Cálculo I                                 Profº. Marcello Santos Chaves       LIMITES TRIGONOMÉTRICOS                     ...
Cálculo I                              Profº. Marcello Santos Chaves                           Tgx                        ...
Cálculo I                                       Profº. Marcello Santos Chaves                                             ...
Cálculo I                                  Profº. Marcello Santos Chaves                                                  ...
Cálculo I                                  Profº. Marcello Santos Chaves                             1 − Cos x11)    Lim f...
Cálculo I                                   Profº. Marcello Santos Chaves                             1 − Cos x12)     Lim...
Cálculo I                             Profº. Marcello Santos Chaves                                                       ...
Cálculo I                                 Profº. Marcello Santos Chaves                           Cos 2 x − Cos 3 x15) Lim...
Cálculo I                           Profº. Marcello Santos Chaves                           1 − 2Cos x + Cos 2 x16) Lim f ...
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Profº Marcelo Santos Chaves Cálculo I (limites trigonométricos)

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Cálculo I: Exercícios Resolvidos - Limites Trigonométricos.

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Profº Marcelo Santos Chaves Cálculo I (limites trigonométricos)

  1. 1. Cálculo I Profº. Marcello Santos Chaves LIMITES TRIGONOMÉTRICOS senx Fundamental: Lim =1 x →0 x Sen 3 x1) Lim f ( x ) = Lim sen 2 x 2) Lim f ( x ) = Lim x →0 x →0 Sen 4 x x→0 x→0 xSolução : Solução : Sen 3 xLim f ( x ) = Lim sen 2 x Lim f ( x ) = Lim x →0 x → 0 Sen 4 xx→0 x→0 x sen 2 x 2 Sen3 xLim f ( x ) = Lim ⋅ xx→0 x→0 x 2 Lim f ( x ) = Lim x →0 x → 0 Sen 4 x sen 2 xLim f ( x ) = Lim 2 ⋅ xx→0 x→0 2x Sen3 x 3 sen 2 x ⋅Lim f ( x ) = Lim 2 ⋅ Lim x 3x→0 x→0 x→0 2x Lim f ( x ) = Lim x →0 x → 0 Sen 4 x 4Lim f ( x) = 2 × 1 ⋅x→0 x 4Lim f ( x ) = 2 Sen 3 xx→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. 2. Cálculo I Profº. Marcello Santos Chaves Tgx Cos θ − 13) Lim f ( x ) = Lim 4) Lim f ( x ) = Lim x→0 x→0 x x →θ x →θ θSolução : Solução : Tgx Cos θ − 1Lim f ( x ) = Lim Lim f ( x ) = Limx →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 + 1Lim f ( x ) = Lim Cosxx →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 ) = LimLim 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 ) = LimLim f ( x) = 1 ⋅ x →θ x →θ θ ⋅ (Cos θ + 1)x →0 Cos 0Lim 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 θx →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. 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 ) = Lim7) 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 ) = LimLim f ( x ) = Lim x →0 x →0 Sen bxx →0 x → 0 Sen 7 x Sen ax Sen 10 x Lim f ( x ) = Lim xLim f ( x ) = Lim x x →0 x → 0 Sen bxx →0 x →0 Sen 7 x x x Sen ax a Sen 10 x 10 ⋅ ⋅ x a x 10 Lim f ( x ) = LimLim f ( x ) = Lim x →0 x → 0 Sen bx bx →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 bxx →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 ⋅1x →0 7 ⋅1 a 10 Lim f ( x ) =Lim f ( x) = x →0 bx →0 7 3 Marcello Santos Chaves Instituto Federal de Educação, Ciência e Tecnologia (IFPA) Belém-PA, Abril de 2011
  4. 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 → −1Solução : Solução : Tg axLim 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)3Lim f ( x ) = Limx →0 x →0 x  x + 1 3 Tg 3  Lim f ( x ) = Lim Sen ax 1 ⋅  4  Lim f ( x ) = Limx →0 x → 0 Cos ax x x → −1 x → −1 3 (x + 1)3 Sen ax 1 aLim 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 − 1Lim f ( x ) = Lim a ⋅ Lim ⋅ Lim 4x →0 x →0 x→0 ax x → 0 Cos ax Se : x → −1∴ u → π 1 Tg uLim f ( x ) = a ⋅ 1 ⋅ Lim f ( x ) = Limx →0 Cos 0 u →π u →π 4u − 1 + 1 1 Tg uLim f ( x ) = a ⋅ 1 ⋅ Lim f ( x ) = Limx →0 1 u →π u →π 4uLim f ( x ) = a Sen ux →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. 5. Cálculo I Profº. Marcello Santos Chaves 1 − Cos x11) Lim f ( x ) = Lim x→0 x→0 xSolução : 1 − Cos xLim f ( x ) = Limx →0 x→0 x − 1 ⋅ (Cos x − 1)Lim f ( x ) = Limx →0 x→0 x Cos x − 1Lim f ( x ) = Lim − 1 ⋅ Limx →0 x→0 x→0 x Cos x − 1 Cos x + 1Lim f ( x ) = Lim − 1 ⋅ Lim ⋅x →0 x→0 x→0 x Cos x + 1 Cos 2 x − 12Lim f ( x ) = Lim − 1 ⋅ Limx →0 x→0 x → 0 x ⋅ (Cos x + 1) Cos 2 x − 1Lim f ( x ) = Lim − 1 ⋅ Limx →0 x→0 x → 0 x ⋅ (Cos x + 1) − Sen 2 xLim f ( x ) = Lim − 1 ⋅ Limx →0 x→0 x→0 x ⋅ (Cos x + 1) − (Sen x ) ⋅ (Sen x )Lim f ( x ) = Lim − 1 ⋅ Limx →0 x→0 x→0 x ⋅ (Cos x + 1) − 1 ⋅ (Sen x ) (Sen x )Lim f ( x ) = Lim − 1 ⋅ Lim ⋅ Limx →0 x→0 x→0 x x → 0 (Cos x + 1)Lim f ( x ) = −1 × (− 1) ⋅ Sen 0x →0 Cos 0 + 1Lim f ( x ) = −1 × (− 1) ⋅ 0x →0 1+1Lim f ( x ) = −1 × (− 1) × 0x →0Lim f ( x ) = 0x →0 5 Marcello Santos Chaves Instituto Federal de Educação, Ciência e Tecnologia (IFPA) Belém-PA, Abril de 2011
  6. 6. Cálculo I Profº. Marcello Santos Chaves 1 − Cos x12) Lim f ( x ) = Lim x→0 x→0 x2Solução : 1 − Cos xLim f ( x ) = Limx →0 x→0 x2 − 1 ⋅ (Cos x − 1)Lim f ( x ) = Limx →0 x→0 x2 Cos x − 1Lim f ( x ) = Lim − 1 ⋅ Limx →0 x→0 x→0 x2 Cos x − 1 Cos x + 1Lim f ( x ) = Lim − 1 ⋅ Lim ⋅x →0 x→0 x→0 x2 Cos x + 1 Cos 2 x − 12Lim f ( x ) = Lim − 1 ⋅ Limx →0 x→0 x→0 x 2 ⋅ (Cos x + 1) Cos 2 x − 1Lim f ( x ) = Lim − 1 ⋅ Lim 2x →0 x→0 x → 0 x ⋅ (Cos x + 1) − Sen 2 xLim f ( x ) = Lim − 1 ⋅ Limx →0 x→0 x→0 x 2 ⋅ (Cos x + 1) − (Sen x ) ⋅ (Sen x )Lim f ( x ) = Lim − 1 ⋅ Limx →0 x→0 x → 0 x ⋅ x ⋅ (Cos x + 1)Lim f ( x ) = Lim − 1 ⋅ Lim (Sen x ) ⋅ Lim (Sen x ) ⋅ Lim − 1x →0 x→0 x→0 x x→0 x x → 0 (Cos x + 1) −1Lim f ( x ) = −1 × 1 × 1 ⋅x →0 Cos 0 + 1 −1Lim f ( x ) = −1 ⋅x →0 1+1  1Lim f ( x ) = −1 ⋅  − x →0  2 1Lim f ( x ) =x →0 2 6 Marcello Santos Chaves Instituto Federal de Educação, Ciência e Tecnologia (IFPA) Belém-PA, Abril de 2011
  7. 7. Cálculo I Profº. Marcello Santos Chaves 6 x − Sen 2 x13) 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 xSolução : Solução :Lim f ( x) = Lim ( x − 3) ⋅ Co sec (πx ) 6 x − Sen 2 xx →3 x →3 Lim f ( x) = Lim x →0 x→0 2 x + 3Sen 4 xLim f ( x) = Lim ( x − 3) ⋅ 1x →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 4xx →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 →0x →3 Lim π ⋅ Lim 6 − 2 ×1 x →3 x →3 (πx − 3π ) Lim f ( x) = 1 x →0 2 + 12 × 1Lim f ( x) = 4x →3 π ⋅1 Lim f ( x) = x →0 14 1Lim f ( x) = 2x →3 π Lim f ( x) = x →0 7 7Marcello Santos ChavesInstituto Federal de Educação, Ciência e Tecnologia (IFPA) Belém-PA, Abril de 2011
  8. 8. Cálculo I Profº. Marcello Santos Chaves Cos 2 x − Cos 3 x15) Lim f ( x) = Lim x →0 x →0 x2Solução : Cos 2 x − Cos 3 xLim f ( x) = Limx →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 2x →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. 9. Cálculo I Profº. Marcello Santos Chaves 1 − 2Cos x + Cos 2 x16) Lim f ( x) = Lim x →0 x →0 x2Solução : 1 − 2Cos x + Cos 2 xLim f ( x) = Limx →0 x →0 x2 1 − 2Cos x + 1 − 2 Sen 2 xLim f ( x) = Limx →0 x →0 x2 2 − 2Cos x − 2 Sen 2 xLim f ( x) = Limx →0 x →0 x2Lim f ( x) = Lim [ ] 2 ⋅ 1 − Cos x − Sen 2 xx →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 xLim f ( x) = Lim 2x →0 x →0 x2  x 4 Sen 2   2Lim f ( x) = Lim  2  − 2 Sen xx →0 x →0 x2 x2  x 4 Sen 2   2Lim f ( x) = Lim  2  − 2 Sen xx →0 x →0 x⋅x x2  x 4 Sen 2   2Lim f ( x) = Lim  2  − 2 Sen xx →0 x →0 2 2 x2 x ⋅ ⋅ x ⋅  2 2  x x 4 Sen 2   2 4 Sen 2   2Lim 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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