Cálculo I: Exercícios Resolvidos - 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