LIMITES I

Prévia do material em texto

Cálculo Diferencial e Integral 
Lista de Exercícios Bs04C3e04 
 
CDCI/CMCD 
(01) Verificar o limite a seguir considerado: 
( )( ) 





=




 −+
++−
−
→
3
44
1223
0u
z4,
2
1
u
)z)uz((
,u2u3.uu2u4lim . 
Verificação: 
Imediatamente, tem-se que: 
( )( ) =




 −+
++−
−
→ u
)z)uz((
,u2u3.uu2u4lim
44
1223
0u
 
( )( ) 




 −+
+++−=
→
−
→ u
)z)uz((lim.ju2u3.uu2u4lim.i
44
0u
1223
0u
rr
. 
Mas, ( )( ) 2
1
)2u3(
)1u2u4(lim
2u3.u
1u2u4.ulim)u2u3(
)uu2u4(lim
2
0u
2
0u2
23
0u
=
+
+−
=
+
+−
=
+
+−
→→→
 
e como )qp).(qp).(qp(qp,Rq,p 2244 ++−=−∈∀ , tem-se que: 
=
++++−+
=
−+
→→ u
)z)uz).((z)uz).((z)uz((lim
u
)z)uz((lim
22
0u
44
0u
 
=+++=
+++
=
→→
)z)uz)((uz2(lim
u
)z)uz)((uz2).(u(lim 22
0u
22
0u
 
32 z4z2.z2 == . 
Logo: ( )( ) =+=




 −+
++−
−
→
j.z4i).2/1(
u
)z)uz((
,u2u3.uu2u4lim 3
44
1223
0u
rr
 






=
3
z4,
2
1
. 
 
(02) Verificar o limite a seguir considerado: 






−=





−+
+−
+−+−
→ 4
3
,2j.)u5uu(
)2u3u(i).2u2u4u2(lim 23
3
23
1u
rr
. 
Verificação: 
Cálculo Diferencial e Integral 
Lista de Exercícios Bs04C3e04 
 
CDCI/CMCD 
=





−+
+−
+−+−
→
j.)u5uu(
)2u3u(i).2u2u4u2(lim 23
3
23
1u
rr
 
.)u5uu(
)2u3u(lim.j)2u2u4u2(lim.i 23
3
1u
23
1u
−+
+−
+−+−=
→→
rr
 
Mas, =−+−=−+−
→→
2u2u4u2lim)2u2u4u2(lim 23
1u
23
1u
 
=−++−+=
→→→→
))2(lim()u2lim())u4(lim()u2lim(
1u1u
2
1u
3
1u
 
=−++−+=
→→→→→→→
)2lim()ulim).(2lim()ulim)).(4(lim()ulim).(2lim(
1u1u1u
2
1u1u
3
1u1u
 
=−+−=
→→→
2)ulim.(2)ulim.(4)ulim.(2
1u
2
1u
3
1u
 
=−+−=
→→→
2)ulim.(2)ulim.(4)ulim.(2
1u
2
1u
3
1u
 
222422)1.(2)1.(4)1.(2 23 −=−+−=−+−= e 
( )( )
( )( ) =−+
−+
=
−+−
−+−
=
+−+
+−
→→→ )3u2u(
)2uu(lim
3u2u.1u
2uu.1ulim)3u5uu(
)2u3u(lim 2
2
1u2
2
1u23
3
1u
 
( )( )
( )( ) 4
3
)3u(
)2u(lim
3u.1u
2u.1ulim
1u1u
=
+
+
=
+−
+−
=
→→
. 
Logo: 





−=





−+
+−
+−+−
→ 4
3
,2j.)u5uu(
)2u3u(i).2u2u4u2(lim 23
3
23
1u
rr
. 
 
(03) Verificar o limite a seguir considerado: 
( ) ( )( ) ( ) ( )( ) =





−
−
−++− −−
→ )5u(
)44(
,u.asenausen,u1ln.1elim
5u
11u
0u
 
))2ln(.2048),acos(,1(= . 
Verificação: 
Como ( ) ( )( ) ( ) ( )( ) =





−
−
−++− −−
→ )5u(
)44(
,u.asenausen,u1ln.1elim
5u
11u
0u
 
Cálculo Diferencial e Integral 
Lista de Exercícios Bs04C3e04 
 
CDCI/CMCD 
( ) ( )( ) ( ) ( )( ) )5u(
)44(lim.ku.asenausenlim.ju1ln.1elim.i
5u
0u
1
0u
1u
0u
−
−
+−+++−=
→
−
→
−
→
rrr
, 
então: 
 (a) ( ) ( )( )
( ) ( )
=
+






−
=
+
−
=+−
→→
−
→ u/1
)eln(
u
0u
u
0u
1u
0u u1ln
1
.
u
1elim
u1ln.
u
u
)1e(limu1ln.1elim
4434421
 
( ) ( ) 1)eln(
1
u1limln
1
u1ln
1lim
u/1
0u
u/10u
==
+
=
+
=
→
→
; 
(b) ( ) ( )( ) =−+=−+
→
−
→ u
))a(sen)au(sen(limu.asenausenlim
0u
1
0u
 
=
++−+
=
→ x
)
2
aua
cos().
2
aau(sen.2
lim
0u
 
=
+
=
→ u
)2/)ua2cos(().2/u(sen.2lim
0u
 
=+=
→→
)2/)ua2cos((lim.
u.
2
2
)2/u(senlim.2
0u0u
 
)acos()acos(.)2/u(
)2/u(senlim)2/a2cos(.)2/u(
)2/u(senlim.
2
2
1
0
2
u0u
===
=
→→
44 344 21
; e 
(c) =−=−=
−
−
→
+
→→ t
)14.(4lim
t
)44(lim)5u(
)44(lim
t5
0t
55t
0t
5u
0u
 
)2ln(.2048)2ln(.4.2)2ln(.4)4ln(.4
t
)14(lim.4 5255
t
0t
5
====
−
=
→
. 
 
(04) Verificar o limite a seguir considerado: 
( ) )1,1,1(k.))bt(sen)at(sen(
)ee(j.
t1ln
)t(seni.
t
)2)tcos(2(lim
btat
20t
−=





−
−
+
+
+
−
→
rrr
. 
Verificação: 
Cálculo Diferencial e Integral 
Lista de Exercícios Bs04C3e04 
 
CDCI/CMCD 
Como ( ) =




−
−
+
+
+
−
→
k.))bt(sen)at(sen(
)ee(j.
t1ln
)t(seni.
t
)2)tcos(2(lim
btat
20t
rrr
 
( ) =−
−
+
+
+
−
=
→→→
k.))bt(sen)at(sen(
)ee(limj.
t1ln
)t(senlimi.
t
)2)tcos(2(lim
btat
0t0t20t
rrr
 
( ) ))bt(sen)at(sen(
)ee(lim.k
t1ln
)t(senlim.j
t
)2)tcos(2(lim.i
btat
0t0t20t
−
−
+
+
+
−
=
→→→
rrr
, então 
(a) =−=−
→→ 20t20t t
)1)t(cos(lim.2
t
)2)tcos(2(lim 
=
+−
−−
=








+−
+−−−
=
→→ )1)t(sen1.(t
)1)t(sen1(lim.2
)1)t(sen1(
)1)t(sen1(
.
t
)1)t(sen1(
lim.2
22
2
0t2
2
2
2
0t
( ) =+−−=+−−= →→ )1)t(sen1( 1.t )t(senlim.21)t(sen1.t )t(senlim.2 2
2
0t22
2
0t
 
1)101(
1
.)1.(2 2 −=
+−
−= ; 
(b) ( ) ( ) ( )
=
+
=
+
=
+ →→→ t1ln.
t
1
.t
)t(senlim
t1ln.
t
t
)t(senlim
t1ln
)t(senlim
0t0t0t
 
( ) ( ) =+=+= →→→ t/10t0tt/10t t1ln
1lim.
t
)t(senlim
t1ln
1
.
t
)t(senlim 
( ) 1)eln(
1
t1limln
1lim
.
t
)t(senlim t/1
0x
0x
1
0t
==
+
=
→
→
=
→
43421
; 
(c) =





 +





 −
−
=
−
−
→→
2
btat
cos.
2
btat
sen.2
)ee(lim))bt(sen)at(sen(
)ee(lim
btat
0t
btat
0t
 
=





 +





 −
−
=
→→
2
btat
cos
1lim.
2
btat
sen2
)ee(lim
0t
btat
0t
 
Cálculo Diferencial e Integral 
Lista de Exercícios Bs04C3e04 
 
CDCI/CMCD 
=
−





 −
−
−
=





 −
−
=
→→
2
)btat(
1
.
2
btat
sen2
2
)btat(
1).ee(
lim
2
btat
sen2
)ee(lim
btat
0t
btat
0t
 
=
−





 −
−
−
=
−





 −
−
−
=
→
−
→
→
→
2
t).ba(
1
.
2
t).ba(
senlim
)btat(
)ee(
.2lim
.
2
1
2
)btat(
1
.
2
btat
senlim
)btat(
)ee(
.2lim
.
2
1
0
2
t).ba(
btat
0t
0t
btat
0t
 
=
−
−
=
−





 −
−
−
=
→
=
→
−
→
)btat(
)ee(
.2lim.
2
1
2
)btat(
2
btat
sen
lim
)btat(
)ee(
.2lim
.
2
1 btat
0t
1
0
2
)btat(
btat
0t
444 3444 21
 
=
−
−
=
−
−
=
→→ )btat(
)ee(lim)btat(
)ee(lim.2.
2
1 btat
0t
btat
0t
 
 
=
−
−
=
−
−
=
→→ t
)ee(lim.)ba(
1
t).ba(
)ee(lim
btat
0x
btat
0t
 
=





−
−
=





−
−
=
→→→ t
elim
t
elim.)ba(
1
t
e
t
elim.)ba(
1 bt
0t
at
0t
btat
0t
 
=





−
−
=
→→ t
)e(lim
t
)e(lim.)ba(
1 tb
0t
ta
0t
 
=




 +−
−
+−
−
=
→→ t
)11)e((lim
t
)11)e((lim.)ba(
1 tb
0t
ta
0t
 
=




 +−
−
+−
−
=
→→ t
)1)1)e(((lim
t
)1)1)e(((lim.)ba(
1 tb
0t
ta
0t
 
Cálculo Diferencial e Integral 
Lista de Exercícios Bs04C3e04 
 
CDCI/CMCD 
=





−
−
−+
−
−
=
→→→→ t
1lim
t
)1)e((lim
t
1lim
t
)1)e((lim.)ba(
1
0t
tb
0t0t
ta
0t
 
=












−
−
−
−
=
=
→
=
→
44 344 2144 344 21
)eln(
tb
0t
)elnta
0t
ba
t
)1)e((lim
t
)1)e((lim.)ba(
1
 
[ ] [ ] =−
−
=−
−
= )eln(.b)eln(.a.)ba(
1)eln()eln(.)ba(
1 ba
 
1)ba(
)ba()ba.()ba(
1
=
−
−
=−
−
= . 
Portanto, de fato, tem-se que: 
( ) )1,1,1(k.))bt(sen)at(sen(
)ee(j.
t1ln
)t(seni.
t
)2)tcos(2(lim
btat
20t
−=





−
−
+
+
+
−
→
rrr
. 
 
(05) Verificar o limite a seguir considerado: 
)0,k,1(
u5
))ucos(1.(3
,
u
)u.k(sen
,
u
)u(tglim
0u
=




 −
→
. 
Verificação: 
De imediato, tem-se que: 
=




 −
→ u5
))ucos(1.(3
,
u
)u.k(sen
,
u
)u(tglim
0u
 
u5
))ucos(1.(3lim.k
u
)u.k(senlim.j
u
)u(tglim.i
0u0u0u
−
++=
→→→
rrr
. 
Observe, entretanto, que: 
(a) 1)ucos(
1lim.1)ucos(
1lim.
u
)u(senlim)ucos(.u
)u(senlim
u
)u(tglim
0u0u0u0u0u
====
→→→→→
; 
(b) k1.k
u.k
)u.k(senlim.k
u
)u.k(sen
.
k
klim
u
)u.k(senlim
0ku0ku0u
====
→→→
; e, 
Cálculo Diferencial e Integral 
Lista de Exercícios Bs04C3e04 
 
CDCI/CMCD 
(c) ==−=−
→→→ u
)2/u(sen.2lim.
5
3
u
))ucos(1(lim.
5
3
u5
))ucos(1.(3lim
2
0u0u0u
 
00.
2
3)2/u(senlim.)2/u(
)2/u(senlim.
5
3
2
2
.u
)2/u(sen.2lim.
5
3
0)2/u(0)2/u(
2
0u
====
→→→
. 
Portanto, tem-se que: 
)0,k,1(k.0j.ki
u5
))ucos(1.(3lim.k
u
)u.k(senlim.j
u
)u(tglim.i
0u0u0u
=++=
−
++=
→→→
rrrrrr
. 
 
(06) Verificar o limite a seguir considerado: 
( ) )2/3),aln(),2(ln(k.
u4
2e2j).u(gcot.a1i).14.(8.u.4lim
u3
)u(tgu11
0u
−=













−
+−+−−−
→
rrr
. 
Verificação: 
Como , então: 
(a) )2ln()4ln(.
2
1
u
)14(lim.
8
4
u
)14(
.
8
4lim)14.(8.u.4lim
4ln(
u
0u
u
0u
u11
0u
==
−
=
−
=−
=
→→
−−
→
4434421
 
(b) ( ) =−−=−=−
→→→ )u(tg
)1a(lim)u(tg
)a1(lim)u(gcot.a1lim
)u(tg
0u
)u(tg
0u
)u(tg
0u
 
)aln(
y
)1a(lim
y
0y
−=
−
−=
→
; e, 
(c) 
2
3)eln(.3.
2
1)eln(.
2
1
u
1)e(lim.
4
2
u4
2e2lim 3
)eln(
3u
0u
u3
0u
3
===





−
=





−
=
→→
44 344 21
. 
Portanto, de fato, tem-se que: 
( ) =












−
+−+−−−
→
k.
u4
2e2j).u(gcot.a1i).14.(8.u.4lim
u3
)u(tgu11
0u
rrr
 
k.
2
3j)).a(ln(i)).2(ln()2/3),aln(),2(ln( rrr +−=−= . 
Cálculo Diferencial e Integral 
Lista de Exercícios Bs04C3e04 
 
CDCI/CMCD 
 
(07) Verificar o limite a seguir considerado: 
j)).35(ln(i.)7ln(.5
)5ln(.7)j)).35ln(t(i).15.()17((lim t71t5
0t
rrrr
+=++−− −
→
. 
Verificação: 
Imediatamente, tem-se que: 
=++−− −
→
)j)).35ln(t(i).15.()17((lim t71t5
0t
rr
 
=+
−
−
=++−−=
→→
−
→
j)).35(ln()17(
)15(lim.i))35ln(t(lim.j)15.()17(lim.i t5
t7
0t0t
t71t5
0t
rrrr
 
=+












−
−
=+












−
−
=
→
→
→
→ j)).35(ln(
t
)1)7((lim
t
)1)5((lim
.ij)).35(ln(
t
)17(lim
t
)15(lim
.i t5
0t
t7
0t
t5
0t
t7
0t rrrr
 
j)).35(ln(i.)7ln(.5
)5ln(.7 rr
+= . 
 
(08) Verificar o limite a seguir considerado: 
( )( ) ( )( )( )=+−+−−−+ −−
→
1532/1234
1b
1b2b.3b4b,2b4.b2b4blim 
j.
3
1i.
2
23)3/1,2/23(
rr
−=−= . 
 
(09) Verificar o limite a seguir considerado: 
( )
( ) ( ) ( ) ( ) ( ) =





+−+−−−−
−−+
−+− −−
−
−
−
→
153131
123
123
1a
1a2a.2.3a4a,a1.3a1,
4aa4a
3a4a3a2lim 
k.
3
2ji.
2
5)3/2,1,2/5(
rrr
−−=−−= . 
 
(10) Verificar o limite a seguir considerado: 
Cálculo Diferencial e Integral 
Lista de Exercícios Bs04C3e04 
 
CDCI/CMCD 
( ) ( )( )( )( )=−+−+−−−− −−−−
→
121131
1t
3t2t.5t1t),)t1(3)t1((lim 
2
ji)2/1,1(
r
r
+−=−= . 
 
(11) Verificar o limite a seguir considerado: 
( ) )2,2/1(j.)t2t3(
)tt2t4(i.)t2t3.(tt2t4lim 12
123
1223
0t
=








+
+−
+++−
−
−
−
→
rr
. 
 
(12) Verificar o limite a seguir considerado: 
( )( ) ( )( ) )8,e,5/2(k).t41ln(.t2j.t21i.1e.1elim 61t/31t5t2
0t
−−
−
→
=++−+−−
srr
. 
 
(13) Verificar o limite a seguir considerado: 
( ) )8/1,e,0(u).)u16(4(,)u(tg31,u.4lim 312/1)u(gcot2
0u
2
−=




−−+ −
→
. 
 
(14) Verificar o limite a seguir considerado: 
( )( ) )2/1,2/1(u/)u(sen)u(tg,u)uu1.(ulim 312/121
0u
=−−++ −−
→
. 
 
(15) Verificar o limite a seguir considerado: 
( )( ) ( )( ) ( )( )( ) )0,2,1()t(sen.)tcos(1,)t(sen)t(tgt,)t(tg.)t(senlim 111
0t
=−+ −−−
→
 
 
(16) Verificar o limite a seguir considerado: 
( ) ( ) ( )( ) )1,0,1())q(sen1)qcos(),2/q(tg.q1,)q(tg1ln).q(g(cotlim 1
0q
=−pi−+ −
→
. 
 
(17) Verificar o limite a seguir considerado: 
Cálculo Diferencial e Integral 
Lista de Exercícios Bs04C3e04 
 
CDCI/CMCD 
( ) ( ) )e,0,0()k.))a(tg31(i.asen.2/asen(lim 3)a(gcot22
0a
2
=++
→
rr
. 
 
(18) Verificar o limite a seguir considerado: 
( )( ) )5,2/1,0()k.)t4(sen.)t5(sen4j).1)t.(cos(t(lim 12
0t
−=+− −−
→
vr
. 
 
(19) Verificar o limite a seguir considerado: 
( ) ( ) 





=





+−+−
→
1,a,
3
2)v1ln(/1e,av1ln.v,
v3
)v(arcsen2lim v1
0v
. 
 
(20) Verificar o limite a seguir considerado: 
[ ] ( ) )0),3ln(.2,e(a2a4a2),13.(a,)aa.(alim 2a21a/121
0a
=





−+−+ −−
→
.