5x
2x
Mudança de variável:
1
ln
1
= * lim
a x"0
x+a
x
ln + 1
a
a
= lim
xS0
x
x
2
1 a + 12
ln
1
2
x
x
a
=
1
1
*1=
a
a
x

5. EXERCÍCIOS DE MATEMÁTICA A 12.° ANO
PROPOSTAS DE RESOLUÇÃO
CAPÍTULO 6
ln(x + 3) (0 )
ln(x + 3)
lim
= lim
xS -2 2x2 + 3x - 2
xS -2 (x + 2)(2x - 1)
Cálculo auxiliar:
0
19.7
= lim
xS -2
19.8
19.9
-2
-4
0
-2
y+1=x+3
§ y=x+2
§ x= y-2
1
1
*1=5
5
ln(3x + 1)
Mudança de variável:
y=x-1
œx + 9 - 3
xS + ?
ln(3x + 1)
= lim(œx + 9 + 3) * lim
xS0
xS0 x + 9 - 9
24 = +? * (+? - 1) = +?
-1
= 2 + 2 * lim
2
ln x
=2+2*0=2
x
31
lim
lim
=
19.20 xS + ?(3x - 2 ln x) (?- ?) xS + ? x 3 -
2 ln x
x
24
1porque
= +? * (3 - 0)
= +?
(0*?)
lim
19.22 xS0 x ln x =
lim e -y(-y)
Mudança de variável:
y = -ln x
y
1
= - lim y
yS + ? e
ey
y
= - lim
1
==0
+?
ln(x + 1)
log2(x + 1)
ln(x + 1)
ln 2
1
19.11 xlim0
= lim
=
* lim
"
x"0
x
x
x
ln 2 x " 0
=
1
1
*1=
ln 2
ln 2
(
]
(?- ?)
=
§ -y = ln x
§ x = e-y
x " 0+ ± y " +?
3 1
lim n * 1 -
n " +?
ln(n)
n
24 = +? * (1 - 0) = +?
1
(?* ?)
lim (2n + 1) *
nS + ?
1 en - 1
*
n
1
n
2n + 1
en - 1
* lim
nS + ?
nS + ?
n
1
n
2n
1
= lim
*1
se n S +?, S 0
n
nS + ? n
1
ln(x + 1)
1
= lim
* lim
1
x"0
x"0
x
1 - ex
-
1
1
=
=1
1 - e-? 1 - 0
=1*
[
lim n - ln(n)
n " +?
1
)
-
20.1
lim
20.2 nS + ?(2n + 1)(en - 1) =
ln(x + 1) (0)
ln(x + 1)
0
= xlim
1
1
"0
x - ex x
x 1 - ex
-
yS + ?
2
ln x
=0
x
ln x
1+0
x
=
=1
1+0
1
1+
x
yS + ?
+
lim
xS + ?
1+
x + ln x
19.21 xS + ?
lim
= lim
xS + ?
x+1
= 6 * 3 * 1 = 18
-
5
1
ln(3x + 1)
ln(3x + 1)
= 6 * lim
= 6 * 3 * lim
x"0
x"0
x
3x
19.12 xlim
"0
2x
x " -? ± y " +?
2x 2 ln x
2x + ln x2
= lim
+
xS + ? x
x
x
§ x=y+1
(œx + 9 - 3)(œx + 9 + 3)
xS0
lim
19.19 xS + ?
3 1x
lim x5
x"1±y"0
ln(3x + 1) * (œx + 9 + 3)
= lim
y = -x § x = -y
x " +?
=2*1=2
19.10 lim
xS0
(?-?)
lim
19.18 xS + ?(2x - x5) =
x " -2 ± y " 0
x-1
x-1
x-1
= lim
= lim
=
1
lnœx xS1 ln x 2 xS1 1
ln x
2
x-1
= 2 * lim
x " 1 ln x
y
= 2 * lim
y " 0 ln(y + 1)
xS1
Mudança de variável:
lim (-y)2e -y
yS + ?
y2
1
= lim y = lim y
yS + ? e
yS + ? e
y2
1
=0
=
+?
Mudança de variável:
ln(x + 1) + 2x
ln(x + 1)
2x
lim
= lim
+ lim
=1+2=3
xS0
xS0
xS0 x
x
x
lim
lim
19.17 xS - ? x2ex =
2
2 -1
ln(x + 3)
1
* lim
2x - 1 xS -2 x + 2
ln(y + 1)
1
* lim
=
y
-5 yS0
=-
(?*0)
3
2
= lim
1
2
x
e
+x
ex + x3
x2
+?
19.13 xS + ? 2
lim
= lim
=
= +?
x + 1 xS +?
1+0
1
1+ 2
x
=2*1=2
1
1
ln(n)
lnœn (? )
ln n 2
2
?
lim
= lim
= lim
20.3 nS + ?
nS + ?
nS + ?
n
n
n
ln(n) 1
1
= * lim
= *0=0
2 n " +? n
2
( )
e -x
x3
1
1
19.14 xS + ? -3 = xS + ? x = xS +? x =
lim
lim
lim
=0
x
e
+?
e
3
x
2 (0*?)
19.15 xS0 x3 * e x =
lim
lim
1y2
yS +?
+
2
3
* ey
8
* ey
y3
ey
= 8 * lim 3
yS +? y
= lim
yS +?
Mudança de variável:
2
2
y=
§ x=
x
y
x " 0+ ± y " +?
CEXMA12 © Porto Editora
= 8 * (+?) = +?
2n
*2-1
?
2n +1 - 3n (? )
2n * 2 - 3n
3n
lim
lim n
lim
20.4 nS + ? n +1
n = nS + ?
n = nS + ?
3 +2
3 *3+2
2n
3+ n
3
2 n
*2-1
3
0*2-1
1
= lim
=
=nS + ?
3+0
3
2 n
3+
3
12
12
x5
1
1
= lim x = lim x =
=0
xS +? e
xS +? e
+?
x5
5 -x (? *0)
lim
19.16 xS + ? x e
47

6. EXERCÍCIOS DE MATEMÁTICA A 12.° ANO
PROPOSTAS DE RESOLUÇÃO
CAPÍTULO 6
1.
T(t) = Ta + (T0 - Ta)e -kt
1.1
2.4 A(t) = 200 §
T0 = 90 8C; T = 60 8C; Ta = 20 8C; t = 10 min
Pág. 143
§ 300 = 200 + 3800e -0,1t ‹ twwuwwv
1 + 19e- 0,1t 0 0
condição universal
100
§ 3800e -0,1t = 100 § e -0,1t =
3800
60 = 20 + (90 - 20)e -k *10 § 40 = 70e -10k
40
4
§ e -10k =
§ -10k = ln
70
7
4
ln
7
§ k=
§ k ) 0,056
-10
12
12
1.2
T(t) = 20 + (90 - 20)e
-0,056 t
§ 35 = 20 + 70e
1 2
1 2
27,5 - 10 = 17,5
1990 + 36,4 = 2026,4
Se o modelo se mantiver válido o número de águias ultrapassará
200 no ano de 2026.
2.5 Dado que C(t) é em milhares, o número de coelhos é dado por
1000 C(t).
1000 C(t)
< 20.
A(t)
1 21 2
10
§ 50 C(t) < A(t)
1 2
§ 50 *
25
300
<
1 + 1,5e0,1t 1 + 19e -0,1t
§ 1250(1 + 19e -0,1t) < 300(1 + 1,5e0,1t)
§ 25(1 + 19e -0,1t) < 6(1 + 1,5e0,1t)
T(t) = 22 + 60e -0,056t, t ≥ 0
§ 25 + 475e -0,1t < 6 + 9 * e0,1t
a) T(0) = 22 + 60e0 = 22 + 60 = 82
§ 9e0,1t - 475e -0,1t - 19 > 0
2
§ 9(e0,1t) - 475 - 19e0,1t > 0
O leite foi servido a uma temperatura de 82 °C.
§ 9y - 19y - 475 > 0
2
b) lim T(t) = lim (22 + 60e -0,056 t) = 22 + 60 * 0 = 22
tS + ?
tS + ?
22 8C. Com o decorrer do tempo a temperatura do leite aproxima-se de 22 8C (provavelmente a temperatura ambiente).
2.
A(t) =
N
25
; C(t) =
1 + 19e -0,1t
1 + 1,5e0,1t
Pág. 144
lim
tS + ?
300
= 30
1 + 19e -0,1t
- 0,1t
-0,1t
1 2 § t=
CEXMA12 © Porto Editora
ln(8,397)
§ t > 21
0,1
lim C(t) = lim
‹ 1 + 19e
00
twwuwwv
condição universal
270
§ 570e -0,1t = 270 § e -0,1t =
570
1192
ln
9
-0,1
± t ) 7,5
1990 + 7,5 = 1997,5
A população de águias duplicou no ano de 1997.
48
§ t>
t " +?
No início de 2005 existiam 57 águias.
9
§ -0,1t = ln
19
2
y = e0,1t 9y - 19y - 475 = 0
§ y ) -6,286 › y ) 8,397
y>0
§ e0,1t > 8,397 § 0,1t > ln (8,397)
2.6 tS + ? A(t) = tS + ?
lim
lim
300
) 57
1 + 19e -0,1*15
§ 300 = 30 + 570e
Cálculo auxiliar:
A partir do ano de 2011 a razão entre o número de coelhos e o
número de águias é inferior a 20.
2.2 2005; t = 2005 - 1990 = 15
2.3 A(t) = 30 §
§ y > 8,397
* e0,1t
1990 + 21 = 2011
N
= 15
1 + 19e0
N
§
= 15 § N = 300
20
2.1 A(0) = 15 §
A(15) =
! !!
Deve ser colocado durante 13,2 min.
1.4
-0,1
1000 C(t)
< 20 § 1000 C(t) < 20 A(t)
A(t)
10
21
§ t=
± t ) 13,2
-0,056
ln
1
+
Como A(t) > 0, A t å R0 , vem:
35 = -15 + (90 + 15)e -0,056 t § 50 = 105e -0,056 t
50
§ e -0,056t =
105
§ -0,056t = ln
1 38 2
ln
± t ) 36,4
Pretende-se resolver a inequação
O copo de leite deve permanecer no local mais 17,5 min.
1.3
1 2 § t=
1
§ -0,1t = ln
38
-0,056 t
§ 15 = 70e -0,056 t
15
§ -0,056t = ln
70
15
ln
70
§ t=
± t ) 27,5
-0,056
300
= 200
1 + 19e -0,1t
300
300
=
= 300
1 + 19e -0,1t 1 + 19 * 0
t " +?
25
25
=
=0
1 + 1,5e0,1t +?
C(t)
0
=
=0
A(t) 300
Com o decorrer do tempo o número de águias tenderá a estabilizar
em 300 unidades.
Como a população de coelhos tende a extinguir-se, se não forem
tomadas medidas poderá estar em causa a base da alimentação da
colónia de águias que, assim, poderá entrar em declínio.