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Cálculo Diferencial e Integral Lista de Exercícios BsN01C3e01 CDCI/CMCD (01) Verificar o limite a seguir considerado: ( ) ( )e,6j.x1i).)y1ln().1e()y.x((lim x2x31 0y 0x =+++−− → → rr . Solução: Imediatamente, tem-se que: ( )=+++−− → → j.x1i).)y1ln().1e()y.x((lim x2x31 0y 0x rr =+++−= → → − → → j.x1limi).)y1ln().1e()y.x((lim x 0y 0x 2x31 0y 0x rr =+++−= → → − → → x 0y 0x 2x31 0y 0x x1lim.j))y1ln().1e()y.x((lim.i rr =++ +− = → → → → x/1 0y 0x x3 0y 0x )x1(lim.j y )y1ln( .2. x )1e(lim.i rr =++ + − = → → → → → → x/1 0y 0x y/1 0y 0x x3 0y 0x )x1(lim.j)y1ln(lim. x )1)e((lim.i.2 rr =++ += → → → → x/1 0y 0x y/1 0y 0x 3 )x1(lim.j)y1(limln).eln(.i.2 rr ( )( ) =++= → → x/1 0y 0x 3 )x1(lim.jeln).eln(.i.2 rr ( ) =+= e.j)eln(.i.2 3 rr =+= e.j)eln(3.i.2 rr )e,6(j.ei.6 =+= rr . Logo, de fato, tem-se que: ( ) ( )e,6j.x1i).)y1ln().1e()y.x((lim x2x31 0y 0x =+++−− → → rr . Cálculo Diferencial e Integral Lista de Exercícios BsN01C3e01 CDCI/CMCD (02) Verificar o limite a seguir considerado: ( ) = ++ + − − → → 3 e4 , 2 1j.))x3(sen.())ycos(1)).(x4(sen(i.)y1ln(.)x(sen.y.x ))xcos(1(lim 3 1)ysec(3 0y 0x rr . Solução: Imediatamente, tem-se que: ( ) = ++ + − − → → j.))x3(sen.())ycos(1)).(x4(sen(i.)y1ln(.)x(sen.y.x ))xcos(1(lim 1)ysec(3 0y 0x rr ( ) =++ + − = − → → → → 44444444 344444444 21 r 44444 344444 21 r II 1)ysec(3 0y 0x I 0y 0x ))x3(sen.())ycos(1)).(x4(sen(lim.j)y1ln(.)x(sen.y.x ))xcos(1(lim.i Calculando I e II, resulta que: (I) = + − → → )y1ln(.)x(sen.y.x ))xcos(1(lim 0y 0x = +− = +− = → → → → → → y )y1ln(lim.)x(sen.x ))xcos(1(lim y )y1ln( .)x(sen.x ))xcos(1(lim 0y 0x 0y 0x 0y 0x =+ + +− = → → → → y/1 0y 0x 0y 0x )y1ln(lim.))xcos(1( ))xcos(1( .)x(sen.x ))xcos(1(lim = + + − = → → → → y/1 0y 0x 2 0y 0x )y1(limln.))xcos(1)).(x(sen.x( ))xcos(1(lim ( ) = + = → → eln.))xcos(1)).(x(sen.x( ))x(sen(lim 2 0y 0x = + = → → 1.))xcos(1).(x( )x(senlim 0y 0x = + = → → → → ))xcos(1( 1lim.)x( )x(senlim 0y 0x 0y 0x Cálculo Diferencial e Integral Lista de Exercícios BsN01C3e01 CDCI/CMCD = + = → → ))xcos(1( 1lim.1 0y 0x 2 1 ))xcos(1( 1lim 0y 0x = + = → → ; (II) ( ) =+ − → → 1)ysec(3 0y 0x ))x3(sen.())ycos(1)).(x4(sen(lim = + = → → 3 )ycos( 1 0y 0x ))ycos(1(.)x3(sen )x4(senlim = + = → → → → 3 )ycos( 1 0y 0x 0y 0x ))ycos(1(lim.)x3(sen )x4(senlim = + = → → → → 3 )ycos( 1 0y 0x Hospital'LPor 0y 0x ))ycos(1(lim.)x3cos(.3 )x4cos(.4lim 43421 ( ) 3 e4 e. )x3cos(lim )x4cos(lim . 3 4 33 0y 0x 0y 0x = = → → → → . Logo, de fato, tem-se que: ( ) = ++ + − − → → 3 e4 , 2 1j.))x3(sen.())ycos(1)).(x4(sen(i.)y1ln(.)x(sen.y.x ))xcos(1(lim 3 1)ysec(3 0y 0x rr . (03) Verificar o limite a seguir considerado: ( ) )0,1,1(k.t5 ))tcos(1.(3j. v )v(tgi. u1ln )u(senlim 0t 0v 0u = − ++ + → → → rrr . Solução: Cálculo Diferencial e Integral Lista de Exercícios BsN01C3e01 CDCI/CMCD Imediatamente, tem-se que: ( ) = − ++ + → → → k. t5 ))tcos(1.(3j. v )v(tgi. u1ln )u(senlim 0t 0v 0u rrr ( ) = − ++ + = → → → → → → → → → k. t5 ))tcos(1.(3limj. v )v(tglimi. u1ln )u(senlim 0t 0v 0u 0t 0v 0u 0t 0v 0u rrr ( ) 444 3444 21 r 43421 r 4434421 r III 0t 0v 0u II 0t 0v 0u I 0t 0v 0u t5 ))tcos(1.(3lim.k v )v(tglim.j u1ln )u(senlim.i −++ + = → → → → → → → → → . Calculando I, II e III, resulta que: (I) ( ) ( ) ( ) = + = + = + → → → → → → → → → u1ln. u 1 .u )u(senlim u1ln. u u )u(senlim u1ln )u(senlim 0t 0v 0u 0t 0v 0u 0t 0v 0u ( ) ( ) =+=+= → → → → → → → → → u/1 0t 0v 0u 0t 0v 0uu/1 0t 0v 0u u1ln 1lim. u )u(senlim u1ln 1 . u )u(senlim ( ) 1)eln( 1 u1limln 1lim . u )u(senlim u/1 0t 0v 0u 0t 0v 0u 1 0t 0v 0u == + = → → → → → → = → → → 43421 ; (II) 1)vcos( 1lim.1)vcos( 1lim. v )v(senlim)vcos(.v )v(senlim v )v(tglim 0t 0v 0u 0t 0v 0u 0t 0v 0u 0t 0v 0u 0t 0v 0u ==== → → → → → → → → → → → → → → → ; (III) ==−=− → → → → → → → → → t )2/t(sen.2lim. 5 3 t ))tcos(1(lim. 5 3 t5 ))tcos(1.(3lim 2 0t 0v 0u 0t 0v 0u 0t 0v 0u Cálculo Diferencial e Integral Lista de Exercícios BsN01C3e01 CDCI/CMCD 00. 2 3)2/t(senlim.)2/t( )2/t(senlim. 5 3 2 2 .t )2/t(sen.2lim. 5 3 0)2/t( 0v 0u 0)2/t( 0v 0u 2 0t 0v 0u ==== → → → → → → → → → . Logo: ( ) )0,1,1(t5 ))tcos(1.(3lim.k v )v(tglim.j u1ln )u(senlim.i III 0t 0v 0u II 0t 0v 0u I 0t 0v 0u = − ++ + → → → → → → → → → 444 3444 21 r 43421 r 4434421 r ; ou seja: ( ) = − ++ + → → → k. t5 ))tcos(1.(3j. v )v(tgi. u1ln )u(senlim 0t 0v 0u rrr ( ) k.0ji)0,1,1(k.t5 ))tcos(1.(3j. v )v(tgi. u1ln )u(senlim )0,0,0()t,v,u( rrrrrr ++== − ++ + = → . (04) Verificar o limite a seguir considerado: ( )( ) )8/1,8/1( y4 y31 ,20x12x.6x5xlim 2 122 2y 2x = − −+− +−+− − → → . Solução: Imediatamente, tem-se que: ( )( ) = − −+− +−+− − → → 2 122 2y 2x y4 y31 ,20x12x.6x5xlim ( )( ) 444 3444 21 r 444444 3444444 21 r II 2 2y 2x I 122 2y 2x y4 y31 lim.j20x12x.6x5xlim.i − −+− ++−+−= → → − → → . Calculando I e II, resulta que: (I) ( )( ) = −− −− =+−+− → → − → → )10x).(2x( )3x).(2x(lim20x12x.6x5xlim 2y 2x 122 2y 2x 8 1 10x 3xlim 2y 2x = − − = → → Cálculo Diferencial e Integral Lista de Exercícios BsN01C3e01 CDCI/CMCD (II) ( )( )( )( ) =+−− +−−−= − −+− → → → → 1y3.y4 1y3.1y3 lim y4 y31 lim 2 2y 2x2 2y 2x ( )( ) ( )( )( ) =+−+− −=+−− −−= →→→→ 1y3y2y2 )y2(lim 1y3.y4 )1y3(lim 2y 2x2 2y 2x ( ) 8/1)1y3.(y2 1lim 2y 2x = +−+ = → → . Logo, de fato, tem-se que: ( )( ) )8/1,8/1( y4 y31 ,20x12x.6x5xlim 2 122 2y 2x = − −+− +−+− − → → . (05) Calcular o limite: ( ) − +−+−−− → k. u4 2e2j).u(gcot.a1i).14.(8.u.4lim u3 )u(tgu11 0u rrr . Solução: Como: )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 , ( ) =−−=−=− →→→ )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, 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 . Tem-se, portanto, que: ( ) = − +−+−−− → k. u4 2e2j).u(gcot.a1i).14.(8.u.4lim u3 )u(tgu11 0u rrr Cálculo Diferencial e Integral Lista de Exercícios BsN01C3e01 CDCI/CMCD k. 2 3j)).a(ln(i)).2(ln()2/3),aln(),2(ln( rrr +−=−= . (06) Verificar o limite a seguir considerado: )2,4/3()j).2v2v4v2(i).2u3u.()3u5uu((lim 233123 1v 1u −=−+−++−+−+ − → → rr . Solução: =−+−++−+−+ − → → )j).2v2v4v2(i).2u3u.()3u5uu((lim 233123 1v 1u rr = −+−+ +−+ +− = → → j).2v2v4v2(i.)3u5uu( )2u3u(lim 2323 3 1v 1u rr =−+−+ +−+ +− = → → → → j).2v2v4v2(limi.)3u5uu( )2u3u(lim 23 1v 1u23 3 1v 1u rr 4444 34444 21 r 4444 34444 21 r II 23 1v 1u I 23 3 1v 1u )2v2v4v2(lim.j)3u5uu( )2u3u(lim.i −+−+ +−+ +− = → → → → Calculando I e II tem-se, respectivamente, que: (I) = +−+ +− → → )3u5uu( )2u3u(lim 23 3 1v 1u ( )( ) ( )( ) =−+ −+ = −+− −+− → → → → )3u2u( )2uu(lim 3u2u.1u 2uu.1ulim 2 2 1v 1u2 2 1v 1u ( )( ) ( )( ) 4 3 )3u( )2u(lim 3u.1u 2u.1ulim 1v 1u 1v 1u = + + = +− +− = → → → → (II) =−+−=−+− → → → → 2v2v4v2lim)2v2v4v2(lim 23 1v 1u 23 1v 1u 22242)2lim()v2lim()v4lim()v2lim( 1v 1u 1v 1u 2 1v 1u 3 1v 1u −=−+−=−++−+= → → → → → → → → Logo: =−+−+ +−+ +− = → → → → 4444 34444 21 r 4444 34444 21 r II 23 1v 1u I 23 3 1v 1u )2v2v4v2(lim.j)3u5uu( )2u3u(lim.i Cálculo Diferencial e Integral Lista de Exercícios BsN01C3e01 CDCI/CMCD )2,4/3(j.2i. 4 32.j 4 3i =+=+= rrrr . (07) Verificar o limite a seguir considerado: ( )( ) ( )( )( ) )x4,2/1(j.a.xaxi.u2u3.uu2u4lim 31441223 0a 0u =−++++− − − → → rr . Solução: Solução: Imediatamente, tem-se que: ( )( ) ( )( )( )=−++++− −− → → j.a.xaxi.u2u3.uu2u4lim 1441223 0a 0u rr ( )( ) ( )( ) =−++++−= − → − → 44444 344444 21 r 4444444 34444444 21 r II 144 )0,0()a,u( I 1223 )0,0()a,u( a.xaxlim.ju2u3.uu2u4lim.i Calculando I e II, resulta que: (I) ( )( ) =++− − → 1223 )0,0()a,u( u2u3.uu2u4lim ( ) ( ) 2 1 )2u3( )1u2u4(lim 2u3.u 1u2u4.ulim 2 )0,0()a,u( 2 )0,0()a,u( = + +− = + +− = →→ (II) Como )yx).(yx).(yx(yx,Ry,x 2244 ++−=−∈∀ , tem-se que: ( )( ) =−+ − → 144 )0,0()a,u( a.xaxlim = ++++−+ = −+ →→ a )x)ax)((x)ax).((x)ax((lim a )x)ax((lim 22 )0,0()a,u( 44 )0,0()a,u( =+++= +++ = →→ )x)ax)((ax2(lim a )x)ax)((ax2).(a(lim 22 )0,0()a,u( 22 )0,0()a,u( 32 x4x2.x2 == . Logo, de fato, tem-se que: ( )( ) ( )( )( ) )x4,2/1(j.a.xaxi.u2u3.uu2u4lim 31441223 0a 0u =−++++− − − → → rr . Cálculo Diferencial e Integral Lista de Exercícios BsN01C3e01 CDCI/CMCD (08) Calcular o limite: )j)).35ln(t(i).15.()17((lim t71t5 0t rr ++−− − → . Solução: 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 += . (09) Verificar o limite a seguir considerado: )1,1()j.))bx(sen)ax(sen( )ee(i.u).2)ucos(2((lim bxax 2 0x 0u −= − − +− − → → rr . Solução: Imediatamente, tem-se que: = − − +− − → → )j.))bx(sen)ax(sen( )ee(i.u).2)ucos(2((lim bxax 2 0x 0u rr 4444 34444 21 r 444 3444 21 r II bxax 0x 0u I 2 0x 0u ))bx(sen)ax(sen( )ee(lim.ju).2)ucos(2(lim.i − − +−= → → − → → . Calculando I e II, resulta que: (I) =−=− → → → → 2 0x 0u2 0x 0u u )1)u(cos(lim.2 u )2)ucos(2(lim Cálculo Diferencial e Integral Lista de Exercícios BsN01C3e01 CDCI/CMCD = +− −− = +− +− −− = → → → → )1)u(sen1.(u )1)u(sen1(lim.2 )1)u(sen1( )1)u(sen1( . u )1usen1(lim.2 22 2 0x 0u2 2 2 2 0x 0u ( ) =+−−=+−−= →→→→ )1)u(sen1( 1 . u )u(senlim.2 1usen1.u )u(senlim.2 2 2 0x 0u22 2 0x 0u 1 2 1 .2 )1)u(sen1( 1lim. u )u(senlim.2 2 0x 0u 2 0x 0u −=−= +− −= → → → → (II) = + − − = − − → → → → 2 bxax cos. 2 bxax sen.2 )ee(lim))bx(sen)ax(sen( )ee(lim bxax 0x 0u bxax 0x 0u = + − − = → → → → 2 bxax cos 1lim. 2 bxax sen2 )ee(lim 0x 0u bxax 0x 0u = − − − − = − − = → → → → 2 )bxax( 1 . 2 bxax sen2 2 )bxax( 1).ee( lim 2 bxax sen2 )ee(lim bxax 0x 0u bxax 0x 0u = − − − − = − − − − = → − → → → → → → → 2 x).ba( 1 . 2 x).ba( senlim )bxax( )ee( .2lim . 2 1 2 )bxax( 1 . 2 bxax senlim )bxax( )ee( .2lim . 2 1 0 2 x).ba( 0u bxax 0x 0u 0x 0u bxax 0x 0u = − − = − − − − = → → = → − → → → )bxax( )ee( .2lim. 2 1 2 )bxax( 2 bxax sen lim )bxax( )ee( .2lim . 2 1 bxax 0x 0u 1 0 2 )bxax( 0u bxax 0x 0u 4444 34444 21 Cálculo Diferencial e Integral Lista de Exercícios BsN01C3e01 CDCI/CMCD = − − = − − = → → → → )bxax( )ee(lim)bxax( )ee(lim.2. 2 1 bxax 0x 0u bxax 0x 0u = − − = − − = → → → → x )ee(lim.)ba( 1 x).ba( )ee(lim bxax 0x 0u bxax 0x 0u = − − = − − = → → → → → → x elim xelim.)ba( 1 x e x elim.)ba( 1 bx 0x 0u ax 0x 0u bxax 0x 0u = − − = → → → → x )e(lim x )e(lim.)ba( 1 xb 0x 0u xa 0x 0u = +− − +− − = → → → → x )11)e((lim x )11)e((lim.)ba( 1 xb 0x 0u xa 0x 0u = +− − +− − = → → → → x )1)1)e(((lim x )1)1)e(((lim.)ba( 1 xb 0x 0u xa 0x 0u = − − −+ − − = → → → → → → → → x 1lim x )1)e((lim x 1lim x )1)e((lim.)ba( 1 0x 0u xb 0x 0u 0x 0u xa 0x 0u = − − − − = = → → = → → 44 344 2144 344 21 )eln( xb 0x 0u )eln xa 0x 0u ba x )1)e((lim x )1)e((lim.)ba( 1 [ ] [ ] =− − =− − = )eln(.b)eln(.a.)ba( 1)eln()eln(.)ba( 1 ba 1)ba( )ba()ba.()ba( 1 = − − =− − = . Logo: 1))bx(sen)ax(sen( )ee(lim bxax 0x 0u = − − → → . Cálculo Diferencial e Integral Lista de Exercícios BsN01C3e01 CDCI/CMCD Portanto: )1,1()j.))bx(sen)ax(sen( )ee(i.u).2)ucos(2((lim bxax 2 0x 0u −= − − +− − → → rr . (10) Verificar o limite a seguir considerado: ( ) ( )( ) ( ) ( )( ) =+−+−++− −− → → → )k.w1ln.1ej.v.asenavseni)).u(gcot).a1(((lim 1w1)u(tg 0w 0v 0u rrr )1,)acos(,)aln((kj)).a(cos(i)).aln(( −=++−= rrr . Solução: Imediatamente, tem-se que: ( ) ( )( ) ( ) ( )( ) =+−+−++− −− → → → )k.w1ln.1ej.v.asenavseni)).u(gcot).a1(((lim 1w1)u(tg 0w 0v 0u rrr ( ) ( )( ) ( ) ( )( ) =+−+−++− − → → → − → → → → → → 4444 34444 21 r 44444 344444 21 r 4444 34444 21 r III 1w 0w 0v 0u II 1 0w 0v 0u I )u(tg 0w 0v 0u w1ln.1elim.kv.asenavsenlim.j)u(gcot).a1(lim.i Calculando I, II e III, resulta que: (I) =− → → → )u(gcot).a1(lim )u(tg 0w 0v 0u = − −= − → → → → → → )u(tg )1a(lim)u(tg )a1(lim )u(tg 0w 0v 0u )u(tg 0w 0v 0u )aln( y )1a(lim y 0w 0v 0y −= − −= → → → ; (II) ( ) ( )( ) =−+ − → → → 1 0w 0v 0u v.asenavsenlim ( ) ( )( ) =−+=−+ → → → − → → → v ))a(sen)av(sen(limv.asenavsenlim 0w 0v 0u 1 0w 0v 0u Cálculo Diferencial e Integral Lista de Exercícios BsN01C3e01 CDCI/CMCD = ++−+ = → → → v ) 2 ava cos(). 2 aav(sen.2 lim 0w 0v 0u = + = → → → v )2/)va2cos(().2/v(sen.2lim 0w 0v 0u =+= → → → → → → )2/)va2cos((lim. v. 2 2 )2/v(senlim.2 0w 0v 0u 0w 0v 0u )acos()acos(.)2/v( )2/v(senlim)2/a2cos(.)2/v( )2/v(senlim. 2 2 1 0w 0)2/v( 0u 0w 0)2/v( 0u === = → → → → → → 44 344 21 ; (III) ( ) ( )( ) =+− − → → → 1w 0w 0v 0u w1ln.1elim ( ) ( )( ) ( ) ( ) = + − = + − =+−= → → → → → → − → → → w/1 )eln( w 0w 0v 0u w 0w 0v 0u 1w 0w 0v 0u w1ln 1 . w 1elim w1ln. w w )1e(limw1ln.1elim 4434421 ( ) ( ) 1)eln( 1 w1limln 1 w1ln 1lim w/1 0w 0v 0u w/1 0w 0v 0u == + = + = → → →→ → → . Portanto, de fato, tem-se que: ( ) ( )( ) ( ) ( )( ) =+−+−++− −− → → → )k.w1ln.1ej.v.asenavseni)).u(gcot).a1(((lim 1w1)u(tg 0w 0v 0u rrr )1,)acos(,)aln((kj)).a(cos(i)).aln(( −=++−= rrr .
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