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Cálculo I
                                 Profº. Marcello Santos Chaves
       LIMITES TRIGONOMÉTRICOS


                                 senx
       Fundamental: Lim               =1
                          x →0     x


                                                                                   Sen 3 x
1)    Lim f ( x ) = Lim
                          sen 2 x                      2)      Lim f ( x ) = Lim
                                                               x →0         x →0   Sen 4 x
      x→0           x→0     x

Solução :                                              Solução :

                                                                             Sen 3 x
Lim f ( x ) = Lim
                     sen 2 x                           Lim f ( x ) = Lim
                                                        x →0           x → 0 Sen 4 x
x→0             x→0     x
                     sen 2 x 2                                               Sen3 x
Lim   f ( x ) = Lim          ⋅                                                  x
x→0             x→0     x       2                      Lim f ( x ) =   Lim
                                                        x →0           x → 0 Sen 4 x
                        sen 2 x
Lim   f ( x ) = Lim 2 ⋅                                                         x
x→0             x→0       2x
                                                                             Sen3 x 3
                               sen 2 x                                               ⋅
Lim   f ( x ) = Lim 2 ⋅ Lim                                                     x      3
x→0             x→0     x→0      2x                    Lim f ( x ) =   Lim
                                                        x →0           x → 0 Sen 4 x 4
Lim   f ( x) = 2 × 1                                                                 ⋅
x→0                                                                             x      4
Lim f ( x ) = 2                                                                 Sen 3 x
x→0                                                                          3⋅
                                                       Lim f ( x ) =   Lim        3x
                                                        x →0           x →0     Sen 4 x
                                                                             4⋅
                                                                                  4x
                                                                                      Sen 3 x
                                                                       Lim 3 ⋅ Lim
                                                                        x →0    x→0     3x
                                                       Lim f ( x ) =
                                                        x →0                          Sen 4 x
                                                                       Lim 4 ⋅ Lim
                                                                       x→0      x→0     4x
                                                                       3 ×1
                                                       Lim f ( x ) =
                                                        x →0           4 ×1
                                                                       3
                                                       Lim f ( x ) =
                                                        x →0           4




                                                               1
       Marcello Santos Chaves
       Instituto Federal de Educação, Ciência e Tecnologia (IFPA)                      Belém-PA, Abril de 2011
Cálculo I
                              Profº. Marcello Santos Chaves

                           Tgx                                                        Cos θ − 1
3)     Lim f ( x ) = Lim                                4)     Lim f ( x ) = Lim
       x→0           x→0    x                                  x →θ            x →θ      θ

Solução :                                               Solução :


                       Tgx                                                     Cos θ − 1
Lim f ( x ) = Lim                                       Lim f ( x ) = Lim
x →0             x→0     x                              x →θ            x →θ    θ
                        Senx                                               Cos θ − 1 Cos θ + 1
                                                        Lim f ( x ) = Lim             ⋅
                                                        x →θ          x →θ      θ       Cos θ + 1
Lim    f ( x ) = Lim Cosx
x →0             x→0      x                                                 Cos θ − 12
                                                                                  2
                                                        Lim f ( x ) = Lim
                        Senx 1                          x →θ          x →θ θ ⋅ (Cos θ + 1)
Lim    f ( x ) = Lim         ⋅
x →0             x → 0 Cosx x
                                                                                 Cos 2θ − 1
                                                        Lim f ( x ) = Lim
Lim    f ( x ) = Lim
                        Senx
                             ⋅
                               1                        x →θ            x →θ   θ ⋅ (Cos θ + 1)
x →0             x→0      x Cosx
                                                                                  − Sen 2θ
                       1                                Lim    f ( x ) = Lim
Lim    f ( x) = 1 ⋅                                     x →θ             x →θ θ ⋅ (Cos θ + 1)
x →0                Cos 0
Lim    f ( x) = 1 × 1                                   Lim    f ( x ) = Lim − 1 ⋅
                                                                                   (Senθ ) ⋅ (Senθ )
x →0                                                    x →θ             x →θ        θ ⋅ (Cos θ + 1)
Lim f ( x ) = 1                                                                     Sen θ     Sen θ
x →0
                                                        Lim    f ( x ) = Lim − 1 ⋅         ⋅
                                                        x →θ             x →θ         θ      Cos θ + 1
                                                                                   Sen 0
                                                        Lim    f ( x ) = −1 × 1 ⋅
                                                        x →θ                      Cos 0 + 1
                                                                               0
                                                        Lim    f ( x ) = −1 ⋅
                                                        x →θ                  1+1
                                                        Lim    f ( x ) = −1 × 0
                                                        x →θ

                                                        Lim f ( x ) = 0
                                                        x →θ




                                                             2
     Marcello Santos Chaves
     Instituto Federal de Educação, Ciência e Tecnologia (IFPA)                        Belém-PA, Abril de 2011
Cálculo I
                                       Profº. Marcello Santos Chaves
                                                                                      Sen 4 x
     5)        Lim f ( x ) = Lim
                                  Sen 9 x               6)      Lim f ( x ) = Lim
                                                                x→0             x→0     3x
               x →0          x →0   x

     Solução :                                           Solução :

                                                                               Sen 4 x
     Lim f ( x ) = Lim
                              Sen 9 x                    Lim f ( x ) = Lim
                                                         x →0             x →0   3x
       x →0              x →0    x
                                                                               1       Sen 4 x 4
     Lim       f ( x ) = Lim
                              Sen 9 x 9
                                      ⋅                  Lim    f ( x ) = Lim ⋅ Lim            ⋅
                                                         x →0             x →0 3 x → 0   x       4
       x →0              x →0    x      9
                                                                               1         Sen 4 x
     Lim       f ( x ) = Lim 9 ⋅
                                 Sen 9 x                 Lim    f ( x ) = Lim ⋅ Lim 4 ⋅
                                                         x →0             x →0 3 x → 0      4x
       x →0              x →0      9x
                                                                               1               Sen 4 x
     Lim       f ( x ) = Lim 9 ⋅ Lim
                                      Sen 9 x            Lim    f ( x ) = Lim ⋅ Lim 4 ⋅ Lim
                                                         x →0             x →0 3 x → 0    x→0     4x
       x →0              x →0    x→0     9x
               f ( x) = 9 ⋅ 1                                             1
     Lim
       x →0
                                                         Lim    f ( x) = ⋅ 4 ⋅ 1
                                                         x →0             3
     Lim f ( x ) = 9                                                      4
       x →0
                                                         Lim    f ( x) =
                                                         x →0             3


                                                                                                   Sen ax
                                Sen 10 x                              8)     Lim f ( x ) = Lim
7)      Lim f ( x ) = Lim                                                    x→0             x→0   Sen bx
          x →0            x→0    Sen 7 x

                                                                      Solução :
Solução :

                                                                                             Sen ax
                    Sen 10 x                                          Lim f ( x ) = Lim
Lim f ( x ) = Lim                                                     x →0            x →0   Sen bx
x →0          x → 0 Sen 7 x
                                                                                            Sen ax
                         Sen 10 x
                                                                      Lim f ( x ) =   Lim     x
Lim     f ( x ) = Lim        x                                        x →0            x → 0 Sen bx
x →0               x →0   Sen 7 x
                                                                                              x
                             x
                                                                                            Sen ax a
                         Sen 10 x 10                                                               ⋅
                                  ⋅                                                           x      a
                             x      10                                Lim f ( x ) =   Lim
Lim     f ( x ) = Lim                                                 x →0            x → 0 Sen bx b
x →0               x →0   Sen 7 x 7                                                                ⋅
                                  ⋅                                                           x      b
                             x      7
                                                                                                   Sen ax
                                  Sen 10 x                                            Lim a ⋅ Lim
                  Lim 10 ⋅ Lim                                                        x→0      x→0    ax
                   x→0        x→0    10 x                             Lim f ( x ) =
Lim     f ( x) =                                                      x →0                         Sen bx
x →0                              Sen 7 x                                             Lim b ⋅ Lim
                     Lim 7 ⋅ Lim                                                       x→0     x→0    bx
                     x→0      x→0    7x
                                                                                      a ⋅1
                  10 ⋅ 1                                              Lim f ( x ) =
Lim     f ( x) =                                                      x →0            b ⋅1
x →0               7 ⋅1
                                                                                      a
                  10                                                  Lim f ( x ) =
Lim     f ( x) =                                                      x →0            b
x →0               7
                                                                      3
              Marcello Santos Chaves
              Instituto Federal de Educação, Ciência e Tecnologia (IFPA)                           Belém-PA, Abril de 2011
Cálculo I
                                  Profº. Marcello Santos Chaves
                                                                                                 x + 1
                         Tg ax                                                             Tg 3       
9)     Lim f ( x ) = Lim                                     10)      Lim f ( x ) = Lim          4 
                                                                                             (x + 1)3
       x→0           x→0   x                                          x → −1        x → −1



Solução :
                                                             Solução :

                    Tg ax
Lim f ( x ) = Lim                                                                       x +1
x →0          x →0    x                                                           Tg 3      
                    Sen ax                                   Lim f ( x ) = Lim          4 
                    Cos ax
                                                             x → −1        x → −1   (x + 1)3
Lim f ( x ) = Lim
x →0          x →0     x                                                                        x + 1
                                                                                      3   Tg 3       
Lim f ( x ) = Lim
                    Sen ax 1
                           ⋅                                                                    4 
                                                             Lim f ( x ) = Lim
x →0          x → 0 Cos ax   x                               x → −1          x → −1       3
                                                                                              (x + 1)3
                    Sen ax 1 a
Lim f ( x ) = Lim          ⋅ ⋅                                                        x +1
x →0          x → 0 Cos ax   x a                                                  Tg       
                                                             Lim f ( x ) = Lim        4 
Lim f ( x ) = Lim a ⋅
                            Sen ax
                                   ⋅
                                       1                     x → −1        x → −1   (x + 1)
x →0             x →0         ax     Cos ax                                  x +1
                                  Sen ax           1         Faça → u =            ∴ x = 4u − 1
Lim f ( x ) = Lim a ⋅ Lim                ⋅ Lim                                  4
x →0             x →0       x→0     ax     x → 0 Cos ax
                                                             Se : x → −1∴ u → π
                         1                                                              Tg u
Lim f ( x ) = a ⋅ 1 ⋅                                        Lim f ( x ) = Lim
x →0                    Cos 0                                u →π             u →π    4u − 1 + 1
                        1                                                             Tg u
Lim f ( x ) = a ⋅ 1 ⋅                                        Lim f ( x ) = Lim
x →0                    1                                    u →π             u →π     4u
Lim f ( x ) = a                                                                       Sen u
x →0
                                                                                      Cos u
                                                             Lim f ( x ) = Lim
                                                             u →π             u →π      4u
                                                                                      Sen u 1
                                                             Lim f ( x ) = Lim               ⋅
                                                             u →π             u →π    Cos u 4u
                                                                                      Sen u          1
                                                             Lim f ( x ) = Lim               ⋅ Lim
                                                             u →π             u →π    Cos u    u →π 4u

                                                                           Sen π 1
                                                             Lim f ( x ) =       ⋅
                                                             u →π          Cos π 4π
                                                                            0 1
                                                             Lim f ( x ) =    ⋅
                                                             u →π          − 1 4π
                                                             Lim f ( x ) = 0
                                                             u →π




                                                               4
       Marcello Santos Chaves
       Instituto Federal de Educação, Ciência e Tecnologia (IFPA)                             Belém-PA, Abril de 2011
Cálculo I
                                  Profº. Marcello Santos Chaves
                             1 − Cos x
11)    Lim f ( x ) = Lim
        x→0           x→0        x

Solução :

                  1 − Cos x
Lim f ( x ) = Lim
x →0          x→0       x
                  − 1 ⋅ (Cos x − 1)
Lim f ( x ) = Lim
x →0          x→0          x
                             Cos x − 1
Lim f ( x ) = Lim − 1 ⋅ Lim
x →0          x→0        x→0     x
                             Cos x − 1 Cos x + 1
Lim f ( x ) = Lim − 1 ⋅ Lim            ⋅
x →0          x→0        x→0     x       Cos x + 1
                               Cos 2 x − 12
Lim f ( x ) = Lim − 1 ⋅ Lim
x →0          x→0       x → 0 x ⋅ (Cos x + 1)


                               Cos 2 x − 1
Lim f ( x ) = Lim − 1 ⋅ Lim
x →0          x→0       x → 0 x ⋅ (Cos x + 1)


                                     − Sen 2 x
Lim f ( x ) = Lim − 1 ⋅ Lim
x →0            x→0         x→0   x ⋅ (Cos x + 1)
                                  − (Sen x ) ⋅ (Sen x )
Lim f ( x ) = Lim − 1 ⋅ Lim
x →0            x→0         x→0     x ⋅ (Cos x + 1)
                              − 1 ⋅ (Sen x )           (Sen x )
Lim f ( x ) = Lim − 1 ⋅ Lim                  ⋅ Lim
x →0            x→0         x→0      x         x → 0 (Cos x + 1)


Lim f ( x ) = −1 × (− 1) ⋅
                            Sen 0
x →0                       Cos 0 + 1

Lim f ( x ) = −1 × (− 1) ⋅
                            0
x →0                      1+1
Lim f ( x ) = −1 × (− 1) × 0
x →0

Lim f ( x ) = 0
x →0




                                                               5
       Marcello Santos Chaves
       Instituto Federal de Educação, Ciência e Tecnologia (IFPA)   Belém-PA, Abril de 2011
Cálculo I
                                   Profº. Marcello Santos Chaves
                             1 − Cos x
12)     Lim f ( x ) = Lim
         x→0           x→0       x2

Solução :

                  1 − Cos x
Lim f ( x ) = Lim
x →0             x→0   x2
                  − 1 ⋅ (Cos x − 1)
Lim f ( x ) = Lim
x →0          x→0          x2
                              Cos x − 1
Lim f ( x ) = Lim − 1 ⋅ Lim
x →0          x→0        x→0    x2
                              Cos x − 1 Cos x + 1
Lim f ( x ) = Lim − 1 ⋅ Lim             ⋅
x →0          x→0        x→0    x2        Cos x + 1
                                     Cos 2 x − 12
Lim f ( x ) = Lim − 1 ⋅ Lim
x →0             x→0         x→0   x 2 ⋅ (Cos x + 1)
                               Cos 2 x − 1
Lim f ( x ) = Lim − 1 ⋅ Lim 2
x →0          x→0       x → 0 x ⋅ (Cos x + 1)


                                       − Sen 2 x
Lim f ( x ) = Lim − 1 ⋅ Lim
x →0             x→0         x→0   x 2 ⋅ (Cos x + 1)
                              − (Sen x ) ⋅ (Sen x )
Lim f ( x ) = Lim − 1 ⋅ Lim
x →0          x→0       x → 0 x ⋅ x ⋅ (Cos x + 1)


Lim f ( x ) = Lim − 1 ⋅ Lim
                              (Sen x ) ⋅ Lim (Sen x ) ⋅ Lim − 1
x →0          x→0       x→0      x       x→0      x     x → 0 (Cos x + 1)

                               −1
Lim f ( x ) = −1 × 1 × 1 ⋅
x →0                         Cos 0 + 1
                    −1
Lim f ( x ) = −1 ⋅
x →0               1+1
                    1
Lim f ( x ) = −1 ⋅  − 
x →0
                    2
              1
Lim f ( x ) =
x →0          2




                                                              6
      Marcello Santos Chaves
      Instituto Federal de Educação, Ciência e Tecnologia (IFPA)            Belém-PA, Abril de 2011
Cálculo I
                             Profº. Marcello Santos Chaves

                                                                                             6 x − Sen 2 x
13) Lim f ( x) = Lim ( x − 3) ⋅ Co sec (πx )                 14) Lim f ( x) = Lim
        x →3          x →3                                           x →0             x →0   2 x + 3Sen 4 x

Solução :                                                    Solução :

Lim f ( x) = Lim ( x − 3) ⋅ Co sec (πx )                                            6 x − Sen 2 x
x →3           x →3
                                                             Lim f ( x) = Lim
                                                             x →0             x→0   2 x + 3Sen 4 x
Lim f ( x) = Lim ( x − 3) ⋅
                                 1
x →3           x →3           Sen (πx )                                                                 2x
                                                                                     6 x − Sen 2 x ⋅
Lim f ( x) = Lim
                    ( x − 3)                                 Lim    f ( x) = Lim                        2x
             x →3 Sen (π − πx )
                                                             x →0             x→0                        4x
x →3
                                                                                    2 x + 3Sen 4 x ⋅

Lim f ( x) = Lim
                     ( x − 3)                                                                            4x
             x →3 Sen (3π − πx )
                                                                                                    Sen 2 x 
x →3                                                                                  6x − 2x ⋅             
                                                                                                    2x 
                     ( x − 3)                                Lim    f ( x) = Lim
                                                             x →0             x→0                     Sen 4 x 
Lim f ( x) = Lim
                     ( x − 3)                                                       2x + 3 ⋅ 4x ⋅             
x →3         x →3 Sen (3π − πx )
                                                                                                      4x 
                     ( x − 3)                                                       6x − 2x ⋅ 
                                                                                                  Sen 2 x 
                                                                                                            
                                1                            Lim    f ( x) = Lim                  2x 
Lim f ( x) = Lim
                x →3 π ⋅ Sen (3π − πx )
                                                             x →0             x→0                  Sen 4 x 
x →3                                                                                2 x + 12 x ⋅            
                         π ⋅ ( x − 3)                                                              4x 
                                                                                                     Sen 2 x 
Lim    f ( x) = Lim
                                1                                                    x ⋅ (6 − 2 ) ⋅          
x →3            x →3 π ⋅ Sen (3π − πx )                                                              2x 
                                                             Lim    f ( x) = Lim
                          (πx − 3π )                         x →0             x→0
                                                                                    x ⋅ (2 + 12 ) ⋅ 
                                                                                                      Sen 4 x 
                                                                                                               
                                1                                                                     4x 
Lim    f ( x) = Lim
                         Sen (πx − 3π )                                                                 Sen 2 x 
                                                                              Lim (6 − 2 ) ⋅ Lim 
x →3            x →3
                      π⋅                                                                                         
                            (πx − 3π )                       Lim    f ( x) =
                                                                               x →0            x →0
                                                                                                        2x 
                                                             x →0                                        Sen 4 x 
                            Lim1                                             Lim (2 + 12 ) ⋅ Lim                 
Lim    f ( x) =              x →3
                               Sen (πx − 3π )                                                            4x 
                                                                              x →0               x →0
x →3
                Lim π ⋅ Lim                                                   6 − 2 ×1
                 x →3    x →3     (πx − 3π )                 Lim    f ( x) =
                  1
                                                             x →0            2 + 12 × 1
Lim    f ( x) =                                                               4
x →3            π ⋅1                                         Lim    f ( x) =
                                                             x →0            14
                1
Lim    f ( x) =                                                              2
x →3           π                                             Lim    f ( x) =
                                                             x →0            7




                                                        7
Marcello Santos Chaves
Instituto Federal de Educação, Ciência e Tecnologia (IFPA)                          Belém-PA, Abril de 2011
Cálculo I
                                 Profº. Marcello Santos Chaves

                           Cos 2 x − Cos 3 x
15) Lim f ( x) = Lim
       x →0         x →0          x2

Solução :


                    Cos 2 x − Cos 3 x
Lim f ( x) = Lim
x →0         x →0           x2
                             2 x + 3x            2 x − 3x 
                    − 2 Sen              ⋅ Sen              
Lim f ( x) = Lim             2                   2 
x →0         x →0                       x2
                             5x           x
                    − 2 Sen   ⋅ Sen  − 
Lim f ( x) = Lim             2            2
x →0         x →0                  2
                                 x
                             5x              x 
                    − 2 Sen   ⋅ − Sen  
                             2               2 
Lim f ( x) = Lim                     2
x →0         x →0                 x
                           5x          x
                    2 Sen   ⋅ Sen  
Lim f ( x) = Lim           2          2
x →0         x →0               2
                              x
                           5x           x
                    2 Sen   Sen  
Lim f ( x) = Lim           2 ⋅         2
x →0         x →0        x              x
                           5x           x                                 5x         x
                    2 Sen   Sen                                    2 Sen   Sen  
Lim f ( x) = Lim           2 ⋅          2  ⇒ Lim f ( x) = Lim             2 ⋅       2 ⇒
x →0         x →0        x              2          x→0          x →0      x             x
                                     x⋅                                              2⋅ 
                                        2                                              2
                             5x  5            x                         5         5x        x
                       Sen   ⋅          Sen                                ⋅ Sen   Sen  
⇒ Lim f ( x) = Lim           2  2⋅            2  ⇒ Lim f ( x) = Lim 2             2 ⋅      2 ⇒
     x →0         x →0        5             x          x →0        x →0       5x          x
                           x⋅                                                 ⋅           
                              2            2                                  2          2
                                        5x                  x
                                 Sen                 Sen  
                       5
⇒ Lim f ( x) = Lim ⋅ Lim                2  ⋅ Lim           2
     x →0         x →0 2 x →0        5x        x →0     x
                                                        
                                     2                 2
                 5
⇒ Lim f ( x) = × 1 × 1
     x →0        2
                 5
⇒ Lim f ( x) =
     x →0        2


                                                               8
       Marcello Santos Chaves
       Instituto Federal de Educação, Ciência e Tecnologia (IFPA)                    Belém-PA, Abril de 2011
Cálculo I
                           Profº. Marcello Santos Chaves

                           1 − 2Cos x + Cos 2 x
16) Lim f ( x) = Lim
       x →0         x →0           x2

Solução :


                    1 − 2Cos x + Cos 2 x
Lim f ( x) = Lim
x →0         x →0               x2
                    1 − 2Cos x + 1 − 2 Sen 2 x
Lim f ( x) = Lim
x →0         x →0                  x2
                    2 − 2Cos x − 2 Sen 2 x
Lim f ( x) = Lim
x →0         x →0                x2

Lim f ( x) = Lim
                      [                   ]
                    2 ⋅ 1 − Cos x − Sen 2 x
x →0         x →0                x2
                                  x            
                    2 ⋅ 2 Sen 2   − Sen 2 x 
                                   2
Lim f ( x) = Lim                    2
                                                  
x →0         x →0                  x
                              x
                    4 Sen 2   − 2 Sen 2 x
Lim f ( x) = Lim             2
x →0         x →0               x2
                              x
                    4 Sen 2                2
Lim f ( x) = Lim              2  − 2 Sen x
x →0         x →0         x2             x2
                              x
                    4 Sen 2                2
Lim f ( x) = Lim              2  − 2 Sen x
x →0         x →0        x⋅x             x2
                                 x
                       4 Sen 2                  2
Lim f ( x) = Lim                 2  − 2 Sen x
x →0         x →0       2        2         x2
                    x ⋅ ⋅ x ⋅ 
                        2        2
                                 x                                       x
                       4 Sen 2                  2
                                                                   4 Sen 2             2
Lim f ( x) = Lim                 2  − 2 Sen x ⇒ Lim f ( x) = Lim          2  − 2 Sen x ⇒
x →0         x →0        x        x        x2      x →0    x→0
                                                                        x
                                                                             2
                                                                                      x2
                    2⋅ ⋅ 2⋅                                     4⋅ 
                        2        2                                 2
                                x
                         Sen 2                         2
⇒ Lim f ( x) = Lim              2  − Lim 2 ⋅ Lim Sen x
                                 2
     x →0         x →0
                            x         x →0      x →0 x2
                            
                           2
⇒ Lim f ( x) = 1 − 2
    x →0

⇒ Lim f ( x) = −1
    x →0
                                                         9
 Marcello Santos Chaves
 Instituto Federal de Educação, Ciência e Tecnologia (IFPA)              Belém-PA, Abril de 2011

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Profº Marcelo Santos Chaves Cálculo I (limites trigonométricos)

  • 1. Cálculo I Profº. Marcello Santos Chaves LIMITES TRIGONOMÉTRICOS senx Fundamental: Lim =1 x →0 x Sen 3 x 1) Lim f ( x ) = Lim sen 2 x 2) Lim f ( x ) = Lim x →0 x →0 Sen 4 x x→0 x→0 x Solução : Solução : Sen 3 x Lim f ( x ) = Lim sen 2 x Lim f ( x ) = Lim x →0 x → 0 Sen 4 x x→0 x→0 x sen 2 x 2 Sen3 x Lim f ( x ) = Lim ⋅ x x→0 x→0 x 2 Lim f ( x ) = Lim x →0 x → 0 Sen 4 x sen 2 x Lim f ( x ) = Lim 2 ⋅ x x→0 x→0 2x Sen3 x 3 sen 2 x ⋅ Lim f ( x ) = Lim 2 ⋅ Lim x 3 x→0 x→0 x→0 2x Lim f ( x ) = Lim x →0 x → 0 Sen 4 x 4 Lim f ( x) = 2 × 1 ⋅ x→0 x 4 Lim f ( x ) = 2 Sen 3 x x→0 3⋅ Lim f ( x ) = Lim 3x x →0 x →0 Sen 4 x 4⋅ 4x Sen 3 x Lim 3 ⋅ Lim x →0 x→0 3x Lim f ( x ) = x →0 Sen 4 x Lim 4 ⋅ Lim x→0 x→0 4x 3 ×1 Lim f ( x ) = x →0 4 ×1 3 Lim f ( x ) = x →0 4 1 Marcello Santos Chaves Instituto Federal de Educação, Ciência e Tecnologia (IFPA) Belém-PA, Abril de 2011
  • 2. Cálculo I Profº. Marcello Santos Chaves Tgx Cos θ − 1 3) Lim f ( x ) = Lim 4) Lim f ( x ) = Lim x→0 x→0 x x →θ x →θ θ Solução : Solução : Tgx Cos θ − 1 Lim f ( x ) = Lim Lim f ( x ) = Lim x →0 x→0 x x →θ x →θ θ Senx Cos θ − 1 Cos θ + 1 Lim f ( x ) = Lim ⋅ x →θ x →θ θ Cos θ + 1 Lim f ( x ) = Lim Cosx x →0 x→0 x Cos θ − 12 2 Lim f ( x ) = Lim Senx 1 x →θ x →θ θ ⋅ (Cos θ + 1) Lim f ( x ) = Lim ⋅ x →0 x → 0 Cosx x Cos 2θ − 1 Lim f ( x ) = Lim Lim f ( x ) = Lim Senx ⋅ 1 x →θ x →θ θ ⋅ (Cos θ + 1) x →0 x→0 x Cosx − Sen 2θ 1 Lim f ( x ) = Lim Lim f ( x) = 1 ⋅ x →θ x →θ θ ⋅ (Cos θ + 1) x →0 Cos 0 Lim f ( x) = 1 × 1 Lim f ( x ) = Lim − 1 ⋅ (Senθ ) ⋅ (Senθ ) x →0 x →θ x →θ θ ⋅ (Cos θ + 1) Lim f ( x ) = 1 Sen θ Sen θ x →0 Lim f ( x ) = Lim − 1 ⋅ ⋅ x →θ x →θ θ Cos θ + 1 Sen 0 Lim f ( x ) = −1 × 1 ⋅ x →θ Cos 0 + 1 0 Lim f ( x ) = −1 ⋅ x →θ 1+1 Lim f ( x ) = −1 × 0 x →θ Lim f ( x ) = 0 x →θ 2 Marcello Santos Chaves Instituto Federal de Educação, Ciência e Tecnologia (IFPA) Belém-PA, Abril de 2011
  • 3. Cálculo I Profº. Marcello Santos Chaves Sen 4 x 5) Lim f ( x ) = Lim Sen 9 x 6) Lim f ( x ) = Lim x→0 x→0 3x x →0 x →0 x Solução : Solução : Sen 4 x Lim f ( x ) = Lim Sen 9 x Lim f ( x ) = Lim x →0 x →0 3x x →0 x →0 x 1 Sen 4 x 4 Lim f ( x ) = Lim Sen 9 x 9 ⋅ Lim f ( x ) = Lim ⋅ Lim ⋅ x →0 x →0 3 x → 0 x 4 x →0 x →0 x 9 1 Sen 4 x Lim f ( x ) = Lim 9 ⋅ Sen 9 x Lim f ( x ) = Lim ⋅ Lim 4 ⋅ x →0 x →0 3 x → 0 4x x →0 x →0 9x 1 Sen 4 x Lim f ( x ) = Lim 9 ⋅ Lim Sen 9 x Lim f ( x ) = Lim ⋅ Lim 4 ⋅ Lim x →0 x →0 3 x → 0 x→0 4x x →0 x →0 x→0 9x f ( x) = 9 ⋅ 1 1 Lim x →0 Lim f ( x) = ⋅ 4 ⋅ 1 x →0 3 Lim f ( x ) = 9 4 x →0 Lim f ( x) = x →0 3 Sen ax Sen 10 x 8) Lim f ( x ) = Lim 7) Lim f ( x ) = Lim x→0 x→0 Sen bx x →0 x→0 Sen 7 x Solução : Solução : Sen ax Sen 10 x Lim f ( x ) = Lim Lim f ( x ) = Lim x →0 x →0 Sen bx x →0 x → 0 Sen 7 x Sen ax Sen 10 x Lim f ( x ) = Lim x Lim f ( x ) = Lim x x →0 x → 0 Sen bx x →0 x →0 Sen 7 x x x Sen ax a Sen 10 x 10 ⋅ ⋅ x a x 10 Lim f ( x ) = Lim Lim f ( x ) = Lim x →0 x → 0 Sen bx b x →0 x →0 Sen 7 x 7 ⋅ ⋅ x b x 7 Sen ax Sen 10 x Lim a ⋅ Lim Lim 10 ⋅ Lim x→0 x→0 ax x→0 x→0 10 x Lim f ( x ) = Lim f ( x) = x →0 Sen bx x →0 Sen 7 x Lim b ⋅ Lim Lim 7 ⋅ Lim x→0 x→0 bx x→0 x→0 7x a ⋅1 10 ⋅ 1 Lim f ( x ) = Lim f ( x) = x →0 b ⋅1 x →0 7 ⋅1 a 10 Lim f ( x ) = Lim f ( x) = x →0 b x →0 7 3 Marcello Santos Chaves Instituto Federal de Educação, Ciência e Tecnologia (IFPA) Belém-PA, Abril de 2011
  • 4. Cálculo I Profº. Marcello Santos Chaves  x + 1 Tg ax Tg 3   9) Lim f ( x ) = Lim 10) Lim f ( x ) = Lim  4  (x + 1)3 x→0 x→0 x x → −1 x → −1 Solução : Solução : Tg ax Lim f ( x ) = Lim  x +1 x →0 x →0 x Tg 3   Sen ax Lim f ( x ) = Lim  4  Cos ax x → −1 x → −1 (x + 1)3 Lim f ( x ) = Lim x →0 x →0 x  x + 1 3 Tg 3   Lim f ( x ) = Lim Sen ax 1 ⋅  4  Lim f ( x ) = Lim x →0 x → 0 Cos ax x x → −1 x → −1 3 (x + 1)3 Sen ax 1 a Lim f ( x ) = Lim ⋅ ⋅  x +1 x →0 x → 0 Cos ax x a Tg   Lim f ( x ) = Lim  4  Lim f ( x ) = Lim a ⋅ Sen ax ⋅ 1 x → −1 x → −1 (x + 1) x →0 x →0 ax Cos ax x +1 Sen ax 1 Faça → u = ∴ x = 4u − 1 Lim f ( x ) = Lim a ⋅ Lim ⋅ Lim 4 x →0 x →0 x→0 ax x → 0 Cos ax Se : x → −1∴ u → π 1 Tg u Lim f ( x ) = a ⋅ 1 ⋅ Lim f ( x ) = Lim x →0 Cos 0 u →π u →π 4u − 1 + 1 1 Tg u Lim f ( x ) = a ⋅ 1 ⋅ Lim f ( x ) = Lim x →0 1 u →π u →π 4u Lim f ( x ) = a Sen u x →0 Cos u Lim f ( x ) = Lim u →π u →π 4u Sen u 1 Lim f ( x ) = Lim ⋅ u →π u →π Cos u 4u Sen u 1 Lim f ( x ) = Lim ⋅ Lim u →π u →π Cos u u →π 4u Sen π 1 Lim f ( x ) = ⋅ u →π Cos π 4π 0 1 Lim f ( x ) = ⋅ u →π − 1 4π Lim f ( x ) = 0 u →π 4 Marcello Santos Chaves Instituto Federal de Educação, Ciência e Tecnologia (IFPA) Belém-PA, Abril de 2011
  • 5. Cálculo I Profº. Marcello Santos Chaves 1 − Cos x 11) Lim f ( x ) = Lim x→0 x→0 x Solução : 1 − Cos x Lim f ( x ) = Lim x →0 x→0 x − 1 ⋅ (Cos x − 1) Lim f ( x ) = Lim x →0 x→0 x Cos x − 1 Lim f ( x ) = Lim − 1 ⋅ Lim x →0 x→0 x→0 x Cos x − 1 Cos x + 1 Lim f ( x ) = Lim − 1 ⋅ Lim ⋅ x →0 x→0 x→0 x Cos x + 1 Cos 2 x − 12 Lim f ( x ) = Lim − 1 ⋅ Lim x →0 x→0 x → 0 x ⋅ (Cos x + 1) Cos 2 x − 1 Lim f ( x ) = Lim − 1 ⋅ Lim x →0 x→0 x → 0 x ⋅ (Cos x + 1) − Sen 2 x Lim f ( x ) = Lim − 1 ⋅ Lim x →0 x→0 x→0 x ⋅ (Cos x + 1) − (Sen x ) ⋅ (Sen x ) Lim f ( x ) = Lim − 1 ⋅ Lim x →0 x→0 x→0 x ⋅ (Cos x + 1) − 1 ⋅ (Sen x ) (Sen x ) Lim f ( x ) = Lim − 1 ⋅ Lim ⋅ Lim x →0 x→0 x→0 x x → 0 (Cos x + 1) Lim f ( x ) = −1 × (− 1) ⋅ Sen 0 x →0 Cos 0 + 1 Lim f ( x ) = −1 × (− 1) ⋅ 0 x →0 1+1 Lim f ( x ) = −1 × (− 1) × 0 x →0 Lim f ( x ) = 0 x →0 5 Marcello Santos Chaves Instituto Federal de Educação, Ciência e Tecnologia (IFPA) Belém-PA, Abril de 2011
  • 6. Cálculo I Profº. Marcello Santos Chaves 1 − Cos x 12) Lim f ( x ) = Lim x→0 x→0 x2 Solução : 1 − Cos x Lim f ( x ) = Lim x →0 x→0 x2 − 1 ⋅ (Cos x − 1) Lim f ( x ) = Lim x →0 x→0 x2 Cos x − 1 Lim f ( x ) = Lim − 1 ⋅ Lim x →0 x→0 x→0 x2 Cos x − 1 Cos x + 1 Lim f ( x ) = Lim − 1 ⋅ Lim ⋅ x →0 x→0 x→0 x2 Cos x + 1 Cos 2 x − 12 Lim f ( x ) = Lim − 1 ⋅ Lim x →0 x→0 x→0 x 2 ⋅ (Cos x + 1) Cos 2 x − 1 Lim f ( x ) = Lim − 1 ⋅ Lim 2 x →0 x→0 x → 0 x ⋅ (Cos x + 1) − Sen 2 x Lim f ( x ) = Lim − 1 ⋅ Lim x →0 x→0 x→0 x 2 ⋅ (Cos x + 1) − (Sen x ) ⋅ (Sen x ) Lim f ( x ) = Lim − 1 ⋅ Lim x →0 x→0 x → 0 x ⋅ x ⋅ (Cos x + 1) Lim f ( x ) = Lim − 1 ⋅ Lim (Sen x ) ⋅ Lim (Sen x ) ⋅ Lim − 1 x →0 x→0 x→0 x x→0 x x → 0 (Cos x + 1) −1 Lim f ( x ) = −1 × 1 × 1 ⋅ x →0 Cos 0 + 1 −1 Lim f ( x ) = −1 ⋅ x →0 1+1  1 Lim f ( x ) = −1 ⋅  −  x →0  2 1 Lim f ( x ) = x →0 2 6 Marcello Santos Chaves Instituto Federal de Educação, Ciência e Tecnologia (IFPA) Belém-PA, Abril de 2011
  • 7. Cálculo I Profº. Marcello Santos Chaves 6 x − Sen 2 x 13) Lim f ( x) = Lim ( x − 3) ⋅ Co sec (πx ) 14) Lim f ( x) = Lim x →3 x →3 x →0 x →0 2 x + 3Sen 4 x Solução : Solução : Lim f ( x) = Lim ( x − 3) ⋅ Co sec (πx ) 6 x − Sen 2 x x →3 x →3 Lim f ( x) = Lim x →0 x→0 2 x + 3Sen 4 x Lim f ( x) = Lim ( x − 3) ⋅ 1 x →3 x →3 Sen (πx ) 2x 6 x − Sen 2 x ⋅ Lim f ( x) = Lim ( x − 3) Lim f ( x) = Lim 2x x →3 Sen (π − πx ) x →0 x→0 4x x →3 2 x + 3Sen 4 x ⋅ Lim f ( x) = Lim ( x − 3) 4x x →3 Sen (3π − πx )  Sen 2 x  x →3 6x − 2x ⋅    2x  ( x − 3) Lim f ( x) = Lim x →0 x→0  Sen 4 x  Lim f ( x) = Lim ( x − 3) 2x + 3 ⋅ 4x ⋅   x →3 x →3 Sen (3π − πx )  4x  ( x − 3) 6x − 2x ⋅   Sen 2 x   1 Lim f ( x) = Lim  2x  Lim f ( x) = Lim x →3 π ⋅ Sen (3π − πx ) x →0 x→0  Sen 4 x  x →3 2 x + 12 x ⋅   π ⋅ ( x − 3)  4x   Sen 2 x  Lim f ( x) = Lim 1 x ⋅ (6 − 2 ) ⋅   x →3 x →3 π ⋅ Sen (3π − πx )  2x  Lim f ( x) = Lim (πx − 3π ) x →0 x→0 x ⋅ (2 + 12 ) ⋅   Sen 4 x   1  4x  Lim f ( x) = Lim Sen (πx − 3π )  Sen 2 x  Lim (6 − 2 ) ⋅ Lim  x →3 x →3 π⋅  (πx − 3π ) Lim f ( x) = x →0 x →0  2x  x →0  Sen 4 x  Lim1 Lim (2 + 12 ) ⋅ Lim   Lim f ( x) = x →3 Sen (πx − 3π )  4x  x →0 x →0 x →3 Lim π ⋅ Lim 6 − 2 ×1 x →3 x →3 (πx − 3π ) Lim f ( x) = 1 x →0 2 + 12 × 1 Lim f ( x) = 4 x →3 π ⋅1 Lim f ( x) = x →0 14 1 Lim f ( x) = 2 x →3 π Lim f ( x) = x →0 7 7 Marcello Santos Chaves Instituto Federal de Educação, Ciência e Tecnologia (IFPA) Belém-PA, Abril de 2011
  • 8. Cálculo I Profº. Marcello Santos Chaves Cos 2 x − Cos 3 x 15) Lim f ( x) = Lim x →0 x →0 x2 Solução : Cos 2 x − Cos 3 x Lim f ( x) = Lim x →0 x →0 x2  2 x + 3x   2 x − 3x  − 2 Sen   ⋅ Sen   Lim f ( x) = Lim  2   2  x →0 x →0 x2  5x   x − 2 Sen   ⋅ Sen  −  Lim f ( x) = Lim  2   2 x →0 x →0 2 x  5x    x  − 2 Sen   ⋅ − Sen    2    2  Lim f ( x) = Lim 2 x →0 x →0 x  5x   x 2 Sen   ⋅ Sen   Lim f ( x) = Lim  2  2 x →0 x →0 2 x  5x   x 2 Sen   Sen   Lim f ( x) = Lim  2 ⋅ 2 x →0 x →0 x x  5x   x  5x   x 2 Sen   Sen   2 Sen   Sen   Lim f ( x) = Lim  2 ⋅  2  ⇒ Lim f ( x) = Lim  2 ⋅ 2 ⇒ x →0 x →0 x 2 x→0 x →0 x  x x⋅  2⋅  2 2  5x  5  x 5  5x   x Sen   ⋅ Sen   ⋅ Sen   Sen   ⇒ Lim f ( x) = Lim  2  2⋅  2  ⇒ Lim f ( x) = Lim 2  2 ⋅ 2 ⇒ x →0 x →0 5  x x →0 x →0  5x   x x⋅     ⋅   2 2  2  2  5x   x Sen   Sen   5 ⇒ Lim f ( x) = Lim ⋅ Lim  2  ⋅ Lim 2 x →0 x →0 2 x →0  5x  x →0  x      2  2 5 ⇒ Lim f ( x) = × 1 × 1 x →0 2 5 ⇒ Lim f ( x) = x →0 2 8 Marcello Santos Chaves Instituto Federal de Educação, Ciência e Tecnologia (IFPA) Belém-PA, Abril de 2011
  • 9. Cálculo I Profº. Marcello Santos Chaves 1 − 2Cos x + Cos 2 x 16) Lim f ( x) = Lim x →0 x →0 x2 Solução : 1 − 2Cos x + Cos 2 x Lim f ( x) = Lim x →0 x →0 x2 1 − 2Cos x + 1 − 2 Sen 2 x Lim f ( x) = Lim x →0 x →0 x2 2 − 2Cos x − 2 Sen 2 x Lim f ( x) = Lim x →0 x →0 x2 Lim f ( x) = Lim [ ] 2 ⋅ 1 − Cos x − Sen 2 x x →0 x →0 x2   x  2 ⋅ 2 Sen 2   − Sen 2 x   2 Lim f ( x) = Lim  2  x →0 x →0 x  x 4 Sen 2   − 2 Sen 2 x Lim f ( x) = Lim 2 x →0 x →0 x2  x 4 Sen 2   2 Lim f ( x) = Lim  2  − 2 Sen x x →0 x →0 x2 x2  x 4 Sen 2   2 Lim f ( x) = Lim  2  − 2 Sen x x →0 x →0 x⋅x x2  x 4 Sen 2   2 Lim f ( x) = Lim  2  − 2 Sen x x →0 x →0 2 2 x2 x ⋅ ⋅ x ⋅  2 2  x x 4 Sen 2   2 4 Sen 2   2 Lim f ( x) = Lim  2  − 2 Sen x ⇒ Lim f ( x) = Lim  2  − 2 Sen x ⇒ x →0 x →0  x  x x2 x →0 x→0  x 2 x2 2⋅ ⋅ 2⋅  4⋅  2 2 2  x Sen 2   2 ⇒ Lim f ( x) = Lim  2  − Lim 2 ⋅ Lim Sen x 2 x →0 x →0  x x →0 x →0 x2   2 ⇒ Lim f ( x) = 1 − 2 x →0 ⇒ Lim f ( x) = −1 x →0 9 Marcello Santos Chaves Instituto Federal de Educação, Ciência e Tecnologia (IFPA) Belém-PA, Abril de 2011