1
Universidade Federal de Alagoas Instituto de Matemtica
Banco de Questes
Calculo 1
(Resoluo)
Organizador: Carlos Alberto Santos Barbosa
Maceio, Brasil Janeiro de 2015
2
Apresentao
O objetivo principal desta resoluo do Banco de Questes (extrado das avaliaes escritas de turmas de Clculo Unificado da Universidade Federal de Alagoas - UFAL) auxiliar no desempenho dos estudantes da disciplina Clculo 1, que durante o processo de conhecimento comeam a inserir-se no aprendizado de modo a melhorar seu desempenho acadmico, mostrando-lhes uma noo de como resolver questes das provas realizadas, visando sempre a clareza e objetividade na obteno dos resultados e suas implicaes referentes ao quesito.
3
Sumrio
2005 ................................................................................................................... 7
1.1 1 Avaliao-21 de Fevereiro de 2005 ..................................................... 7
2007 ................................................................................................................. 12
2.1 1 Prova-15 de Setembro de 2007.......................................................... 12
2.2 2 Prova-05 de Outubro de 2007 ............................................................ 16
2.3 1 Prova-06 de Outubro de 2007 ............................................................ 20
2.4 4 Prova-01 de Dezembro de 2007......................................................... 24
2.5 4 Prova-06 de Dezembro de 2007......................................................... 29
2.6 Reavaliao da 2 mdia-07 de Dezembro de 2007 ............................... 35
2008 ................................................................................................................. 41
3.1 1 Prova-14 de Maro de 2008 ............................................................... 41
3.2 2 VPA-12 de Abril de 2008 .................................................................... 44
3.3 3 Avaliao-17 de Maio de 2008 ........................................................... 47
3.4 4 Prova-14 de Junho de 2008 ............................................................... 53
3.5 Reavaliao da 2 mdia-21 de Junho de 2008 ..................................... 59
3.6 VPA 1-12 de Setembro de 2008 ............................................................. 62
3.7 VPA 1-13 de Setembro de 2008 ............................................................. 67
3.8 2 Prova-03 de Outubro de 2008 ............................................................ 72
3.9 2 Avaliao-04 de Outubro de 2008 ...................................................... 75
3.10 3 Prova-01 de Novembro de 2008 ....................................................... 79
3.11 Reposio da 1 Mdia-13 de Dezembro de 2008 ............................... 83
3.12 Reavaliao da 2 mdia-13 de Dezembro de 2008 ............................. 88
3.13 Prova Final-18 de Dezembro de 2008 .................................................. 94
2009 ............................................................................................................... 104
4.1 1 Avaliao-21 de Maro de 2009 ....................................................... 104
4.2 2 Avaliao-17 de Abril de 2009 .......................................................... 108
4.3 2 Avaliao-18 de Abril de 2009 .......................................................... 112
4.4 3 Avaliao-16 de Maio de 2009 ......................................................... 115
2010 ............................................................................................................... 119
5.1 1 Prova-03 de Setembro de 2010........................................................ 119
5.2 1 Prova-04 de Setembro de 2010........................................................ 123
5.3 2 Avaliao-07 de Outubro de 2010 .................................................... 128
4
5.4 2 Avaliao-08 de Outubro de 2010 .................................................... 132
5.5 3 Avaliao-12 de Novembro de 2010 ................................................ 135
5.6 3 Avaliao-13 de Novembro de 2010 ................................................ 139
5.7 4 Avaliao-10 de Dezembro de 2010 ................................................ 144
5.8 4 Avaliao-11 de Dezembro de 2010 ................................................ 150
5.9 Reavaliao AB1-17 de Dezembro de 2010 ......................................... 156
5.10 Reavaliao AB2-17 de Dezembro de 2010 ....................................... 160
5.11 Reavaliao AB2-17 de Dezembro de 2010 ....................................... 163
5.12 Avaliao Final-21 de Dezembro de 2010 .......................................... 167
2011.1 ............................................................................................................ 173
6.1 1 Avaliao-25 de Maro de 2011 ....................................................... 173
6.2 1 Avaliao-26 de Maro de 2011 ....................................................... 178
6.3 2 Avalio-15 de Abril de 2011 ............................................................ 182
6.4 2 Avaliao-16 de Abril de 2011 .......................................................... 185
6.5 3 Avaliao-20 de Maio de 2011 ......................................................... 188
6.6 3 Avaliao-21 de Maio de 2011 ......................................................... 191
6.7 4 Avaliao-17 de Junho de 2011 ....................................................... 194
6.8 4 Avaliao-18 de Junho de 2011 ....................................................... 199
6.9 Reavaliao AB1-22 de Junho de 2011................................................ 204
6.10 Reavaliao AB1-25 de Junho de 2011.............................................. 209
6.11 Reavaliao AB2-22 de Junho de 2011.............................................. 215
6.12 Reavaliao AB2-25 de Junho de 2011.............................................. 220
6.13 Avaliao Final-01 de Julho de 2011 .................................................. 226
2011.2 ............................................................................................................ 234
7.1 1 Avaliao-02 de Setembro de 2011 ................................................. 234
7.2 1 Avaliao-03 de Setembro de 2011 ................................................. 237
7.3 2 Avaliao-30 de Setembro de 2011 ................................................. 241
7.4 2 Avaliao-01 de Outubro de 2011 .................................................... 244
7.5 3 Avaliao-04 de Novembro de 2011 ................................................ 247
7.6 3 Avaliao-05 de Novembro de 2011 ................................................ 250
7.7 4 Avaliao-02 de Dezembro de 2011 ................................................ 254
7.8 4 Avaliao-03 de Dezembro de 2011 ................................................ 260
7.9 Reavaliao AB1-09 de Dezembro de 2011 ......................................... 266
7.10 Reavaliao AB1-10 de Dezembro de 2011 ....................................... 270
7.11 Reavaliao AB2-09 de Dezembro de 2011 ....................................... 274
7.12 Reavaliao AB2-10 de Dezembro de 2011 ....................................... 278
7.13 Avaliao Final-16 de Dezembro de 2011 .......................................... 283
5
2015.1 ............................................................................................................ 292
11.1 1 Prova 11 de Abril de 2015 ........................................................... 292
11.2 2 Prova 08 de Maio de 2015........................................................... 299
11.3 2 Prova 09 de Maio de 2015........................................................... 306
11.4 3 Prova 16 de Outubro de 2015 ..................................................... 315
11.5 3 Prova 17 de Outubro de 2015 ..................................................... 324
11.6 4 Prova 13 de Novembro de 2015 .................................................. 332
11.7 4 Prova 14 de Novembro de 2015 .................................................. 343
11.8 Reavaliao da 1 Mdia 27 de Novembro de 2015 ........................ 353
11.9 Reavaliao da 1 Mdia 28 de Novembro de 2015 ........................ 364
11.10 Reavaliao da 2 Mdia 27 de Novembro de 2015 ...................... 371
11.11 Reavaliao da 2 Mdia 28 de Novembro de 2015 ...................... 380
11.12 Avaliao Final 03 de Dezembro de 2015 ..................................... 391
2015.2 ............................................................................................................ 412
12.1 1 Prova 12 de Fevereiro de 2016 ................................................... 412
12.2 1 Prova 13 de Fevereiro de 2016 ................................................... 418
12.3 2 Prova 11 de Maro de 2016 ........................................................ 427
12.4 2 Prova 12 de Maro de 2016 ........................................................ 434
12.5 3 Prova 08 de Abril de 2016 ........................................................... 440
12.6 3 Prova 09 de Abril de 2016 ........................................................... 448
12.7 4 Prova 06 de Maio de 2016........................................................... 457
12.8 4 Prova 07 de Maio de 2016........................................................... 468
12.9 Reavaliao da 1 Mdia 20 de Maio de 2016 ................................ 477
12.10 Reavaliao da 1 Mdia 21 de Maio de 2016............................... 487
12.11 Reavaliao da 2 Mdia 20 de Maio de 2016............................... 495
12.12 Reavaliao da 2 Mdia 21 de Maio de 2016............................... 504
12.13 Avaliao Final 27 de Maio de 2016 .............................................. 513
2016.1 ............................................................................................................ 535
13.1 1 Prova 22 de Julho de 2016 .......................................................... 535
13.2 1 Prova 23 de Julho de 2016 .......................................................... 543
13.3 2 Prova 19 de Agosto de 2016 ....................................................... 551
13.4 2 Prova 20 de Agosto de 2016 ....................................................... 559
13.5 3 Prova 23 de Setembro de 2016 ................................................... 567
13.6 3 Prova 24 de Setembro de 2016 ................................................... 577
13.7 4 Prova 21 de Outubro de 2016 ..................................................... 585
13.8 4 Prova 22 de Outubro de 2016 ..................................................... 591
13.9 Reavaliao da 1 Mdia 27 de Outubro de 2016 ........................... 599
6
13.10 Reavaliao da 1 Mdia 29 de Outubro de 2016 ......................... 607
13.11 Reavaliao da 2 Mdia 27 de Outubro de 2016 ......................... 613
13.12 Reavaliao da 2 Mdia 29 de Outubro de 2016 ......................... 622
13.13 Avaliao Final 04 de Novembro de 2016 ..................................... 629
2016.2 ............................................................................................................ 650
14.1 1 Prova 17 de Fevereiro de 2017 ................................................... 650
14.2 1 Prova 18 de Fevereiro de 2017 ................................................... 658
14.3 2 Prova 24 de Maro de 2017 ........................................................ 666
14.4 2 Prova 25 de Maro de 2017 ........................................................ 674
14.5 3 Prova 28 de Abril de 2017 ........................................................... 682
14.6 3 Prova 29 de Abril de 2017 ........................................................... 690
14.7 4 Prova 19 de Maio de 2017........................................................... 697
14.8 4 Prova 20 de Maio de 2017........................................................... 705
14.9 Reavaliao da 1 Mdia 26 de Maio de 2017 ................................ 713
14.10 Reavaliao da 1 Mdia 27 de Maio de 2017............................... 721
14.11 Reavaliao da 2 Mdia 26 de Maio de 2017............................... 728
14.12 Reavaliao da 2 Mdia 27 de Maio de 2017............................... 737
14.13 Avaliao Final 02 de Junho de 2017 ............................................ 745
7
Captulo 1 2005
1.1 1 Avaliao-21 de Fevereiro de 2005
1.
() =3 1
53 202 + 15
() =( 1)(2 + + 1)
5( 1)( 3)
() , () = {0, 1, 3}. :
= :
lim+
() = lim
() =
= 0; = 1 = 3 . : 0+, > 0. , 5 > 0.
lim0+
() = lim0+
( 1)
1
(2 + + 1)
1
50+
( 1) 1
( 3) 3
= lim0+
( 1)(2 + + 1)
1
5( 1)( 3) 0+
=
, = 0 ().
: 1, 1. , 1 0 , ,( 1)
( 1)= 1, 1.
lim1+
() = lim1+
( 1)(2 + + 1)
5( 1)( 3)= lim1+
(2 + + 1)
3
5( 3) 10
= 3
10
lim1
() = lim1
( 1)(2 + + 1)
5( 1)( 3)= lim1
(2 + + 1)
3
5( 3) 10
= 3
10
, = 1 (). : 3+, > 3. , 3 > 0.
lim3+
() = lim3+
( 1)
2
(2 + + 1)
13
515
( 1) 2
( 3) 0+
= lim0+
( 1)(2 + + 1)
26
5( 1)( 3) 0+
= +
, = 3 ().
8
:
= :
lim+
() = lim
() =
lim+
() = lim+
3 1
53 202 + 15= lim+
3 (1 13)
3 (5 20 +
152)= lim+
1 13
5 20 +
152
=lim+
1 lim+
13
lim+
5 lim+
20 + lim+
152
=1 0
5 0 + 0=1
5.
, = 1 5 ().
lim
() = lim
3 1
53 202 + 15= lim
3 (1 13)
3 (5 20 +
152)= lim
1 13
5 20 +
152
=lim
1 lim
13
lim
5 lim
20 + lim
152
=1 0
5 0 + 0=1
5.
.
, = 0 , = 3 = 1 5 ().
2.
) lim2
+ 2 2
3 8= lim
2
( + 2 2)
3 8.( + 2 + 2)
( + 2 + 2)= lim2
2
(3 8)( + 2 + 2)=
lim2
( 2)
( 2)(2 + 2 + 4)( + 2 + 2)= lim2
1
(2 + 2 + 4) 12
( + 2 + 2) 4
=1
48
: 2, 2. , 2 > 0 , ,( 2)
( 2)= 1, 2.
) lim0+
(1
1
||) . : || = {
, 0, < 0
; 0+, > 0. , || = .
lim0+
(1
1
||) = lim
0+(1
1
) = lim
0+(0
) = lim
0+0 = 0.
) lim010 (
50
3 )
:
1 (50
3 ) 1
10 :
10 10 (50
3 )
10
9
() = 10, () = 10 (50
3 ) () =
10. ,
() () () , lim
0() = 0 lim
0() = 0, , lim
0() = lim
0() ,
() () () lim0() = lim
0() ,
lim0() = lim
0() = lim
0() = 0. ,
lim010 (
50
3 ) = 0
3.
() = {2 4
2+ 10, < 2
23 , 2
() = {
( 2)( + 2)
2+ 10, < 2
23 , 2
: < 2 2. , 2 0 , ,( 2)
( 2)= 1, 2.
, () :
() = { + 2 + 10, < 2
23 , 2
() = { + 12, < 2
23 , 2
) () , , . , () (, 2) (2, +).
() = 2, .
: 2+, > 2. , () = 23
lim2+
() = lim2+
(23 ) = lim2+
23 lim2+
= 16 2 = 14.
: 2, < 2. , () = + 12.
lim2
() = lim2+
( + 12) = lim2+
+ lim2+
12 = 2 + 12 = 14.
, lim2() = 14.
10
(2) = 2.23 2 = 16 2 = 14. lim
2() = (2), () = 2.
, () . ) (2) = 14, (3). , : (3) = 2.33 3 = 54 3 = 51.
() = 15. () [2, 3] (2) < () < (3), (2, 3) () = 15.
4. () = | 1|
| 1| = { 1, 1( 1), < 1
() = {2 , 1
2 + , < 1
) () = 1. (1) = 12 1 = 1 1 = 0. lim1+
() = lim1+
(2 ) = lim1+
2 lim1+
= 1 1 = 0.
lim1
() = lim1
(2 + ) = lim1
2 + lim1
= 1 + 1 = 0.
lim1+
() = lim1
() , lim1() = 0.
, , lim
1() = (1). , () = 1.
+(1) = lim1+
() (1)
1= lim1+
2
1= lim1+
( 1)
( 1)= lim1+
= 1.
(1) = lim
1
() (1)
1= lim1
2 +
1= lim1
( 1)
( 1)= lim1
= 1.
= 1 , , + (1)
(1), () = 1. ) (2, 2) (). (2) = 22 2 = 4 2 = 2.
1 . ( , ) () :
= ( )
11
, = (2). , 2 = (2)( 2)
(2) = lim2
() (2)
2= lim2
2 2
2= lim
2
( + 1)( 2)
( 2)= lim2( + 1) = 3.
, (2, 2) :
2 = 3( 2) 2 = 3 6 = 3 4
12
Captulo 2 2007
2.1 1 Prova-15 de Setembro de 2007
1. () = | 3|
() = { 3, 3( 3), < 3
) () = , ,
1) () ; 2) lim
() ;
3) () = lim
().
= 3 : 1) (3) = |3 3| = |0| = 0. 2) lim
3() . , :
lim3+
() = lim3+
( 3) = lim3+
lim3+
3 = 3 3 = 0.
lim3
() = lim3+
( + 3) = lim3+
+ lim3+
3 = 3 + 3 = 0.
, lim3+
() = lim3
() lim3() = 0.
3)(3) = lim3() = 0.
, = 3.
= 3 :
+(3) = lim
3+
() (3)
3= lim
3+
3 0
3= lim3+
( 3)
( 3)= 1.
(3) = lim
3
() (3)
3= lim
3
( 3) 0
3= lim3
( 3)
( 3)= 1.
+(3)
(3) = 3.
) = + 2 2 + 2 = 0 () = 3 2 + 2. (2) (0). (2) = (2)3 (2)2 + 2 = 8 4 + 2 = 10. (0) = 03 02 + 2 = 0 0 + 2 = 2. () = 0. , , [2,0], (2) < () < (0), (2, 0) () = 0. 2.
13
2.
() = 22 + 7 , :
() = lim0
( + ) ()
, = ():
() = lim
() ()
, (1) . , :
(1) = lim1
() (1)
1
lim1
() (1)
1= lim1
22 + 7 3
1= lim1
(22 + 7 3)
1.(22 + 7 + 3)
(22 + 7 + 3)=
lim1
22 + 7 9
( 1)(22 + 7 + 3)= lim
1
( 1)(2 + 9)
( 1)(22 + 7 + 3)= lim1
(2 + 9)
(22 + 7 + 3)=
2 + 9
9 + 3=11
6.
() = 1 :
(1) = (1)( 1)
3 =11
6( 1)
6 18 = 11 11 6 = 11 + 7
=11
6 +
7
6
3.
) lim6
(3
4) =
3
2, () = 3
4
= 6, lim6() =
3
2
> 0 > 0 |() 3
2| <
0 < | 6| < .
|() 3
2| < |3
43
2| < |
3
2
4| <
1
4| 6| < | 6| < 4
| 6| < , = 4.
! ( ). > 0 , 0 < | 6| < ,
|3
2
4| <
1
4| 6| 3. , = 3
lim3
() = lim3
(3 +
2) = lim
33 + lim
3
2= 3 +
2
2= 4
: 3, < 3. , = 2.
5. = 256 162
) () =
= () = 256 32 () = 256 32
(6) = 256 32 6 = 256 192 = 64 / ) () = 0 256 32 = 0 32 = 256 = 8 ) (8) = 256 8 32 8 (8) = 2.048 256 (8) = 1.792
16
2.2 2 Prova-05 de Outubro de 2007
1. 2 + 2 = 4
: 2 + 2. = 0
=
: = .
= , , , = 75.
= 75.
75 = (30 + 45) =30 + 45
1 30. 45=
1
3+ 1
1 1
3
=
1 + 3
3
3 1
3
=3 + 1
3 1
2 + 2 = 4 :
= ( )
. 2 = . +
2
. = . + 2 +
2 . = . + 4
= +
4
, .
=
1
3 + 1
3 1
=1 3
1 + 3
1 3
1 + 3= 165. ,
.
165 = (120 + 45) =120 + 45
1 120. 45=3 + 1
1 + 3=1 3
1 + 3
, 15 .
2. ) 3. ( + ) + 107 = :
(3. ( + )) +
(107) =
(2)
3( + ) + 3.1 +
1 + ( + )2+ 0 = 2 + 2
17
3( + ) +3 + 3
1 + ( + )2= 2 + 2
3( + )[1 + ( + )2] + 3 + 3 = 2[1 + ( + )2] + 2[1 + ( + )2] [2(1 + ( + )2) 3] = 3( + )[1 + ( + )2] + 3 2[1 + ( + )2]
=3( + )[1 + ( + )2] + 3 2[1 + ( + )2]
2[1 + ( + )2] 3
) () = log3 52+1 . ().
= 2 + 1 = 5 = () = log3 :
=
.
.
= (2). 5 ln(5) .
1
. ln(3)
=2. 5
2+1 ln(5)
52+1 ln(3)
() = 2.ln(5)
ln(3)
3. () = 2() + 2() () = 2cos() + 2. (). cos() [0,2] () . ( [0, 2] () = 0) () = 0 2 cos() + 2. (). cos() = 0 2 cos() [1 + ()] = 0
{2cos() = 01 + () = 0
:
cos() = 0 =
2 =
3
2
: 1 + 2() = 0
() = 1 =3
2
, [0, 2] ()
, : (
2, 3) , (
3
2,1).
(
2) = 2 (
2) + 2 (
2) = 2 + 12 = 2 + 1 = 3. (
2, 3) ;
(3
2) = (
3
2) + 2 (
3
2) = 1 + (1)2 = 1 + 1 = 0. (
3
2, 0) ;
4.
18
) =1
() + cos() . = 0.
=1
(0) + cos(0)=
1
0 + 1=1
1= 1.
=(cos() ())
(() + cos())2=() cos()
[() + cos()]2
(0) =(0) cos(0)
[(0) + cos(0)]2=
0 1
(0 + 1)2=1
1= 1.
(0,1) : 1 = 1( 0) 1 = = + 1
) () = 3(4). ().
= 4; = () ; = 3; = () = 2().
, :
=
.
.
.
= (43).
1
1 + 2. 3 ln(3) . 2(). cos()
= () =43. 3(
4). ln(3). (2)
1 + 8
() =43. 3(
4). ln(3). (2. 3(4))
1 + 8
5. = 9 2. (2, 1) (2,1) = 9 2. ! :
= ( ) ( , ) , .
1 = ( 2) 1 = 2( 2)
9 2 1 = 22 + 4 2 4 + 8 = 0 = 16 32 = 16.
. , = 9 2 (2,1). , !
: = 9 3, (2,1). :
1 = (2). ( 2)
19
1 = (3.22)( 2) 1 = 12( 2) 1 = 12 + 24 = 12 + 25
, = (12).
20
2.3 1 Prova-06 de Outubro de 2007
1.
) () = . (1
) (
1
) . () .
= : lim
+() = lim
() = .
lim+
() = lim+
(. (1
) (
1
)) . =
1
, : +, 0.
lim0
(()
()) = lim
0
()
lim0
cos() = 1 1 = 0.
, = 0 ().
lim
() = lim
(. (1
) (
1
)) . =
1
, : , 0.
lim0
(()
()) = lim
0
()
lim0
cos() = 1 1 = 0.
, = 0 ().
)
, . arccos( + ) = 2.
:
arccos( + ) + ((1 + )
1 ( + )2) 0 = 2.
arccos( + ) [1 ( + )2] = 2. [1 ( + )2]
[21 ( + )2 + ] = arccos( + ) [1 ( + )2]
=arccos( + ) [1 ( + )2]
21 ( + )2 +
2. = 22 1 . (4, 13). = 4 :
= ( ) 13 = 4( 4)
22 1 13 = 42 16
21
22 16 + 14 = 0 2 8 + 7 = 0 = 64 28 = 36
=8 6
2 = 7 = 1
= 1, = 4 , :
13 = 4( 4) 13 = 4 16 = 4 3
= 7, = 28 , :
13 = 28( 4) 13 = 28 112 = 28 99
3.
() = 3 + || () =3
4||
4 ; (0) (0) ,
( )(0) .
+(0) = lim
0+
() (0)
0= lim
0+
4 0
= lim0+
4
= 4.
(0) = lim
0
() (0)
0= lim
0
2 0
= lim0
2
= 2.
+(0)
(0) , = 0.
+(0) = lim
0+
() (0)
0= lim
0+
2 0
= lim0+
2=1
2.
(0) = lim
0
() (0)
0= lim
0
0
= lim0
= 1.
+(0)
(0) , = 0.
( )():
(()) = 3. () + |()| =9
43||
4+ |3
4||
4|
(()) = {
9
43
4+3
4
4, 0
9
4+3
4 (3
4+
4) , < 0
(()) = {2, 02, < 0
, (()) = 2.
( )+(0) = lim
0+
( )() ( )(0)
0= lim
0+
64 +
24 0
= lim0+
2
= 2.
( )(0) = lim
0
( )() ( )(0)
0= lim
0
3 0
= lim0
2
= 2.
( )
+(0) = ( )
(0) , = 0.
( )(0) = 2.
22
4.
) =() 1
sec() . = 0.
= 0 =(0) 1
sec(0)=0 1
1= 1. (0, 1)
=2(). sec() [() 1] sec() . ()
()
=3() sec() . 2() + sec() . ()
()
=2() 2() + ()
() ; 2() + 1 = 2() 2() 2() = 1
=1 + ()
() (0) =
1 + (0)
(0)=1 + 0
1= 1.
, = 0 : + 1 = 1( 0) + 1 = = 1
) () = [cos()] . ().
= cos() ; = ; = (); = () = cos() :
=
.
.
.
= (()).
1
2. (2()). (())
= () = (()).1
2cos(). (cos()). ( (cos()))
() =() (cos()) cos()
2cos()
) () = + . ().
= + ; = () = ; :
=
.
= (1 +
1
2) (
1
2)
= (1 +
1
2)(
1
2 + )
5.
) () = 5((2)). ().
= 2; = (); = (); = () = 5 . :
23
=
.
.
.
= (2). (cos()).
1
1 + 2. 5 . ln(5)
= () = (2)(cos(2))1
1 + 2(2). 5(
(2)). ln(5)
() =2. cos(2) . 5(
(2)). ln(5)
1 + 2(2)
) = . + 5 6 = 0. = ; = . : + 5 6 = 0 0 , 2 + 5 6 = 0 = 25 + 24 = 49
=5 7
2 = 1 = 6.
, 1 6.
24
2.4 4 Prova-01 de Dezembro de 2007
1. ) 4 + 4 + 1000 = 0 , , . () = 4 + 4 + 1000. , , . () 1 , . [, ], , (, ) :
() =() ()
=0 0
= 0
() = 43 + 4 () = 0 43 + 4 = 0 3 = 1 = 1 , : < 1 < , , 1 () = 0. ! () . , () , , . ) [2, 5] 1 () 4 (2, 5), 3 (5) (2) 12.
[2, 5] (2, 5), [2,5] :
() =(5) (2)
5 2
:
1 (5) (2)
5 2 4
1 (5) (2)
3 4
3 (5) (2) 12 2. () = 2 + (1) = 0 (1) = 7 ; , : (2) = 2 () = 2 + (1) = 2 + = 0 () (1) = + = 7 () (2) = 4 + 2 = 2 ()
{2 + = 0 + = 7
4 + 2 = 2 {2 + = 0 = 7 = 2
= 2; = 9 ; = 18
, () = 92 + 18 2
3. ) () = + + :
25
() = 2 + () = 2 () > 0 2 > 0 > 0. , () > 0 . , . . > 0. () < 0 2 < 0 < 0. , () < 0 . , . . < 0.
) lim0+
( + 1)() = . .
lim0+
( + 1)() = lim0+
ln(+1)()
= lim0+
().ln(+1) = lim0+
ln(+1)1
() =
lim0+
ln(+1)1
() .
lim0+
ln( + 1)
1()
= lim0+
ln( + 1)
()= lim0+
+ 1()
=
+ 1
lim0+
ln(+1)1
() = +1 =
+ 1= = ( + 1) = =
+ 1
4. () = (I) :
() = 0 . 2= 0 ;
2 0 x
() = 0 2 = 0 = 0. (0, 0) . (II) :
() = 2. 2+ 2(2).
2
() = 2(23 + 2)
() = . 2(22 + 2)
Obs: 2> 0 ,
(), : 0 + + + + +++++ + (1) + + + + + + 1 (22 + 2)
+++++++++ ++++++++++ + 2
++++ (1) 0 + + + 1 () = . 2(22 + 2)
, () (,1) (0, 1) (1, 0) (1,+). () : () = 0 () . , = 1; = 0 = 1.
(1) =1
; (0) = 0 ; (1) =
1
.
26
, (1,1
) , (0, 0) (1,
1
).
()
() = (2). 2(23 + 2) +
2(62 + 2)
() = 2(44 42 62 + 2)
() = 2(44 102 + 2)
(): = 2 42 10 + 2 = 0 = 100 32 = 68
=10 68
8 =
10 68
8;
() = 44 102 + 2 () = () . ,
: 1 ()
(,10 + 68
8) (
10 68
8,10 68
8) (
10 + 68
8,+)
2 ()
(10 + 68
8,
10 68
8) (
10 68
8,10 + 68
8)
() :
() = 0, , = 10 68
8.
():
: ! , () =2
2 () = . ,
, . : = : lim
+() = lim
() =
lim+
() = lim+
2
2 = lim
+
2
(2)2 = lim
+
1
2 = 0.
27
lim
() = lim
2
2 = lim
2
(2)2 = lim
1
2 = 0.
, = 0 ().
5. ) 2 = 1 . (0, 1). (, ) , :
= 2 + ( + 1)2 = 2 + 2 + 2 + 1 = 22 + 2 + 2
() =4 + 2
222 + 2 + 2=
2 + 1
22 + 2 + 2
() = 0 2 + 1 = 0 = 1
2
, = 1
2 :
2 1
4= 1 2 =
5
4 =
5
2.
(0, 1) (5
2,1
2) (
5
2,1
2).
) + = 0 () = 32 (0) = 0. . () = 32. () = 3 + , . (0) = 0, = 0. , () = 3. = 1, () = 1. , (1, 1).
() = 3 () =1
44 + ,
. (1) = 1, : 1
4(1)4 + = 1 = 1
1
4 =
3
4.
28
, () =1
44 +
3
4.
29
2.5 4 Prova-06 de Dezembro de 2007
1. )
() = 73 4 + 3 + 72() + , () = (). () :
[. ] = . 1 = 1 2
. 1
1 2= 1 . 3 2 = 1. , :
{ 3
2= 0
= 1
=3
2 =
1
=2
3.
: 2
33 2 .
[. ] = 3 4 . 1 = 3 4
. 1
3 4= 1 . +1 4 = 1. ,
{ +1
4= 0
= 1
= 1
4 = 4.
: (7). [41 4 ] = 281 4 .
[. ] = 3 = 3 = 3.
: 3. 7. (). : .
, () =2
33 2 + 281 4 + 3 + 7. () + + .
, , () = (). ) () = 3 + 4(), (0) = 0 () = 0.
(). () = 3 4 cos() + , .
:
() = 3 4() + + , .
(0) = 0 30 4(0) + . 0 + = 0 3 0 + 0 + = 0 = 3.
() = 0 3 4() + . + 3 = 0 3 0 + . + 3 = 0 . = 3 3
=3 3
.
, () = 3 4() +3 3
+ 3.
30
2.
:
=
= =
( )
.
= 2 = 2.
( ) =
(2 3).
.
() =
(2 32); () = 0 2 32 = 0 ;
, =2
3;
: = 0 !
=2
3, =
( 23 )
=
(3)
=
3
= 1 = 1 : =2
3 =
1
3.
= . 2. = . (
2
3)2
. (1
3) =
4
9 .
3.
) lim2
(cos )cos = lim2
ln(cos)cos
= lim2
cos.ln(cos) = lim2
ln(cos)sec =
lim2
ln(cos)sec
.
:
lim2
ln(cos )
sec = lim2
cos
sec . = lim2
sec . = lim2
1
sec = lim2
( cos ) = 0.
,
lim2
(cos )cos = lim2
ln(cos)sec
= 0 = 1.
) lim
(1 +3
+5
2)
= lim
ln(1+
3+52)
= lim
.ln(1+
3+52)= lim
ln(1+3+52)
1 =
lim
ln(1+3+52)
1 .
31
:
lim
ln (1 +3 +
52)
1
= lim
(32103)
(1 +3 +
52)
12
= lim
2(
32103)
1 +3 +
52
= lim
3 +10
1 +3 +
52
=
3 + 0
1 + 0 + 0=3
1= 3.
,
lim
(1 +3
+5
2)
= lim
ln(1+3+52)
1 = 3.
4. () = 3 + 2, = 0 = 1.
:
= lim
()
=1
=1 0
=1
; =
. , :
= lim
(
)1
=1
= lim
1
(
3
3+2
2)
=1
= lim
1
[1
33
=1
+1
22
=1
]
1 : 2
=1
=(22 + 3 + 1)
6
2 : 3
=1
=( + 2 + 1)
4
= lim
1
[1
3(2 + 2 + 1)
4+1
2(22 + 3 + 1)
6]
= lim
1
[(2 + 2 + 1)
4+(22 + 3 + 1)
6]
32
= lim
1
[6(2 + 2 + 1) + 4(22 + 3 + 1)
24]
= lim
142 + 24 + 10
242
= lim
142
2+242+102
242
2
= lim
14 +24 +
102
24
= 14 + 0 + 0
24
=14
24=7
12.
5. () =3 + 1
2 ;
) ; , , , . , 2 0 0. () = {0}. ) ; , = 0 ; () = 0 3 + 1 = 0 (2) (1):
(2) =8 + 2 + 1
4=
5
4 (1) =
1 + 1 + 1
1= 1.
, , [2, 1], (2) < 0 < (1), (2,1) () = 0. , () = 2 = 1. ) ;
() =(32 1)2 (3 + 1)2
4
() =34 2 24 + 22 2
4
() =4 + 2 2
4 ; 0 :
() =3 + 2
3
() = 0 3 + 2
3= 0 3 + 2 = 0
2,1, 1 2 :
(2) =8 2 2
8=10
8 ; (1) =
1 1 2
1= 4 ;
33
(1) =1 + 1 2
1= 0 ; (2) =
8 + 2 2
8= 1.
, = 1 () = 0. () : 3 + 2 = ( 1)(2 + + 2) , , . , () ( = 1).
(1) =13 1 + 1
12= 1. (1, 1).
) ; () (): 1 + + + +++ ++++++ ( 1) +++++++++ ++++++++++ ++++++ (2 + + 2) 0 + + + ++++++++ ++++++ 3
+++++++0 1 + + + +++++++++ () =3+2
3
: () (, 0) (1,+) () (0, 1) ) ;
() =(32 + 1)3 (3 + 2)32
6
() =35 + 3 35 33 + 62
6
() =23 + 62
6 ; 0
() =2 + 6
4
() (): +++++++++ ++++ 3 (2 + 6) +++++++ + 0 + + + + ++++++ ++++++ 4
++++++++ 0 + + + + 3 () =2+6
4
: () (, 0) (0,3) () (3, +) ) ; () = 0, . , = 3.
(3) =33 3 + 1
32=25
9
(3,25
9)
) : : = :
lim+
() = lim
() =
= 0 :
34
lim0+
() = lim0+
3 + 1
2= lim0+
3 + 1
1
20+
= +
, = 0 ().
: = :
lim+
() = lim
() =
lim+
() = lim+
3 + 1
2= lim+
3 1
2= lim+
6
2= lim+
3 = +
lim
() = lim
3 + 1
2= lim
3 1
2= lim
6
2= lim
3 =
, (). : = +
lim
[() ( + )] = 0
() =3 + 1
2=
1
2
() = 1
2
lim
[() ] = lim
( + 1
2) = lim
2+12
2
2
= lim
1 +
12
1=
lim
[1
+1
2] = lim
1
+ lim
1
2= 0 + 0 = 0
, = ().
35
2.6 Reavaliao da 2 mdia-07 de Dezembro de 2007
1. , () . :
=
=
, , 8/ . ,
= 8
, , . ,
= 8
:
=
.
:
= 1/ = 0,1/
=2
3. :
=20
5=> =
4. :
=3
48 ,
=32
48=2
16. ,
= 0,1
16=>
0,1
16.
= 8
0,12
16 . = 16 :
= 8
0,1(16)2
16= (8 1,6)3/
2. ) = = 1. = 1 : 1 = 1 1 = . (1, 1). , : ln = ln . ln = . ln , :
. ln + .1
= ln + .
36
. ln .
= ln
. [ln
] = ln
=ln
ln
(1, 1) :
=ln1
11
ln 1 11
=0 1
0 1=1
1= 1.
, (1, 1) 1: 1 = 1( 1) 1 = 1
= ) = cosh 1. = () = 1 () = 1 ()
() =
2 ; ():
2= 1
= 2
1
2 = 0
2 2 1 = 0 = :
2 2 1 = 0 : = , > 0 . , : = 4 + 4 = 8
=2 22
2 = 1 + 2 = 1 2 ( < 0)
! : =
1 + 2 =
= ln(1 + 2)
, = ln(1 + 2):
= cosh((1 + 2))
=(1+2) + (1+2)
2
=
1 + 2 +1
1 + 22
=1 + 22 + 2 + 1
2 + 22=4 + 22
2 + 22=2 + 2
1 + 2= 2.
, (ln(1 + 2) , 2 ).
37
3.
) () = . 2 (2, 1). .
, , (2) = 1 (2) = 0. , (2) = 2. 4 = 1
() = . 2+ 22.
2
(2) = . 4 + 8. 4 = 0 ; 4 0 .
{ 2. 4 = 1
+ 8 = 0 , :
(1 + 8) = 0
= 0 1 + 8 = 0 = 1
8; = 0, () = 0 . ! ,
(2) = 1.
, = 1
8. :
2. 1 2 = 1 2
= 1 2 = =
1
2;
() =1
2. .
2 8
) lim0(1 + ) .
=1
, , 0 .
lim0(1 + ) = lim
(1 +
1
)
= [lim
(1 +1
)
]
=
: lim
(1 +1
)
=
4. = 20 . = 2( + ) = 20 + = 10 = 10 () = . 2. = . 2. (10 ) = (102 3) () = (20 32) () = 0 :
20 32 = 0 = 0 =20
3 . = 0 !
= 10 = 10 20
3=10
3.
, =20
3 =
10
3.
5. =22
9 2. () = =
22
9 2
(): () = ; 9 2 0 () = {3, 3}.
38
() : :
(0) =2.02
9 02=0
9= 0. (0, 0)
: () = 0 22 = 0 = 0. (0, 0) (): : = :
lim+
() = lim
() =
= 3 = 3 :
lim3+
() = lim3+
22
9 2;
: (3) + ++ + + ++ (3) (9 2)
1: 3+, > 3 . , 9 2 > 0 9 2 0+.
0+, > 0. ,
lim3+
() = lim3+
22
9 2= +
lim3
() = lim3
22
9 2;
2: 3, < 3 . , 9 2 < 0 9 2 0.
0, > 0. ,
lim3
() = lim3
22
9 2= .
, = 3 ().
lim3+
() = lim3+
22
9 2;
3: 3+, > 3 . , 9 2 < 0 9 2 0.
0, > 0. ,
lim3+
() = lim3+
22
9 2=
lim3
() = lim3
22
9 2;
2: 3, < 3 . , 9 2 > 0 9 2 0+.
0+, > 0. ,
lim3
() = lim3
22
9 2= +.
, = 3 ().
39
: = :
lim+
() = lim
() =
lim+
() = lim+
22
9 2= lim
+
22
2
9 2
2
= lim+
2
92 1
=2
1= 2.
lim
() = lim
22
9 2= lim
22
2
9 2
2
= lim
2
92 1
=2
1= 2.
, = 2 (). () :
() =4. (9 2) 22. (2)
(9 2)2
() =36 43 + 43
(9 2)2
() =36
(9 2)2;
() (), :
0 + + + +++++ ++ 36 ++++ (3) + + + + + +++(3) + + + + + (9 2)2
(3) 0 + + + (3) + + + + + () =36
(92)2
() :
() (0, 3) (3, +) () (,3) (3, 0) () :
, () = 0, , = 0 (0) = 0. (0, 0). () :
() =36(9 2)2 36(2)(2)(9 2)
(9 2)4
() =36(9 2) + 1442
(9 2)3
() =1082 + 324
(9 2)3
() =108(2 + 3)
(9 2)3
40
(), : +++++++++ ++++++++++ ++ 108(2 + 3) (3) + + + ++ +(3) (9 2)3
(3) + + + ++ +(3) () = 108(2+3)
(92)3
() :
() (3, 3) () (,3) (3,+)
(), ; () = 0.
, () = =2
9 :
41
Captulo 3 2008
3.1 1 Prova-14 de Maro de 2008
1.
() =1
2 ; (2, 1 4 )
( ) () = 2.
() = lim0
( + ) ()
= lim0
1( + )2
12
= lim0
2 ( + )2
2( + )2
=
lim0
2 2 2. 2
[2( + )2]= lim
0
2. 2
[2( + )2]= lim0
(2 )
[2( + )2]=
lim0
2
2( + )2=2
4.
() =2
4=2
3
(2) =2
(2)3=2
8=1
4.
: = ( )
1
4=1
4( + 2)
=1
4 +
3
4
2.
) lim0
( )
= lim0
( )cos( )
= lim0
( )
[cos( )]=
lim0
. cos . cos
[cos( )]= lim0
[cos( )]= lim0
1
[cos( )]=
1
cos()
=1
1= 1.
) lim
(3 + 2)2
(2 1)3(3 1)= lim
6 + 44 + 42
(83 122 + 6 1)(3 1)=
lim
6 + 44 + 42
86 125 + 64 93 + 122 6 + 1=
lim
6 + 44 + 42
6
86 125 + 64 93 + 122 6 + 16
=
lim
1 +42+44
8 12 +
6293+12465+16
=1 + 0 + 0
8 0 + 0 0 + 0 0 + 0=1
8.
42
) lim2
sec + cos
cos = lim2
cos
1cos
+ cos
cos = lim2
1 + cos2
cos2 =
lim2
1 + 1 2
1 2= lim
2
2
1 2= lim2
(1 )
(1 )(1 + )=
lim2
(1 + )=
1
(1 + 1)=1
2.
3. [, ] : (, ),
1) () () ; 2) lim
+() lim
+() ;
3) lim+ () = () lim+ () =() , :
1) () () ; 2) lim
() lim
() ;
3) lim
() = () lim
() = ();
, : (2) = 7 ; (2) = 2 ; (5) = 4 (5) = 5.
) lim5
2() 25
() 5= lim5
[() 5][() + 5]
() 5= lim5
[() + 5] =
lim5
() + lim5
5 = 5 + 5 = 10.
) ( )() = () () ( )(2) = (2) (2) = 2 7 = 5. ( )(5) = (5) (5) = 5 4 = 1. ) 1 , , ( )() [2, 5], , ( )(2) < 0 < ( )(5). , (2, 5) ( )() = 0. ,
( )() = 0 () () = 0 () = ().
4.
() = {3
2 + 5, 2,
2 + + (1 ), > 2
= 2. (2) = lim
2+() ,
43
(2). ,
(2) = 3
2(2) + 5 = 3 + 5 = 2 ()
lim2+
() = lim2+
[2 + + (1 )] = 4 + 2 1 = 2 5 ()
, :
2 5 = 2 2 = 7 =7
2.
() , , , , = 2. , . , (, 2) (2,+). , = 7 2 , = 2 , , (,+). , .
5. () = + 1
1.
, ! () = { | 0, 1}
: = :
lim+
() = lim
() =
= 1 :
lim1+
() = lim1+
+ 1
1= + ; : 1 0+
: 1+ > 1 , , > 1 1 > 0; , = 1 ().
: = :
lim+
() = lim
() =
() , = 1, lim
() . , +.
lim+
() = lim+
+ 1
1= lim
+
(1 +1
)
(1 1
)= lim+
1 +1
1 1
=1 + 0
1 0=1
1= 1.
, = 1 ().
44
3.2 2 VPA-12 de Abril de 2008
1.
() = 1 + +2
2+3
3++
10
10
) () = 0 + cos + . cos + . cos + + 9. cos () = cos (1 + + 2 + 3 ++ 9)
) (
6) = cos
6(1 +
6+ 2
6+ 3
6++ 9
6)
(. ) =
6,
=1
2.
. :
=1(1
)
1
=1(1
1210)
1 12
=
210 1210
12
=2(1024 1)
1024=1023
512
(
6) =
3
21023
512=10233
1024
)() = cos (1 + + 2 + 3 ++ 9)
(1 + + 2 + 3 ++ 9) =1 10
1
() = cos 1 10
1 = 1
, : 1 10 = 1 (9 1) = 0
{ = 0
9 1 = 0 = 0 =
2;
(0,
2) ,
() = cos .
2.
() = {32, 1 + , > 1.
) = 1 :
45
(1) = lim1+
() , (1).
(1) = 3.12 = 3.1 = 3 lim1+
() = lim1+
[ + ] = +
, : + = 3 () ) = 1. = 1. = 1, :
(1) = lim
1
() (1)
1= lim1
3 3
1= lim1
3( 1)( + 1)
1= lim1
[3 + 3] = 6
+(1) = lim
1+
() (1)
1= lim1+
+ 3
1= lim1+
+ (3 ) 3
1=
lim1+
( 1)
1= lim1+
= .
= 1 . ,
(1) =
+(1) = 6. , () = 3.
3.
) +3
22 =
15
2
, : 2 + 3 = 0 (3 ) = 2
= 2
3
) . ,
= 1
= 3
2 ; (2, 1) :
=2 3
1 + 4=
5
5= 1.
, (2, 1) : 1 = 1( + 2) 1 = 2 = 1
= 1 .
2 ( 1) +3
2( 1)2 =
15
2
2 + 2 + +3
2(2 + 2 + 1) =
15
2
42 + 2 + 32 + 6 + 3 = 15 72 + 8 12 = 0 = 64 + 336 = 400
=8 20
14 =
12
14 = 2 . = 2 (2,1)
, , =12
14=6
7.
46
= 1
= 6
7 1 =
13
7.
, , , (6
7,13
7).
4.
) () = 102++1 . (0, 10).
() = 102++1. (2 + 1). ln(10)
(0) = 10. ln 10 (0, 10):
10 = 10. ln 10 ( 0) = (10. ln 10) + 10
) () = [()]2 ; (1) = (1) = 1.
() = 1.
(1) = [(1)]2= 1
2= . (1, ).
() = 2(). (). [()]2
(1) = 2(1). (1). [(1)]2
(1) = 2.1.1. = 2
(1, ) = 2( 1) = 2 2 = 2
5.
) () = + ; =
4.
(
4) =
4+
4= 2 + 1. : (
4, 2 + 1)
() = . 2
(
4) = 2 2
(
4, 2 + 1):
(2 + 1) = (2 + 2) (
4)
= (2 2) +(2 + 2)
4 + 2 + 1
= (2 2) +2( + 4) + 2( + 2)
4
) () = (cos(2 + 1)) . (). = 2 + 1 ; = cos ; = () = 2 , :
=
.
.
= () = (2). (()). 2. cos
() = 2. (2 + 1). (2) () = 2. (2 + 1). (2cos(2 + 1))
47
3.3 3 Avaliao-17 de Maio de 2008
1. ) () = ln ; , (). ln () = ln ln ln () = (ln )2 ()
()= 2. ln .
1
() = 2ln . ln . 1 () = 2ln 1. ln ) = cosh 1. :
, ,
4= 1.
, , 1. = () = 1 () = 1 ()
() =
2 ; ():
2= 1
= 2
1
2 = 0
2 2 1 = 0 = :
2 2 1 = 0 : = , > 0 . , : = 4 + 4 = 8
=2 22
2 = 1 + 2 = 1 2 ( < 0)
! : =
1 + 2 =
= ln(1 + 2)
, = ln(1 + 2):
= cosh((1 + 2))
=(1+2) + (1+2)
2
=
1 + 2 +1
1 + 22
=1 + 22 + 2 + 1
2 + 22=4 + 22
2 + 22=2 + 2
1 + 2= 2.
, (ln(1 + 2) , 2 ).
48
2.
, : 1,8
=15
15 =
27
15
= 1,5/
, :
=
.
() =27
(15 )2. 1,5
= 9
() =27
(15 9)2. 1,5
() =27
36. 1,5
() =3
4. 1,5 =
4,5
4/
3.
) 8,013
.
8,013
. :
() = 3 () =
1
33
.
8,01, 83
, (8 + 0,01). :
= (). = ( + ) () , :
( + ) () (). (8 + 0,01) = 8 = 0,01, :
(8 + 0,01) (8) 1
383
. (0,01)
(8,01) 830,01
12
(8,01) 0,01
12+ 2
49
(8,01) 24,01
12
) (0, +) () > 0 (, 0) () < 0 . () = (2). . () . () = 2. (2) (2) > 0 . , () () , 2. () > 0 > 0 . , (0, +) () < 0 < 0. , (, 0).
4. () = 4
462+ , ,
(1, 1), .
() = (3 12 + ). 4
462+
, ; () = 0 () . (), , () . , () = 0.
: 4
462+ > 0 , .
(1) = (11 + ). 146
(1) = (11 + ). 146+
> 11, , (1) > 0 (1) > 0. , (1, 1); < 11, , (1) < 0 (1) < 0. , (1,1); 11 < < 11, , (1) < 0 < (1) , (1, 1), (1,1) () = 0. , ().
() (1,1) 11 < < 11. . (1,1) () = 0. () = 3 12 + , :
50
= 2 4(2 12) = 32 + 48. (1) = 0 = 4 = 16 ; 11 < < 11. , (1,1) = 0. (2) > 0 , > 4 < 4 , , > 16 < 16. . (3) < 0, () () = 0 , , = .
, , (1, 1) , 11 < < 11.
5.
) () =
( 2)2 ;
(1) ; () = {2} (2) ; (0) = 0 () = 0 = 0. (0, 0). (3); : = :
lim+
() = lim
() =
= 2 :
lim2+
() = lim2+
( 2)2= lim2+
2
( 2)2 0+
=
: 2+, > 2 , , 2 > 0 2 0+.
< 0,
0+ .
, = 2 (). : = :
lim+
() = lim
() =
lim+
() = lim+
( 2)2= lim
+
2 4 + 4= lim+
2 4 + 4=
lim+
2
2
242+42
= lim+
1
0
1 40
+420
=0
1 0 + 0=0
1= 0.
51
, = 0 (). (4) ;
() =( 2)2 + . 2( 2)
( 2)4
() = + 2 + 2
( 2)3
() = + 2
( 2)3
() ():
(2) + + + + + ++++++ +++ ( + 2) (2) + + + + + + ( 2)3
++++++ (2) (2) + + + + + () = ( + 2) ( 2)
, :
() (,2) (2,+) () (2, 2) (5) ;
() =( 2)3 3( + 2)( 2)2
( 2)6
() = 2 3 6
( 2)4
() =2 8
( 2)4
():
++++++ (4) (2 8) +++++++++ +++++ (2) + ++ + + + ( 2)4
++++++ (4) (2) () = (2 8) ( 2)4
, :
() (,4) () (4,2) (2,+).
, .
(4,1
9) !
52
) !
53
3.4 4 Prova-14 de Junho de 2008
1.
) lim+
[
+ 1]
: = + 1 , : + , +.
lim+
[
+ 1]
= lim+
[ 1
]1
= lim+
[1 1
]1
= lim+
[(1 1
)
(1 1
)1
] =
= ; +,
lim
[(1 +1
)
(1 +1
)1
] = lim
[(1 +1
)
]
1
lim
(1 +1
)1
= 1 1 =1
lim
(1 +1
)
= .
) lim1[
11
ln ] = lim
1[. ln + 1
( 1). ln ] = lim
1[(ln 1) + 1
( 1). ln ] ;
:
lim1[ln 1 + 1
ln + 1 1
] = lim1[
ln
ln + 1 1
] = lim1[
1
1 +
12
] = lim1[
1
+ 12
] = lim1[
+ 1]
=1
1 + 1=1
2.
2. () = 32 = 3 . . () = 32 () = 3 + , . = 3 , . , () = 3; 32 = 3 = 1 = 1, = 3. , (1, 3) (1) = 13 + = 1 + = 3 = 2. () = 3 + 2 = 1, = 3. , (1,3) (1) = (1)3 + = 1 + = 3 = 2 () = 3 2.
3. 2 + 2 = 25 ; = (2, 0) = (2, 0) : ):
(, ) , : (, 25 2)
, :
= ( + 2)2 + 25 2 = 4 + 29
54
= ( 2)2 + 25 2 = 4 + 29
= 4 + 29 + 4 + 29
() =1
24 + 29. 4 +
1
24 + 29. (4)
() =2
4 + 29
2
4 + 29
() = 0, : 2
4 + 29
2
4 + 29= 0 4 + 29 = 4 + 29
4 + 29 = 4 + 29 4 + 29
4 + 29= 1 4 + 29 = 4 + 29 = 0.
= 0, = 5. , = (0,5) = (0,5) .
= 4.0 + 29 + 4.0 + 29 = 29 + 29 = 229 ! 5 5. () = (). , ! 1: (5,5) . (5) = (5).
(5) = 4.5 + 29 + 4.5 + 29 = 49 + 9 = 7 + 3 = 10 < 229
, = 229 ! ): , , , 3 , . , = 5. = 5, = 0 , , = 5 = 0. (5) = 10 (5) = 10.
4. ) () = 2 ; ; = 1 = 5.
() < 0 < 2, = 1 = 2:
55
=1
2(2 1). ((2) (1)) =
1
2. (1). (0 (1)) =
1
2.
= 2 = 5:
=1
2(5 2)((5) (2)) =
1
2. (3). (3 0) =
9
2.
, = 1 = 5 :
15 = 1
2+9
2=10
2= 5 .
) ( 2) 5
1
= [1
22 2|
1
5
] =1
2(5)2 2. (5)
1
2(1)2 + 2. (1) =
25
2 10
1
2+ 2
=24
2 8 = 12 8 = 4.
: , !
) ( 2) 2
0
;
=2 0
=2
; =
2
; () = 2
( 2) 2
0
= lim
()
=1
= lim
()
=1
= lim
2
(
2
2)
=1
= lim
2
[
2
=1
2
=1
]
= lim
2
[2
=1
2]
= lim
2
[2
.( + 1)
2 2]
= lim
(2( + 1)
4)
= lim
(2
+2
4)
= lim
(2
2)
= 2.
5.
() =3
1 2;
() =2(3 2)
(1 2)2;
56
() =2(2 + 3)
(1 2)3
() : () = {1, 1}; () : (0) = 0 ; () = 0 3 = 0 = 0. (0, 0). () : : = :
lim+
() = lim
() =
= 1 = 1 :
lim1+
() = lim1+
3
1 2= lim1+
3
1
(1 + ) 0+
(1 ) 2
=
, = 1 ().
lim1+
() = lim1+
3
1 2= lim1+
3
1
(1 + ) 2
(1 ) 0
=
, = 1 ().
: = :
lim+
() = lim
() =
lim+
() = lim+
3
1 2= lim
+
32
2= lim+
3
2= .
lim
() = lim
3
1 2= lim
32
2= lim+
3
2= +.
, ().
: = +
lim
[() ( + )] = 0
() =3
1 2= +
1 2
() () =
1 2
lim
[() ()] = lim
1 2= lim
1
2= 0.
, = (). () :
57
() =2(3 2)
(1 2)2
(), : +++++++++ +++++ 0 + ++ + ++++++++ 2
(3) + + + + ++++++++ + (3) (3 2)
+++++++++ + (1) + + + + (1) + + + ++ +++ (1 2)2
(3) + + + (1) + +0 + +(1) + +(3) () = 2(32)
(12)2
, :
() (3,1) (1,1) (1, 3)
() (,3) (3,+).
(() = 0 () )
() = 1 = 1. , = 1 = 1 , , !
() = 0 = 3 = 0.
(3) =(3)
3
1 (3)2 =
33
2=
33
2. (3,
33
2)
(3) =(3)
3
1 (3)2 =
33
2=33
2. (3,
33
2)
() :
() =2(2 + 3)
(1 2)3
() (), : 0 + + + ++ ++++++++ 2 +++++++++ ++++++++++ ++++++++ (2 + 3) (1) + + + + + +++++(1) (1 2)3
++++++ (1) 0 + + + + + (1) () =2(2+3)
(12)3
, :
() (,1) (0, 1) () (1, 0) (1,+)
. , = 0 . . (0,0)
58
:
59
3.5 Reavaliao da 2 mdia-21 de Junho de 2008
1.
) lim
() = lim
() = lim
() = lim
. ()
()= 1.
lim
. ()
()= lim
() + . ()
()= lim
() + . ()
()=
lim
() + () + . ()
()= lim
2() + . ()
()=
lim
2()
()+ lim
. ()
()= 2 + 1 = 3.
) lim
( +
)
= 4 . .
= : : , .
lim
( +
)
= lim
( + 2
)+
= lim
(1 +2
)+
;
= 2 :
lim
(1 +2
)+
= lim
(1 +1
)2+
= lim
(1 +1
)2
(1 +1
)
=
lim
(1 +1
)2
lim
(1 +1
)
:
lim
(1 +1
)2
= lim
[(1 +1
)
]
2
= 2.
lim
(1 +1
)
= 1.
lim
( +
)
= 2 = 4 2 = ln 4 =1
2ln 4 = ln 41 2 = ln 2
: .
2.
() = 2 ; [0, 1].
() = (2 1)2
: 2 > 0, . , ()
(2 1).
60
() > 0 >1
2 ; () = 0 =
1
2 () < 0 0 , . : 164 0 3 > 0 0. , > 0 164 + 3 + 2 > 0 164 + 3 < 0 < 0 ?
1)164 + 3 = 0 = 0 = 3
16
3
() = 164 + 3 = ( + 3
16
3
)(162 163
16
3
+ 9
256
3
)
2 , , . , .
64
, ! () , 2, 164 + 3 + 2 > 0 . () = 643 + 3
() = 0 643 + 3 = 0 = 33
4
(33
4) = 16. (
33
4)
4
+ 3(33
4) =
933
16 ; 2.
, () = . : ().
: = :
lim+
() = lim
() =
lim+
() = lim+
9
164 + 3 + 24 = lim+
9||
164 + 3 + 24
||
;
: || = 2 = 44
. +, || = .
lim+
9||
164 + 3 + 24
||
= lim+
9
164 + 3 + 24
44
= lim+
1 9
16 +33+24
4
=
1 0
16 + 0 + 04 =
1
164 =
1
2.
, =1
2 ().
lim
() = lim
9
164 + 3 + 24
= lim
9||
164 + 3 + 24
||
;
: || = 2 = 44
. , || = .
lim
9||
164 + 3 + 24
||
= lim
9
164 + 3 + 24
44
= lim
1 +9
16 +33+24
4
=
1 + 0
16 + 0 + 04 =
1
164 =
1
2.
, = 1
2 ().
: () , () = 0.
65
) =1
2 3 + 2 ; = () =
1
( 1)( 2)
: = :
lim+
() = lim
() =
= 1 = 2 (), .
lim1+
() = lim1+
1
( 1)( 2)= lim1+
1
( 1) 0+
( 2) 1
=
: 1+ , > 1. , 1 > 0.
lim1
() = lim1
1
( 1)( 2)= lim1
1
( 1) 0
( 2) 1
= +
: 1 , < 1. , 1 < 0.
, = 1 .
lim2+
() = lim2+
1
( 1)( 2)= lim1+
1
( 1) 1
( 2) 0+
= +
: 2+ , > 2. , 2 > 0.
lim2
() = lim2
1
( 1)( 2)= lim2
1
( 1) 1
( 2) 0
=
: 2 , < 2. , 2 < 0.
, = 2 . , , = 1 = 2 .
4.
) lim2
| 2|
3 8 ; : | 2| = {
2, 2( 2), < 2
.
lim2+
| 2|
3 8= lim2+
2
3 8= lim2+
( 2)
( 2)(2 + 2 + 4)= lim2+
1
(2 + 2 + 4)=1
12
: 2+, > 2. , | 2| = 2
lim2+
| 2|
3 8= lim2+
2
3 8= lim2+
( 2)
( 2)(2 + 2 + 4)= lim2+
1
(2 + 2 + 4)=
1
12
: 2, < 2. , | 2| = ( 2).
lim2+
| 2|
3 8 lim2
| 2|
3 8, lim
2
| 2|
3 8 .
66
) lim0+
(3 2
2) = 0
1
2 1
21 22 21
1
2 2
2 2
3
2
3 2
2 2
3
() =3
2 , () =
3 2
2 () = 2
3 . , :
() () () , lim
0() = 0 lim
0() = 0, , lim
0() = lim
0() ,
() () () lim0() = lim
0() ,
lim0() = lim
0() = lim
0() = 0. ,
lim0+
(3 2
2) = 0
5.
=1
. = . > 0.
: = (,1
).
= 1
2 . () =
1
2.
= ( )
1
=
1
2( )
= 1
2 +
2
:
= (0,2
) = (2, 0)
, :
=1
2( ) ( ) =
1
2(2) (
2
) = 2 .
67
3.7 VPA 1-13 de Setembro de 2008
1.
() = {
2 + 1 , 1
2
2+ 4, 1 < 4
2 , > 4
, , , . , (, 1) (1,4) (4,+). , , = 1 = 4. = , ,
1) () ; 2) lim
() ;
3) lim
() = ()
= 1: 1) (1) = 1 + 1 = 1 + 1 = 2.
2) lim1+
() = lim1+
(2
2+ 4) = lim
1+2
2+ lim1+
4 = 1
2+ 4 =
7
2.
lim1
() = lim1
(2 + 1) = lim1
2 + lim1
1 = 1 + 1 = 2.
lim1+
() lim1
() lim1() .
, = 1.
= 4:
1) (4) = 42
2+ 4 = 8 + 4 = 4.
2) lim4+
() = lim4+
2 = 2.
lim4
() = lim4
(2
2+ 4) = lim
4
2
2+ lim4
4 = 8 + 4 = 4.
lim4+
() lim4
() lim4() .
, = 4.
= 1 = 4 , , . ) ():
68
2.
) () =5 3
2 ; , , ().
() = lim0
( + ) ()
= lim0
5 3( + )( + ) 2
5 3 2
=
lim0
( 2)[5 3( + )] [( + ) 2](5 3)( 2)[( + ) 2]
=
lim0
5 32 3 10 + 6 + 6 [5 + 5 10 32 3 + 6]
( 2)[( + ) 2]=
lim0
5 32 3 10 + 6 + 6 5 5 + 10 + 32 + 3 6
( 2)[( + ) 2]=
lim0
( 2)[( + ) 2]= lim0
1
( 2)[( + ) 2]=
1
( 2)2
() =1
( 2)2.
) (3, 4)
(3) =1
(3 2)2=1
12= 1.
, : = ( ) (4) = 1( 3) + 4 = 3 = 7
3.
) () =
2 + 1 ; .
.
69
2 + 1 > 0 2 + 1 > 0 ; 2 > 1 ; : 2 > 0 . + 1 > 0 . , () = . , , , !
: = :
lim+
() = lim
() =
lim+
() = lim+
2 + 1= lim+
||
2 + 1||
= lim+
2 + 1
2
= lim+
1
1 +12
=
1
1 + 0=1
1= 1. : +, || = . || = 2.
, = 1 ().
lim
() = lim
2 + 1= lim
||
2 + 1||
= lim
2 + 1
2
= lim+
1
1 +12
=
1
1 + 0=1
1= 1. : +, || = . || = 2.
, = 1 ().
) lim
(2 + 2 ) = lim
(2 + 2 ) (2 + + 2 )
(2 + + 2 )=
lim
2 + 2 +
2 + + 2 = lim
2
2 + + 2 ;
lim+
2
2 + + 2 = lim+
2||
2 + ||
+2 ||
= lim+
2
2 +
2+2
2
=
lim+
2
1 +1 +
1 1
=2
1 + 0 + 1 0=
2
1 + 1=2
2= 1.
lim
2
2 + + 2 = lim
2||
2 + ||
+2 ||
= lim
2
2 +
2+2
2
=
lim
2
1 +1 +
1 1
=2
1 + 0 + 1 0=2
1 + 1=2
2= 1.
, lim
(2 + 2 ) .
70
4.
) lim
2 2
| |= lim
( )( + )
| | ; : | | = {
, ( ), <
.
lim+
( )( + )
| |= lim+
( )( + )
( )= lim+
( + ) = 2.
lim
( )( + )
| |= lim
( )( + )
( )= lim+
( + ) = 2.
, , lim
2 2
| | .
) lim2
6 2
3 1= lim2
6 2
3 1(6 + 2)
(6 + 2)(3 + 1)
(3 + 1)=
lim2
( 2)(3 + 1)
( 2)(6 + 2)= lim
2
(3 + 1)
(6 + 2)=3 2 + 1
6 2 + 2=1 + 1
4 + 2=2
4=1
2.
5. ) |() + 4| < 2( 3)4 ; lim
3().
: 2( 3)4 < () + 4 < 2( 3)4
2( 3)4 4 < () < 2( 3)4 4 () = 2( 3)4 4 () = 2( 3)4 4, :
() < () < () , lim
3() = 0 lim
3() = 0, , lim
3() = lim
3() ,
() < () < () lim0() = lim
0() ,
lim3() = lim
3() = lim
3() = 4. ,
lim3() = 4
) () =1
2 + 5 + 6 ; () =
1
( + 2)( + 3) .
() = {2,3}. , , , .
= :
lim+
() = lim
() =
lim2+
() = lim2+
1
( + 2)( + 3)= lim2+
1
( + 2) 0+
( + 3) 1
= +
: 2+, > 2 + 2 > 0.
lim2
() = lim2
1
( + 2)( + 3)= lim2+
1
( + 2) 0
( + 3) 1
=
71
: 2, < 2 + 2 < 0.
, = 2 ().
lim3+
() = lim3+
1
( + 2)( + 3)= lim3+
1
( + 2) 1
( + 3) 0+
=
: 3+, > 3 + 3 > 0.
lim3
() = lim3
1
( + 2)( + 3)= lim3+
1
( + 2) 1
( + 3) 0
= +
: 3, < 3 + 3 < 0.
, = 3 ().
72
3.8 2 Prova-03 de Outubro de 2008
1. ) () = ln|cos sec(3) . cotg(3)| . ().
() =[cos sec(3) . cotg(3)]
cos sec(3) . cotg(3)
() =3 cos sec(3) . cotg2(3) 3 cos sec3(3)
cos sec(3) . cotg(3)
() =3cos sec(3)[cotg2(3) + cos sec(3)]
cos sec(3) . cotg(3) ; 1 + cotg2() = cos sec()
() =3[2 cos sec2(3) 1]
cotg(3) .
) sen + cos 2 = sen ; (0, 4 ). , : sen + . cos + 2. ( sen 2) = cos . (. cos 2 cos 2) = cos sen
=cos sen
. cos 2 sen 2
, :
=cos 0 sen
4
2. sen2
=1
22
2=2 2
4.
:
4=2 2
4( 0)
=2 2
4 +
4 =
(2 2) +
4
2. () = + 1(2 )5
( + 3)4 ; (0) .
ln () = ln + 1(2 )5
( + 3)4
ln () = ln + 1 + ln(2 )5 ln( + 3)4
ln () =1
2ln( + 1) + 5 ln(2 ) 4 ln( + 3)
()
()=1
2
1
( + 1)+ 5
(1)
(2 ) 4
1
( + 3)
() = () [1
2( + 1)
5
(2 )
4
( + 3)]
() = + 1(2 )5
( + 3)4[
1
2( + 1)
5
(2 )
4
( + 3)]
(0) =0 + 1(2 0)5
(0 + 3)4[
1
2(0 + 1)
5
(2 0)
4
(0 + 3)] =
1(2)5
34[1
25
24
3] =
32
81[2
4
3] =
32
81 (
10
3) =
320
243.
73
3. ) = log3(log2 ) .
(2). = log2 , = log3 . ,
=
=
1
. ln 21
. ln 3
= () =1
(. log2 ) ln(2). ln(3) ; log2 =
ln
ln 2 ( !)
() =1
. ln . ln 3 () =
1
ln . ln 3 .
(2) =1
ln 4 . ln 3.
) = ||2
. .
ln = 2. ln||
= (2).
2. ln|| +
2.1
= . 2[2. ln|| +
1
]
=
= ||
2
. 2[2. ln|| +
1
].
4. ) = senh() ; = 1.
= cosh() = +
2 .
= 1 +
2= 1
+ = 2 + 2 = 0
+1
2 = 0
2 2 + 1 = 0 ; = . : > 0, > 0 . 2 2 + 1 = 0 ( 1)2 = 0 = 1 ( ). = 1 = ; = ln 1 = 0.
, (0, senh(0)) = (0, 0).
) () = 2arctg 2. (1).
= 2 = arctg . , () = 2. , :
=
() = (2) 1
1 + 2 2 ln(2)
74
() =2. 2arctg
2. . ln(2)
1 + 4
(1) =2. 2arctg(1). 1. ln(2)
1 + 14=2. 2arctg(1). 1. ln(2)
1 + 1= 2.2
4 . ln(2)
2= 2
4 . ln(2)
5.
) () = sen + cos + tg ; (0) = 2, () = 0 (
6) = 4.
(0) = . sen(0) + . cos(0) + . tg(0) (0) = ; (0) = 2 = 2.
() = . cos 2. sen + . sec2 () = . cos 2. sen + . sec2 () = + ; () = 0 + = 0 ()
() = . sen 2. cos + (2. sec2 . tg )
(
6) = . sen
6 2. cos
6+ (2. sec2
6. tg
6)
(
6) =
2 3 + (2.
4
3.3
3)
(
6) =
2 3 +
83
9 ; (
6) = 4 =
2 3 +
83
9 = 4 ()
() = . (), :
2 3 +
83
9 = 4 (
1
2+83
9) = 4 + 3 (
9 + 163
18) = 4 + 3
=18(4 + 3)
163 9 ;
, = =18(4 + 3)
163 9 = 2.
) lim1
sen( 1)
2 + 2= lim1
sen( 1)
( 1)( + 2);
= 1 , : : 1, 0.
lim1
sen( 1)
( 1)( + 2)= lim0
sen()
( + 3)= lim0
sen()
.
1
( + 3)= lim0
sen()
lim0
1
( + 3)
= 1 1
3=1
3.
, lim1
sen( 1)
2 + 2=1
3.
75
3.9 2 Avaliao-04 de Outubro de 2008
1.
=cos2 tg4
(2 + 1)3 ; .
ln = lncos2 tg4
(2 + 1)3
ln = ln(cos2 ) + ln(tg4 ) ln(2 + 1)3 ln = 2. ln(cos ) + 4. ln(tg ) 3. ln(2 + 1)
= 2.
sen
cos + 4.
sec2
tg 3
2. 2
(2 + 1)
= [2. tg + 4(cotg + tg ) 6. 2
(2 + 1)]
= [2. tg + 4(cotg + tg ) 6. 2
(2 + 1)]
= [2. tg + 4. cotg 6. 2
(2 + 1)]
= 2 [tg + 2. cotg 3. 2
(2 + 1)]
= 2 cos2 tg4
(2 + 1)3 [tg + 2. cotg
3. 2
(2 + 1)].
2. 22 + = 2 ; = 0 ? , : 22 + 22 + + = 0 (22 + ) = (22 + )
= 22 +
22 + =
(2 + 1)
(2 + 1) =
; = 0 = 0.
= 0 : 2. 02 + . 0 = 2 0 = 2 ( !), 0 2.
, = 0 , , = 5.
3.
) = 3cos(2) ; = 1 0
2.
= 1 3cos(2) = 1 3cos(2) = 30 cos(2) = 0.
2 =
2 2 2 =
3
2 2
=
4 =
3
4 .
0
2, , 1 , =
4.
= 2 = cos , = 3 .
76
, :
=
= 2. ( sen ). 3. ln(3)
= 2. 3cos(2). sen(2) . ln(3)
(
4) = 2. 3cos(
2). sen (
2) . ln(3)
(
4) = 2. 30. 1. ln(3)
(
4) = 2. ln(3)
(
4, 1):
= ( )
1 = 2. ln(3) (
4)
= 2. ln(3) +
2ln(3) + 1
) = (ln )sen(2) ;
.
ln = sen(2) . ln(ln )
= 2. cos(2). ln(ln ) + sen(2) .
1
.1
ln
= [2. cos(2). ln(ln ) +sen(2)
ln ]
= (ln )sen(2) [2. cos(2). ln(ln ) +sen(2)
ln ].
4.
) = arcsen (1 2) ;
, 0 < < 1.
= 1 2 = . = arcsen . , :
=
= (2) 1
2
1
1 2
=
1 2 1 1 + 2
=
1 2 2 ; : 2 = ||
=
||1 2
) () =log
2 ; (1).
() =[log ].
2 [2]. log
(2)2
77
() =
1 ln
2 (2). log
4
() =1 2 log
3. ln (1) =
1 2 log 1
13. ln =1 2.0
ln =1
ln .
5.
) = tgh() ; (0,0).
( ) . ,
=1
= []. tgh() + . [tgh()]
= tgh() + . (1
2) . sech2()
(0, 0), :
(0) = tgh(0) + . (1
20) . sech2() ;
= 0.
, = tgh() + . (1
2) . sech2()
> 0. , +(0) .
lim0+
= lim0+
[tgh() + . (1
2) . sech2()] = lim
0+tgh() + lim
0+
. sech2()
2
! .
lim0+
. sech2()
2; 0+ > 0. ,
= .
lim0+
. sech2()
2= lim0+
. sech2()
2=0.1
2= 0.
, +(0) = 0. ,
(0) . (0) .
, (0, 0) , , !
:
+(0), , ,
= 0, , .
) () = arccos ( 3
3) ; (
1
2).
() = (1 2) 1
1 ( 3
3 )2
78
() = (1 2)
1 ( 3
3 )2
= (1 2)
1 2 +24
3 6
9
= 3(1 2)
9 92 + 64 6.
(1
2) =
(1 1 4 )
1 121 576=
24. (3 4 )
455=
18
455.
ln () = ln3(1 2)
9 92 + 64 6
ln () = ln(3 + 32) ln9 92 + 64 6
ln () = ln(3 + 32) 1
2ln(9 92 + 64 6)
()
()=
6
3 + 321
2(65 + 243 18)
(9 92 + 64 6)
() = () [2
(1 2)(65 + 243 18)
2(9 92 + 64 6)]
(1
2) = (
1
2) . [2 (
12)
(1 14)(6.
132 + 24.
18 18.
12)
2 (9 9.14 + 6.
116
164)]
(1
2) =
18
455[4
3
(316 + 3 9)
2 (9 94 +
38
164)]
(1
2) =
18
455[4
3(9916)
(45532 )
]
(1
2) =
18
455[4
3+198
455]
(1
2) =
24
455
3564
455455.
79
3.10 3 Prova-01 de Novembro de 2008
1.
) () =1
+ 13 = 7.
() = () + ()( )
(7) =1
7 + 13 =
1
83 =
1
2.
() =
13
1
( + 1)23
( + 13
)2 =
1
3( + 1) + 13 ;
(7) = 1
3(7 + 1)7 + 13 =
1
3. (8). 83 =
1
48.
:
() =1
21
48( 7)
() = 1
48 +
31
48
) ln(1,08) .
:
() = ln () =1
1,08, ln 1 = 0. , (1 + 0,08). :
= (). = ( + ) () , :
( + ) () (). (1 + 0,08) = 1 = 0,08, :
(1 + 0,08) (1) 1
1. (0,08)
(1,08) 0 0,08 (1,08) 0,08
2. ) () = 2 sen + cos(2) . . () = 0 () .
() = 2 cos 2 sen(2) () = 2 cos 4 sen . cos () = 2 cos (1 2 sen )
() = 0 {2 cos = 0
1 2 sen = 0
80
2 cos = 0 ; =
2 =
2 . [, ].
1 2 sen = 0 sen =1
2 ; =
6 =
5
6.
, :
2,
6,
2 5
6.
) , [, ]. 1) (, ); 2) ; 3) 1 2 , .
1) (
2) = 2 sen (
2) + cos () = 2 1 = 3.
(
6) = 2 sen
6+ cos
3= 1 +
1
2=3
2.
(
2) = 2 sen
2+ cos = 2 1 = 1.
(5
6) = 2 sen
5
6+ cos
5
3= 1 +
1
2=3
2.
2) () = 2sen + cos 2 = 0 + 1 = 1. () = 2 sen() + cos(2) = 0 + 1 = 1.
3) , , (
6) = (
5
6) =
3
2
(
2) = 3.
3.
) () = + 3
2 ; [0,5] () =
(5) (0)
5 0.
() =
83 +
32
5=25
30=5
6.
() = 2 3
( 2)2=
5
( 2)2
:
5
( 2)2=5
6 ( 2)2 = 6 ( 2) = 6 .
, . ? : 1) [, ]; 2) (, );
(, ) () =() ()
.
1) () = + 3
2 [0, 5], , = 2
81
. , = 2.
2) () = + 3
2 (0,5), , () =
5
( 2)2
2, 2 (0,5).
, [0,5] () =(5) (0)
5 0
!
) () = + 9 + 33 8 . (0) = 8 (1) = 35. () = 0. , , [0, 1] , (0) < 0 < (1), , (0, 1) () = 0. , (0,1). 2 , , () = () = 0, . , , (, ) () = 0. () = 32 + 18 + 33 () = 0 32 + 18 + 33 = 0 2 + 6 + 11 = 0 = 36 44 = 8. < 0, () . , () .
4.
= 1 + ;
= 4/ = 3
= 1 = 2 13
.
= 3 1 + 3 = 3 1 + 3 = 9 3 = 8 = 2.
=
=1
3(2)
1
(2 1)23
4
=
8
3(2 1)23
=
8.3
3(32 1)23
=
8
83
=8
4= 2/
, , :
() = 1 + ()3 () =1
2(3()2. ()).
1
1 + ()3
82
4 =1
2(3. (2)2. ()).
1
1 + 23
4 =1
2(12()).
1
3
24 = 12. () () = 2/
5. :
= 42 =4
33
= 42/
= 0,1/
.
=
4 = 8. (0,1)
=5
=
= 42. (0,1)
= 4.
25
2. (0,1)
=10
/
83
3.11 Reposio da 1 Mdia-13 de Dezembro de 2008
1. ) = cosh ; = . = cosh = cosh = 1 = 0. : (0, ). = senh . cosh . : . , ,
= 1
=
1
senh . cosh
= 0, = 1
0 ? ? ?
. , . =
= ( )
= 1
0( 0)
( )0 = = 0.
) =32 + 2
4 3 ; .
= ().
() = {3
4}.
: : = :
lim+
() = lim
() =
. , = 3 4 :
lim34
+() = lim
34
+
32 + 2
4 3= lim34
+
32 + 2
59 4
4 3 0+
= +
: 3 4+, > 3 4 3 4 > 0.
lim34
() = lim34
32 + 2
4 3= lim34
32 + 2
59 4
4 3 0
=
: 3 4, < 3 4 3 4 < 0.
, =3
4 .
: =
84
: lim+
() = lim
() =
lim+
() = lim+
32 + 2
4 3= lim+
32 + 2||
4 3||
= lim+
32 + 2
2
4 3
= lim+
3 +22
4 3
=
3 + 0
4 0=3
4.
: + || = ; || = 2.
, =3
4 ().
lim
() = lim
32 + 2
4 3= lim
32 + 2||
4 3||
= lim
32 + 2
2
4 3
= lim
3 +22
4 +3
=
3 + 0
4 + 0=
3
4.
: || = ; || = 2.
, = 3
4 ().
2.
) () = . || ; () = {2, 0
2, < 0.
, : + > 0, :
() = lim0
( + ) ()
= lim0
( + )2 2
= lim0
2 + 2 + 2 2
= lim0
2 + 2
= lim0
(2 + )
= lim0
(2 + ) = 2.
+ < 0, :
() = lim0
( + ) ()
= lim0
( + )2 + 2
= lim0
2 2 2 + 2
= lim0
2 2
= lim0
(2 )
= lim0
(2 ) = 2.
, ():
() = {2, 02, < 0
() , , , (, 0) (0,+). = 0. ,
+(0) = 2.0 = 0
(0) = 2. (0) = 0
+(0) =
(0) = 0 , ,
.
85
) lim+
(2 + 25 ) = lim+
(2 + 25 )(2 + 25 + )
(2 + 25 + )=
lim+
2 + 25 2
(2 + 25 + )= lim+
25
2 + 25 + = lim+
25||
2 + 25 + ||
=
lim+
25
2 + 25
2+
= lim+
25
1 +25 + 1
=25
1 + 0 + 1=25
1 + 1=25
2.
: + || = ; || = 2.
3.
) () = { + , < 2
22 1, 2.
= , , :
1) () ; 2) lim
() ;
3) lim
() = ().
= 2, : 1) (2) = 2. (2)2 1 = 8 1 = 7. 2) lim
2() = lim
2( + ) = 2 + .
: = 2, lim2+
() = (2).
3) lim2
() = (2) 2 + = 7 ()
= 2. ) = 2:
+(2) = lim
2+
() (2)
2= lim2+
22 1 7
2= lim2+
22 8
2= lim2+
2(2 4)
2=
lim2+
2( 2)( + 2)
( 2)= lim2+
2( + 2) = 2(2 + 2) = 8.
(2) = lim
2
() (2)
2= lim2
+ 7
2= lim2
+ 7 2 7
2=
lim2
( 2)
( 2)= lim2
= .
= 2 +(2) =
(2). ,
= 8. () : 2. + = 7 16 + = 7 = 9. : = 8 = 9.
4. ) + = 6 ; (3,3). , : 32 + 32 = 6 + 6
86
2 + 2 = 2 + 2 . (2 2) = 2 2
=2
2; (3,3) =
2.3 3
3 2.3=6 9
9 6=3
3= 1.
: = ( ) 3 = 1( 3) 3 = + 3 = + 6
) ? ( = 0? ) = 0 2 2 = 0 2 2 0
2 = 2 =2
2.
, :
3 + (2
2)
3
= 6 (2
2)
3 +6
8= 33 23 =
6
8 6 163 = 0 3(3 16) = 0
,
{ 3 = 0
3 16 = 0 = 0 = 16
3 .
= 0 : 03 + 3 = 6.0. 3 = 0 = 0. (0, 0). : (0, 0) 2 2. , .
= 163
:
(163
)3+ 3 = 6. 16
3. 3 616
3 + 16 = 0
2 2. 2 = 2.
, = 2163
. , :
(2163
)
3
6163
(2163
) + 16 = 0
2163
(2163
) 6163
(2163
) + 16 = 0
2163
(2163
6163
) + 16 = 0
4163 216
3+ 16 0.
, 2 2. ,
= 163
.
=2
2 =
1623
2.
87
, (163
,1623
2).
5.
) = sen(tgsen ) ; .
= sen , = = tg , = sen , :
=
= (cos ).1
2. (sec2 ). (cos )
=cos . sec2(sen ) . cos(tg sen )
2sen
)
() = 232
(1). = 2, = 3 . () = 2. , :
=
() = (2) 3 ln(3) 2 ln(2)
() = (2) 32 ln(3) 23
2
ln(2)
(1) = 2 31 ln(3) 231 ln(2)
(1) = 48 ln(3) ln(2)
88
3.12 Reavaliao da 2 mdia-13 de Dezembro de 2008
1.
, : (2)2 = 2 + 2, = . ;
= 2/
, : 2 + 2 = 100
()2 + ()2 = 100 2(). () + 2(). () = 0
() = (). ()
()
: () = 2/ () = (). ()
() = (). () + (). ()
() = 2() + 2()2
()
() = 2100 ()2 + 2()2
100 ()2
() = 6 :
= () = 2100 36 + 2
36
100 36= 264 +
72
64= 16 + 9 = 7/
2.
89
, , . , :
=
= =
( )
=1
32. =
1
32.
( )
=
(2 3)
3
: !
() =
3(2 32) ; () = 0 :
2 32 = 0
(2 3) = 0 =2
3.
=2
3 , :
=( )
= =
( 23)
=
3=
3.
, =2
3 =
1
3.
=1
3. 2. =
1
3 (42
9) . (
3) =
42
81 .
3.
) ( + 5)
= [1
22 + 5|
] =1
2 + 5
1
22 5 =
2 2
2+ 5( ).
) () = 3 2 + 1 . (0) (2): (0) = 03 02 + 0 1 = 1. (2) = 23 22 + 2 1 = 5. , (0) < 0 < (2), , () = 0 (0) (2).
90
[0, 2], 0 2 () = 0. , () = 2 + 1 (0, 2). 2 . (, ), , (, ) () = 0. () = 32 2 + 1 ; = 4 12 = 8. , () . , () = 0 , , () .
: 3 . = 1 (). , () = ( 1)(2 + 1). , , . = 1, () = 0.
4.
() =3
( 1)2 ; !
1) : () = {1}. 2) : (0, 0). 3) :
() =32( 1)2 3. 2. ( 1)
( 1)4=32( 1) 23
( 1)3=3 32
( 1)3=2( 3)
( 1)3;
() = 2( 3)
( 1)3
() (), : ++++ 0 + + + + ++ + + + + + +++ ++ 2 3 + + + + + ++ ( 3) 1 + + + + + ++++++ ++ ( 1)3 ++++ 0 + + 1 3 + + + + + ++ () = ( 3)/( 1)
, : (, 1) (3,+) (1, 3) 4) : () = 0 () . () = 0 = 0 = 3.
(0) = 0; (3) =33
(3 1)2=27
4. (0, 0) (3,
27
4).
() = 1. , = 1 , , . 5) :
() =(32 6)( 1)3 (3 32). 3. ( 1)2
( 1)6
() =(32 6)( 1) 33 + 92
( 1)4
91
() =6
( 1)4
() () :
0 + ++ + +++ ++++ 6 +++++++++ +++++++ 1 + + + +++ ( 1)4 0 + ++ + 1 + + + ++ + () = 6/( 1)4
, : (0, 1) (1,+) (, 0) 6) : () = 0 = 0. (0) = 0 . (0, 0). 7) : : = :
lim+
() = lim
() =
= 1. = 1 :
lim1+
() = lim1+
3
( 1)2= lim1+
3
1
( 1)2 0+
= +
: 1+ , > 1 1 > 0 ( 1)2 > 0
lim1
() = lim1
3
( 1)2= lim1+
3
1
( 1)2 0+
= +
: 1 , < 1 1 < 0 ( 1)2 > 0. , = 1 ().
: = :
lim+
() = lim
() =
lim+
() = lim+
3
( 1)2= lim