Gabarito - Lista 2 - Cálculo I (4)

Prévia do material em texto

Gabarito - Lista 2
a) 2
b) 3
c) 4
d) 6
e) 7
f) 8
g) 9
h) 10
i) 12
j) 14
l) 15
m) 17
n) 18
o) 20
p) 22
q) 24
1
a)
𝑥 0
lim
→
𝑡𝑔 (𝑥)
𝑥
𝑥 0
lim
→
𝑡𝑔 (𝑥)
𝑥 =
𝑥 0
lim
→
𝑠𝑒𝑛 (𝑥)
𝑥 * 𝑐𝑜𝑠(𝑥) =
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) * (
1
𝑐𝑜𝑠 (𝑥) )
(usando o Limite Fundamental)
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) 
● se x > 0
𝑠𝑒𝑛 (𝑥) ≤ 𝑥 ≤ 𝑡𝑔 (𝑥) ⇒ 1 ≤ 𝑥𝑠𝑒𝑛 (𝑥) ≤
1
𝑐𝑜𝑠 (𝑥) ⇒ 1 ≥
𝑠𝑒𝑛 (𝑥)
𝑥 ≥ 𝑐𝑜𝑠 (𝑥)
Já que 1=1 e cos (x) = 1
𝑥 0
lim
→ 𝑥 0
lim
→
Pelo Teorema do Confronto,
𝑥 0+
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) = 1
● se x < 0
𝑠𝑒𝑛 (𝑥) ≥ 𝑥 ≥ 𝑡𝑔 (𝑥) ⇒ 1 ≥ 𝑥𝑠𝑒𝑛 (𝑥) ≥
1
𝑐𝑜𝑠 (𝑥) ⇒ 1 ≤
𝑠𝑒𝑛 (𝑥)
𝑥 ≤ 𝑐𝑜𝑠 (𝑥)
Já que 1=1 e cos (x) = 1
𝑥 0
lim
→ 𝑥 0
lim
→
Pelo Teorema do Confronto,
𝑥 0−
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) = 1
Então, já que
𝑥 0+
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) =
𝑥 0−
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) = 1, 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) = 1
𝑥 0
lim
→
 ( 1𝑐𝑜𝑠 (𝑥) ) = 
1
𝑐𝑜𝑠 (0) = 1
𝑆𝑒 
𝑥 𝑎
lim
→
 𝑓(𝑥) 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚, 𝑒𝑛𝑡ã𝑜 
𝑥 𝑎
lim
→
 ( 𝑓(𝑥) * 𝑔(𝑥)) =
𝑥 𝑎
lim
→
 𝑓(𝑥) *
𝑥 𝑎
lim
→
 𝑔(𝑥) 
𝑥 0
lim
→
𝑡𝑔 (𝑥)
𝑥 =
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) * (
1
𝑐𝑜𝑠 (𝑥) ) =
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) *
𝑥 0
lim
→
 ( 1𝑐𝑜𝑠 (𝑥) ) = 1 * 1 = 1 
2
Logo 1
𝑥 0
lim
→
𝑡𝑔 (𝑥)
𝑥 =
_____________________________________________________________________
3
b)
𝑥 0
lim
→
𝑥
𝑠𝑒𝑛 (𝑥)
(usando o Limite Fundamental)
● se x > 0
𝑠𝑒𝑛 (𝑥) ≤ 𝑥 ≤ 𝑡𝑔 (𝑥) ⇒ 1 ≤ 𝑥𝑠𝑒𝑛 (𝑥) ≤
1
𝑐𝑜𝑠 (𝑥)
Já que 1=1 e = 1
𝑥 0
lim
→ 𝑥 0
lim
→
1
𝑐𝑜𝑠 (𝑥)
Pelo Teorema do Confronto,
𝑥 0+
lim
→
 ( 𝑥𝑠𝑒𝑛 (𝑥) ) = 1
● se x < 0
𝑠𝑒𝑛 (𝑥) ≥ 𝑥 ≥ 𝑡𝑔 (𝑥) ⇒ 1 ≥ 𝑥𝑠𝑒𝑛 (𝑥) ≥
1
𝑐𝑜𝑠 (𝑥)
Já que 1=1 e = 1
𝑥 0
lim
→ 𝑥 0
lim
→
1
𝑐𝑜𝑠 (𝑥)
Pelo Teorema do Confronto,
𝑥 0−
lim
→
 ( 𝑥𝑠𝑒𝑛 (𝑥) ) = 1
Então, já que
𝑥 0+
lim
→
 ( 𝑥𝑠𝑒𝑛 (𝑥) ) =
𝑥 0−
lim
→
 ( 𝑥𝑠𝑒𝑛 (𝑥) ) = 1,
𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 
𝑥 0
lim
→
 ( 𝑥𝑠𝑒𝑛 (𝑥) ) = 1
_____________________________________________________________________
4
c)
𝑥 0
lim
→
𝑠𝑒𝑛 (3𝑥)
𝑥
● 𝑠𝑒𝑛 (3𝑥) = 𝑠𝑒𝑛 (2𝑥) * 𝑐𝑜𝑠 (𝑥) + 𝑠𝑒𝑛 (𝑥) * 𝑐𝑜𝑠 (2𝑥) 
= 2 * 𝑠𝑒𝑛(𝑥) * 𝑐𝑜𝑠2 (𝑥) + 𝑠𝑒𝑛(𝑥) * 𝑐𝑜𝑠2 (𝑥) − 𝑠𝑒𝑛3(𝑥)
= 3 * 𝑠𝑒𝑛(𝑥) * 𝑐𝑜𝑠2 (𝑥) − 𝑠𝑒𝑛3(𝑥)
= 3 * 𝑠𝑒𝑛(𝑥) * (1 − 𝑠𝑒𝑛2 (𝑥)) − 𝑠𝑒𝑛3(𝑥)
= 3 * 𝑠𝑒𝑛(𝑥) − 4 * 𝑠𝑒𝑛3(𝑥)
= 𝑠𝑒𝑛 (𝑥) * (3 − 4 * 𝑠𝑒𝑛2(𝑥))
𝑥 0
lim
→
𝑠𝑒𝑛 (3𝑥)
𝑥 =
𝑥 0
lim
→
𝑠𝑒𝑛 (𝑥) *(3−4*𝑠𝑒𝑛2(𝑥))
𝑥 =
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) * (3 − 4 * 𝑠𝑒𝑛
2(𝑥))
(usando o Limite Fundamental)
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) 
● se x > 0
𝑠𝑒𝑛 (𝑥) ≤ 𝑥 ≤ 𝑡𝑔 (𝑥) ⇒ 1 ≤ 𝑥𝑠𝑒𝑛 (𝑥) ≤
1
𝑐𝑜𝑠 (𝑥) ⇒ 1 ≥
𝑠𝑒𝑛 (𝑥)
𝑥 ≥ 𝑐𝑜𝑠 (𝑥)
Já que 1=1 e cos (x) = 1
𝑥 0
lim
→ 𝑥 0
lim
→
Pelo Teorema do Confronto,
𝑥 0+
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) = 1
● se x < 0
𝑠𝑒𝑛 (𝑥) ≥ 𝑥 ≥ 𝑡𝑔 (𝑥) ⇒ 1 ≥ 𝑥𝑠𝑒𝑛 (𝑥) ≥
1
𝑐𝑜𝑠 (𝑥) ⇒ 1 ≤
𝑠𝑒𝑛 (𝑥)
𝑥 ≤ 𝑐𝑜𝑠 (𝑥)
Já que 1=1 e cos (x) = 1
𝑥 0
lim
→ 𝑥 0
lim
→
Pelo Teorema do Confronto,
𝑥 0−
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) = 1
Então, já que
𝑥 0+
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) =
𝑥 0−
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) = 1,
𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) = 1
5
𝑥 0
lim
→
(3 − 4 * 𝑠𝑒𝑛2(𝑥)) = 3 − 4 * 𝑠𝑒𝑛2(0) = 3 − 4 * 0 = 3
𝑆𝑒 
𝑥 𝑎
lim
→
 𝑓(𝑥) 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚, 𝑒𝑛𝑡ã𝑜 
𝑥 𝑎
lim
→
 ( 𝑓(𝑥) * 𝑔(𝑥)) =
𝑥 𝑎
lim
→
 𝑓(𝑥) *
𝑥 𝑎
lim
→
 𝑔(𝑥) 
𝑥 0
lim
→
𝑠𝑒𝑛 (3𝑥)
𝑥 =
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) * (3 − 4 * 𝑠𝑒𝑛
2(𝑥)) =
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) *
𝑥 0
lim
→
(3 − 4 * 𝑠𝑒𝑛2(𝑥)) = 1 * 3 = 3
Logo
𝑥 0
lim
→
𝑠𝑒𝑛 (3𝑥)
𝑥 = 3
_____________________________________________________________________
6
d)
𝑥 π
lim
→
𝑠𝑒𝑛 (𝑥)
𝑥−π
(usando o Limite Fundamental)
● se 𝑥 > π ⇒ 𝑥 − π > 0
𝑠𝑒𝑛 (𝑥 − π) ≤ 𝑥 − π ≤ 𝑡𝑔 (𝑥 − π) ⇒ 1 ≤ 𝑥−π𝑠𝑒𝑛 (𝑥−π) ≤
1
𝑐𝑜𝑠 (𝑥−π) ⇒ 1 ≥
𝑠𝑒𝑛 (𝑥−π)
𝑥−π ≥ 𝑐𝑜𝑠 (𝑥 − π)
obs.: 𝑠𝑒𝑛 (𝑥 − π) = 𝑠𝑒𝑛 (𝑥) * 𝑐𝑜𝑠 (π) − 𝑠𝑒𝑛 (π) * 𝑐𝑜𝑠 (𝑥) = − 𝑠𝑒𝑛(𝑥)
1 ≥ 𝑠𝑒𝑛 (𝑥−π)𝑥−π ≥ 𝑐𝑜𝑠 (𝑥 − π) ⇒ 1 ≥
−𝑠𝑒𝑛 (𝑥)
𝑥−π ≥ 𝑐𝑜𝑠 (𝑥 − π) ⇒ − 1 ≤
𝑠𝑒𝑛 (𝑥)
𝑥−π ≤− 𝑐𝑜𝑠 (𝑥 − π)
Já que -1=-1 e = -1
𝑥 π
lim
→ 𝑥 π
lim
→
− 𝑐𝑜𝑠 (𝑥 − π)
Pelo Teorema do Confronto,
𝑥 π+
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥−π ) =− 1
● se x < π ⇒ 𝑥 − π < 0
𝑠𝑒𝑛 (𝑥 − π) ≥ 𝑥 − π ≥ 𝑡𝑔 (𝑥 − π) ⇒ 1 ≥ 𝑥−π𝑠𝑒𝑛 (𝑥−π) ≥
1
𝑐𝑜𝑠 (𝑥−π) ⇒ 1 ≤
𝑠𝑒𝑛 (𝑥−π)
𝑥−π ≤ 𝑐𝑜𝑠 (𝑥 − π)
obs.: 𝑠𝑒𝑛 (𝑥 − π) = 𝑠𝑒𝑛 (𝑥) * 𝑐𝑜𝑠 (π) − 𝑠𝑒𝑛 (π) * 𝑐𝑜𝑠 (𝑥) = − 𝑠𝑒𝑛(𝑥)
1 ≤ 𝑠𝑒𝑛 (𝑥−π)𝑥−π ≤ 𝑐𝑜𝑠 (𝑥 − π) ⇒ 1 ≤
−𝑠𝑒𝑛 (𝑥)
𝑥−π ≤ 𝑐𝑜𝑠 (𝑥 − π) ⇒ − 1 ≥
𝑠𝑒𝑛 (𝑥)
𝑥−π ≥− 𝑐𝑜𝑠 (𝑥 − π)
Já que -1=-1 e cos (x ) = -1
𝑥 π
lim
→ 𝑥 π
lim
→
− − π
Pelo Teorema do Confronto,
𝑥 π−
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥−π ) =− 1
Então, já que
𝑥 π+
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥−π ) =
𝑥 π−
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥−π ) =− 1,
𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 
𝑥 π
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥−π ) = − 1
_____________________________________________________________________
7
e)
𝑥 0
lim
→
𝑥2
𝑠𝑒𝑛 (𝑥)
(usando o Limite Fundamental)
● se x > 0
𝑠𝑒𝑛 (𝑥) ≤ 𝑥 ≤ 𝑡𝑔 (𝑥) ⇒ 1 ≤ 𝑥𝑠𝑒𝑛 (𝑥) ≤
1
𝑐𝑜𝑠 (𝑥) ⇒ 𝑥 ≤
𝑥2
𝑠𝑒𝑛 (𝑥) ≤
𝑥
𝑐𝑜𝑠 (𝑥)
Já que x=0 e = 0
𝑥 0
lim
→ 𝑥 0
lim
→
𝑥
𝑐𝑜𝑠 (𝑥)
Pelo Teorema do Confronto,
𝑥 0+
lim
→
 ( 𝑥
2
𝑠𝑒𝑛 (𝑥) ) = 0
● se x < 0
𝑠𝑒𝑛 (𝑥) ≥ 𝑥 ≥ 𝑡𝑔 (𝑥) ⇒ 1 ≥ 𝑥𝑠𝑒𝑛 (𝑥) ≥
1
𝑐𝑜𝑠 (𝑥) ⇒ 𝑥 ≥
𝑥2
𝑠𝑒𝑛 (𝑥) ≥
𝑥
𝑐𝑜𝑠 (𝑥)
Já que x=0 e = 0
𝑥 0
lim
→ 𝑥 0
lim
→
𝑥
𝑐𝑜𝑠 (𝑥)
Pelo Teorema do Confronto,
𝑥 0−
lim
→
 ( 𝑥
2
𝑠𝑒𝑛 (𝑥) ) = 0
Então, já que
𝑥 0+
lim
→
 ( 𝑥
~2
𝑠𝑒𝑛 (𝑥) ) =
𝑥 0−
lim
→
 ( 𝑥
2
𝑠𝑒𝑛 (𝑥) ) = 0,
𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 
𝑥 0
lim
→
 ( 𝑥
2
𝑠𝑒𝑛 (𝑥) ) = 0
_____________________________________________________________________
8
f)
𝑥 0
lim
→
3𝑥2
𝑡𝑔 (𝑥) * 𝑠𝑒𝑛 (𝑥)
3𝑥2
𝑡𝑔 (𝑥) * 𝑠𝑒𝑛 (𝑥) =
3𝑥2* 𝑐𝑜𝑠 (𝑥)
 𝑠𝑒𝑛2 (𝑥)
= ( 𝑥
2
 𝑠𝑒𝑛2 (𝑥)
) * (3 * 𝑐𝑜𝑠(𝑥)) = ( 𝑥
 𝑠𝑒𝑛 (𝑥)
) * ( 𝑥
 𝑠𝑒𝑛 (𝑥)
) * (3 * 𝑐𝑜𝑠(𝑥))
(usando o Limite Fundamental)
𝑥 0
lim
→
𝑥
 𝑠𝑒𝑛 (𝑥)
● se x > 0
𝑠𝑒𝑛 (𝑥) ≤ 𝑥 ≤ 𝑡𝑔 (𝑥) ⇒ 1 ≤ 𝑥𝑠𝑒𝑛 (𝑥) ≤
1
𝑐𝑜𝑠 (𝑥)
Já que 1=1 e = 1
𝑥 0
lim
→ 𝑥 0
lim
→
1
𝑐𝑜𝑠 (𝑥)
Pelo Teorema do Confronto,
𝑥 0+
lim
→
 ( 𝑥𝑠𝑒𝑛 (𝑥) ) = 1
● se x < 0
𝑠𝑒𝑛 (𝑥) ≥ 𝑥 ≥ 𝑡𝑔 (𝑥) ⇒ 1 ≥ 𝑥𝑠𝑒𝑛 (𝑥) ≥
1
𝑐𝑜𝑠 (𝑥)
Já que 1=1 e = 1
𝑥 0
lim
→ 𝑥 0
lim
→
1
𝑐𝑜𝑠 (𝑥)
Pelo Teorema do Confronto,
𝑥 0−
lim
→
 ( 𝑥𝑠𝑒𝑛 (𝑥) ) = 1
Então, já que
𝑥 0+
lim
→
 ( 𝑥𝑠𝑒𝑛 (𝑥) ) =
𝑥 0−
lim
→
 ( 𝑥𝑠𝑒𝑛 (𝑥) ) = 1,
𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 
𝑥 0
lim
→
 ( 𝑥𝑠𝑒𝑛 (𝑥) ) = 1
𝑥 0
lim
→
3 * 𝑐𝑜𝑠 (𝑥) = 3 * 𝑐𝑜𝑠 (0) = 3
𝑆𝑒 
𝑥 𝑎
lim
→
 𝑓(𝑥) 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚, 𝑒𝑛𝑡ã𝑜 
𝑥 𝑎
lim
→
 ( 𝑓(𝑥) * 𝑔(𝑥)) =
𝑥 𝑎
lim
→
 𝑓(𝑥) *
𝑥 𝑎
lim
→
 𝑔(𝑥) 
𝑥 0
lim
→
3𝑥2
𝑡𝑔 (𝑥) * 𝑠𝑒𝑛 (𝑥) =
𝑥 0
lim
→
 ( 𝑥
2
 𝑠𝑒𝑛2 (𝑥)
) * (3 * 𝑐𝑜𝑠(𝑥)) =
𝑥 0
lim
→
 ( 𝑥
2
 𝑠𝑒𝑛2 (𝑥)
) *
𝑥 0
lim
→
 (3 * 𝑐𝑜𝑠(𝑥))
9
𝑥 0
lim
→
3𝑥2
𝑡𝑔 (𝑥) * 𝑠𝑒𝑛 (𝑥) =
𝑥 0
lim
→
 ( 𝑥
 𝑠𝑒𝑛 (𝑥)
) *
𝑥 0
lim
→
 ( 𝑥 𝑠𝑒𝑛 (𝑥) ) *
𝑥 0
lim
→
 (3 * 𝑐𝑜𝑠(𝑥)) = 1 * 1 * 3 = 3
Logo
𝑥 0
lim
→
3𝑥2
𝑡𝑔 (𝑥) * 𝑠𝑒𝑛 (𝑥) = 3
_____________________________________________________________________
g)
𝑥 0
lim
→
𝑡𝑔 (3𝑥)
𝑠𝑒𝑛 (4𝑥)
𝑡𝑔 (3𝑥)
𝑠𝑒𝑛 (4𝑥) =
𝑠𝑒𝑛(3𝑥)
𝑐𝑜𝑠(3𝑥) * 𝑠𝑒𝑛 (4𝑥) =
𝑠𝑒𝑛 (𝑥) *𝑐𝑜𝑠 (2𝑥) + 𝑠𝑒𝑛 (2𝑥) * 𝑐𝑜𝑠 (𝑥)
(𝑐𝑜𝑠 (2𝑥)*𝑐𝑜𝑠 (𝑥) − 𝑠𝑒𝑛 (2𝑥)*𝑠𝑒𝑛 (𝑥))* (2* 𝑠𝑒𝑛 (2𝑥) * 𝑐𝑜𝑠 (2𝑥))
= 𝑠𝑒𝑛 (𝑥) *(𝑐𝑜𝑠
2 (𝑥) − 𝑠𝑒𝑛2(𝑥)) + 2*𝑠𝑒𝑛 (𝑥)* 𝑐𝑜𝑠2 (𝑥)
(𝑐𝑜𝑠 (2𝑥)*𝑐𝑜𝑠 (𝑥) − 𝑠𝑒𝑛 (2𝑥)*𝑠𝑒𝑛 (𝑥))* (2* 𝑠𝑒𝑛 (2𝑥) * 𝑐𝑜𝑠 (2𝑥))
= 𝑠𝑒𝑛 (𝑥) *(𝑐𝑜𝑠
2 (𝑥) − 𝑠𝑒𝑛2(𝑥)) + 2*𝑠𝑒𝑛 (𝑥)* 𝑐𝑜𝑠2 (𝑥)
(𝑐𝑜𝑠 (2𝑥)*𝑐𝑜𝑠 (𝑥) − 𝑠𝑒𝑛 (2𝑥)*𝑠𝑒𝑛(𝑥))* (4* 𝑠𝑒𝑛 (𝑥)*𝑐𝑜𝑠(𝑥) * 𝑐𝑜𝑠 (2𝑥))
= (𝑐𝑜𝑠
2 (𝑥) − 𝑠𝑒𝑛2(𝑥)) + 2* 𝑐𝑜𝑠2 (𝑥)
(𝑐𝑜𝑠 (2𝑥)*𝑐𝑜𝑠 (𝑥) − 𝑠𝑒𝑛 (2𝑥)*𝑠𝑒𝑛 (𝑥))* (4*𝑐𝑜𝑠(𝑥) * 𝑐𝑜𝑠 (2𝑥))
𝑥 0
lim
→
𝑡𝑔 (3𝑥)
𝑠𝑒𝑛 (4𝑥) =
𝑥 0
lim
→
(𝑐𝑜𝑠2 (𝑥) − 𝑠𝑒𝑛2(𝑥)) + 2* 𝑐𝑜𝑠2 (𝑥)
(𝑐𝑜𝑠 (2𝑥)*𝑐𝑜𝑠 (𝑥) − 𝑠𝑒𝑛 (2𝑥)*𝑠𝑒𝑛 (𝑥))* (4*𝑐𝑜𝑠(𝑥) * 𝑐𝑜𝑠 (2𝑥)) =
= (𝑐𝑜𝑠
2 (0) − 𝑠𝑒𝑛2(0)) + 2* 𝑐𝑜𝑠2 (0)
(𝑐𝑜𝑠 (0)*𝑐𝑜𝑠 (0) − 𝑠𝑒𝑛 (0)*𝑠𝑒𝑛 (0))* (4*𝑐𝑜𝑠(0)* 𝑐𝑜𝑠 (0)) =
(1 − 0) + 2*1
(1*1− 0*0)* (4*1*1) =
3
4
Logo
𝑥 0
lim
→
𝑡𝑔 (3𝑥)
𝑠𝑒𝑛 (4𝑥) =
3
4
_____________________________________________________________________
10
h)
𝑥 0
lim
→
1−𝑐𝑜𝑠 (𝑥)
𝑥
● 1−𝑐𝑜𝑠 (𝑥)𝑥 =
1 − 𝑐𝑜𝑠2 (𝑥)
𝑥*(1+𝑐𝑜𝑠 (𝑥)) =
𝑠𝑒𝑛2(𝑥)
𝑥*(1+𝑐𝑜𝑠 (𝑥)) = (
1
(1+𝑐𝑜𝑠 (𝑥)) ) * (
𝑠𝑒𝑛2(𝑥)
𝑥 )
𝑥 0
lim
→
1−𝑐𝑜𝑠 (𝑥)
𝑥 =
𝑥 0
lim
→
( 1(1+𝑐𝑜𝑠 (𝑥)) ) * (
𝑠𝑒𝑛2(𝑥)
𝑥 )
𝑥 0
lim
→
𝑠𝑒𝑛2(𝑥)
𝑥
● se x > 0
𝑠𝑒𝑛 (𝑥) ≤ 𝑥 ≤ 𝑡𝑔 (𝑥) ⇒ 1 ≤ 𝑥𝑠𝑒𝑛 (𝑥) ≤
1
𝑐𝑜𝑠 (𝑥) ⇒ 1 ≥
𝑠𝑒𝑛 (𝑥)
𝑥 ≥ 𝑐𝑜𝑠 (𝑥)
1 ≤ 𝑥𝑠𝑒𝑛 (𝑥) ≤
1
𝑐𝑜𝑠 (𝑥) ⇒ 1 ≥
𝑠𝑒𝑛 (𝑥)
𝑥 ≥ 𝑐𝑜𝑠 (𝑥) ⇒ 𝑠𝑒𝑛(𝑥) ≥
𝑠𝑒𝑛2 (𝑥)
𝑥 ≥ 𝑐𝑜𝑠 (𝑥) * 𝑠𝑒𝑛(𝑥)
Já que sen(x)=0 e cos (x)*sen(x) = 0
𝑥 0
lim
→ 𝑥 0
lim
→
Pelo Teorema do Confronto,
𝑥 0+
lim
→
 ( 𝑠𝑒𝑛
2 (𝑥)
𝑥 ) = 0
● se x < 0
𝑠𝑒𝑛 (𝑥) ≥ 𝑥 ≥ 𝑡𝑔 (𝑥) ⇒ 1 ≥ 𝑥𝑠𝑒𝑛 (𝑥) ≥
1
𝑐𝑜𝑠 (𝑥) ⇒ 1 ≤
𝑠𝑒𝑛 (𝑥)
𝑥 ≤ 𝑐𝑜𝑠 (𝑥)
1 ≥ 𝑥𝑠𝑒𝑛 (𝑥) ≥
1
𝑐𝑜𝑠 (𝑥) ⇒ 1 ≤
𝑠𝑒𝑛 (𝑥)
𝑥 ≤ 𝑐𝑜𝑠 (𝑥) ⇒ 𝑠𝑒𝑛(𝑥) ≤
𝑠𝑒𝑛2 (𝑥)
𝑥 ≤ 𝑐𝑜𝑠 (𝑥) * 𝑠𝑒𝑛(𝑥)
Já que =0 e cos (x) = 0
𝑥 0
lim
→
𝑠𝑒𝑛(𝑥)
𝑥 0
lim
→
* 𝑠𝑒𝑛(𝑥)
Pelo Teorema do Confronto,
𝑥 0−
lim
→
 ( 𝑠𝑒𝑛
2 (𝑥)
𝑥 ) = 0
Então, já que
𝑥 0+
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) =
𝑥 0−
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) = 0,
𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) = 0
11
𝑥 0
lim
→
1
(1+𝑐𝑜𝑠 (𝑥)) =
1
(1+𝑐𝑜𝑠 (0)) =
1
2
𝑆𝑒 
𝑥 𝑎
lim
→
 𝑓(𝑥) 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚, 𝑒𝑛𝑡ã𝑜 
𝑥 𝑎
lim
→
 ( 𝑓(𝑥) * 𝑔(𝑥)) =
𝑥 𝑎
lim
→
 𝑓(𝑥) *
𝑥 𝑎
lim
→
 𝑔(𝑥) 
𝑥 0
lim
→
1−𝑐𝑜𝑠 (𝑥)
𝑥 =
𝑥 0
lim
→
( 1(1+𝑐𝑜𝑠 (𝑥)) ) * (
𝑠𝑒𝑛2(𝑥)
𝑥 ) =
𝑥 0
lim
→
( 1(1+𝑐𝑜𝑠 (𝑥)) ) *
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) =
1
2 * 0 = 0
Logo
𝑥 0
lim
→
1−𝑐𝑜𝑠 (𝑥)
𝑥 = 0
_____________________________________________________________________
12
i)
𝑥 π2
lim
→
1−𝑠𝑒𝑛 (𝑥)
2𝑥−π
● 1−𝑠𝑒𝑛 (𝑥)2𝑥−π =
12−𝑠𝑒𝑛2 (𝑥)
(2𝑥−π)(1+𝑠𝑒𝑛 (𝑥)) =
𝑐𝑜𝑠2(𝑥)
2𝑥−π *
1
1+𝑠𝑒𝑛 (𝑥)
𝑥 π/2
lim
→
1−𝑠𝑒𝑛 (𝑥)
2𝑥−π =
𝑥 π/2
lim
→
 ( 𝑐𝑜𝑠
2(𝑥)
2𝑥−π *
1
1+𝑠𝑒𝑛 (𝑥) )
𝑥 π/2
lim
→
 11+𝑠𝑒𝑛 (𝑥) = 
1
1+𝑠𝑒𝑛 (π/2) =
1
2
𝑥 π/2
lim
→
 𝑐𝑜𝑠
2(𝑥)
2𝑥−π
● se 𝑥 > π/2 ⇒ 𝑥 − π2 > 0
𝑠𝑒𝑛 (𝑥 − π2 ) ≤ 𝑥 −
π
2 ≤ 𝑡𝑔 (𝑥 −
π
2 ) ⇒ 1 ≤
𝑥− π2
𝑠𝑒𝑛 (𝑥− π2 )
≤ 1
𝑐𝑜𝑠 (𝑥− π2 )
obs.: 𝑠𝑒𝑛 (𝑥 − π2 ) = 𝑠𝑒𝑛 (𝑥) * 𝑐𝑜𝑠 (π/2) − 𝑠𝑒𝑛 (π/2) * 𝑐𝑜𝑠 (𝑥) = − 𝑐𝑜𝑠(𝑥)
1 ≤
𝑥− π2
𝑠𝑒𝑛 (𝑥− π2 )
≤ 1
𝑐𝑜𝑠 (𝑥− π2 )
⇒ 2 ≤ 2𝑥−π−𝑐𝑜𝑠(𝑥) ≤
2
𝑐𝑜𝑠 (𝑥− π2 )
⇒ − 2 ≤ 𝑐𝑜𝑠(𝑥)2𝑥−π ≤
−𝑐𝑜𝑠 (𝑥− π2 )
2
− 2 ≤ 𝑐𝑜𝑠(𝑥)2𝑥−π ≤
−𝑐𝑜𝑠 (𝑥− π2 )
2 ⇒ − 2𝑐𝑜𝑠 (𝑥) ≤
𝑐𝑜𝑠2(𝑥)
2𝑥−π ≤
−𝑐𝑜𝑠 (𝑥− π2 )*𝑐𝑜𝑠(𝑥)
2
Já que = 0 e = 0
𝑥 π/2
lim
→
− 2𝑐𝑜𝑠 (𝑥)
𝑥 π 
lim
→
−𝑐𝑜𝑠 (𝑥− π2 )*𝑐𝑜𝑠(𝑥)
2
Pelo Teorema do Confronto,
𝑥 (π/2)+
lim
→
 ( 𝑐𝑜𝑠
2(𝑥)
𝑥−(π/2) ) = 0
● se 𝑥 < π/2 ⇒ 𝑥 − π2 > 0
13
𝑠𝑒𝑛 (𝑥 − π2 ) ≥ 𝑥 −
π
2 ≥ 𝑡𝑔 (𝑥 −
π
2 ) ⇒ 1 ≥
𝑥− π2
𝑠𝑒𝑛 (𝑥− π2 )
≥ 1
𝑐𝑜𝑠 (𝑥− π2 )
obs.: 𝑠𝑒𝑛 (𝑥 − π2 ) = 𝑠𝑒𝑛 (𝑥) * 𝑐𝑜𝑠 (π/2) − 𝑠𝑒𝑛 (π/2) * 𝑐𝑜𝑠 (𝑥) = − 𝑐𝑜𝑠(𝑥)
1 ≥
𝑥− π2
𝑠𝑒𝑛 (𝑥− π2 )
≥ 1
𝑐𝑜𝑠 (𝑥− π2 )
⇒ 2 ≥ 2𝑥−π−𝑐𝑜𝑠(𝑥) ≥
2
𝑐𝑜𝑠 (𝑥− π2 )
⇒ − 2 ≥ 𝑐𝑜𝑠(𝑥)2𝑥−π ≥
−𝑐𝑜𝑠 (𝑥− π2 )
2
− 2 ≥ 𝑐𝑜𝑠(𝑥)2𝑥−π ≥
−𝑐𝑜𝑠 (𝑥− π2 )
2 ⇒ − 2𝑐𝑜𝑠 (𝑥) ≥
𝑐𝑜𝑠2(𝑥)
2𝑥−π ≥
−𝑐𝑜𝑠 (𝑥− π2 )*𝑐𝑜𝑠(𝑥)
2
Já que = 0 e = 0
𝑥 π/2
lim
→
− 2𝑐𝑜𝑠 (𝑥)
𝑥 π 
lim
→
−𝑐𝑜𝑠 (𝑥− π2 )*𝑐𝑜𝑠(𝑥)
2
Pelo Teorema do Confronto,
𝑥 (π/2)−
lim
→
 ( 𝑐𝑜𝑠
2(𝑥)
𝑥−(π/2) ) = 0
Então, já que
𝑥 (π/2)+
lim
→
 ( 𝑐𝑜𝑠
2(𝑥)
𝑥−(π/2) ) =
𝑥 (π/2)−
lim
→
 ( 𝑐𝑜𝑠
2(𝑥)
𝑥−(π/2) ) = 0,
𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒
𝑥 (π/2)
lim
→
 ( 𝑐𝑜𝑠
2(𝑥)
𝑥−(π/2) ) = 0
𝑆𝑒 
𝑥 𝑎
lim
→
 𝑓(𝑥) 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚, 𝑒𝑛𝑡ã𝑜 
𝑥 𝑎
lim
→
 ( 𝑓(𝑥) * 𝑔(𝑥)) =
𝑥 𝑎
lim
→
 𝑓(𝑥) *
𝑥 𝑎
lim
→
 𝑔(𝑥) 
𝑥 π/2
lim
→
1−𝑠𝑒𝑛 (𝑥)
2𝑥−π =
𝑥 π/2
lim
→
 ( 𝑐𝑜𝑠
2(𝑥)
2𝑥−π *
1
1+𝑠𝑒𝑛 (𝑥) ) =
𝑥 π/2
lim
→
 ( 𝑐𝑜𝑠
2(𝑥)
2𝑥−π ) *
𝑥 π/2
lim
→
 11+𝑠𝑒𝑛 (𝑥) =
1
2 * 0 = 0
_____________________________________________________________________
14
j)
𝑥 0
lim
→
𝑥 * 𝑠𝑒𝑛 (1/𝑥)
𝑃𝑒𝑙𝑜 𝐶𝑜𝑟𝑜𝑙á𝑟𝑖𝑜 𝑑𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝐶𝑜𝑛𝑓𝑟𝑜𝑛𝑡𝑜
𝐽á 𝑞𝑢𝑒 
𝑥 0
lim
→
 𝑥 = 0 𝑒 𝑠𝑒𝑛 (1/𝑥) é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑙𝑖𝑚𝑖𝑡𝑎𝑑𝑎 
𝑥 0
lim
→
𝑥 * 𝑠𝑒𝑛 (1/𝑥) = 0
_____________________________________________________________________
15
l)
𝑥 𝑝
lim
→
𝑡𝑔(𝑥−𝑝)
𝑥2−𝑝2
 , 𝑝 ≠ 0
𝑥 𝑝
lim
→
𝑡𝑔(𝑥−𝑝)
𝑥2−𝑝2
=
𝑥 𝑝
lim
→
𝑡𝑔(𝑥−𝑝)
𝑥 −𝑝
* 1
𝑥 + 𝑝
𝑥 𝑝
lim
→
1
𝑥 + 𝑝
= 12 𝑝
𝑥 𝑝
lim
→
𝑡𝑔(𝑥−𝑝)
𝑥 −𝑝
=
𝑦 0
lim
→
𝑡𝑔(𝑦)
𝑦 ==
𝑦 0
lim
→
𝑠𝑒𝑛 (𝑦)
𝑥 * 𝑐𝑜𝑠(𝑦) =
𝑦 0
lim
→
 ( 𝑠𝑒𝑛 (𝑦)𝑦 ) * (
1
𝑐𝑜𝑠 (𝑦) )
𝑦 0
lim
→
 ( 𝑠𝑒𝑛 (𝑦)𝑦 ) 
● se y > 0
𝑠𝑒𝑛 (𝑦) ≤ 𝑦 ≤ 𝑡𝑔 (𝑦) ⇒ 1 ≤ 𝑦𝑠𝑒𝑛 (𝑦) ≤
1
𝑐𝑜𝑠 (𝑦) ⇒ 1 ≥
𝑠𝑒𝑛 (𝑦)
𝑦 ≥ 𝑐𝑜𝑠 (𝑦)
Já que 1=1 e cos (y) = 1
𝑦 0
lim
→ 𝑦 0
lim
→
Pelo Teorema do Confronto,
𝑦 0+
lim
→
 ( 𝑠𝑒𝑛 (𝑦)𝑦 ) = 1
● se y < 0
𝑠𝑒𝑛 (𝑦) ≥ 𝑦 ≥ 𝑡𝑔 (𝑦) ⇒ 1 ≥ 𝑦𝑠𝑒𝑛 (𝑦) ≥
1
𝑐𝑜𝑠 (𝑦) ⇒ 1 ≤
𝑠𝑒𝑛 (𝑦)
𝑦 ≤ 𝑐𝑜𝑠 (𝑦)
Já que 1=1 e cos (y) = 1
𝑦 0
lim
→ 𝑦 0
lim
→
Pelo Teorema do Confronto,
𝑦 0−
lim
→
 ( 𝑠𝑒𝑛 (𝑦)𝑦 ) = 1
Então, já que
𝑦 0+
lim
→
 ( 𝑠𝑒𝑛 (𝑦)𝑦 ) =
𝑦 0−
lim
→
 ( 𝑠𝑒𝑛 (𝑦)𝑦 ) = 1,
𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 
𝑦 0
lim
→
 ( 𝑠𝑒𝑛 (𝑦)𝑦 ) = 1
16
𝑦 0
lim
→
 ( 1𝑐𝑜𝑠 (𝑦) ) = 
1
𝑐𝑜𝑠 (0) = 1
𝑆𝑒 
𝑥 𝑎
lim
→
 𝑓(𝑥) 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚, 𝑒𝑛𝑡ã𝑜 
𝑥 𝑎
lim
→
 ( 𝑓(𝑥) * 𝑔(𝑥)) =
𝑥 𝑎
lim
→
 𝑓(𝑥) *
𝑥 𝑎
lim
→
 𝑔(𝑥) 
𝑥 𝑝
lim
→
𝑡𝑔(𝑥−𝑝)
𝑥2−𝑝2
=
𝑥 𝑝
lim
→
𝑡𝑔(𝑥−𝑝)
𝑥 −𝑝
* 1
𝑥 + 𝑝
=
𝑦 0
lim
→
𝑡𝑔 (𝑦)
𝑦 *
𝑥 𝑝
lim
→
1
𝑥 + 𝑝
=
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) * (
1
𝑐𝑜𝑠 (𝑥) ) *
𝑥 𝑝
lim
→
1
𝑥 + 𝑝
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) * (
1
𝑐𝑜𝑠 (𝑥) ) *
𝑥 𝑝
lim
→
1
𝑥 + 𝑝
 =
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) *
𝑥 0
lim
→
 ( 1𝑐𝑜𝑠 (𝑥) ) *
𝑥 𝑝
lim
→
1
𝑥 + 𝑝
 = 1 * 1 * 12 𝑝 =
1
2 𝑝 
Logo
𝑥 𝑝
lim
→
𝑡𝑔(𝑥−𝑝)
𝑥2−𝑝2
= 12 𝑝
_____________________________________________________________________
17
m)
𝑥 𝑝
lim
→
𝑠𝑒𝑛(𝑥2−𝑝2)
𝑥 −𝑝
𝑥 𝑝
lim
→
𝑠𝑒𝑛(𝑥2−𝑝2)
𝑥 −𝑝
=
𝑥 𝑝
lim
→
𝑠𝑒𝑛(𝑥2−𝑝2)*(𝑥+𝑝)
𝑥2 −𝑝2
=
𝑥 𝑝
lim
→
𝑠𝑒𝑛(𝑥2−𝑝2)
𝑥
2
−𝑝2
* (𝑥 + 𝑝)
𝑥 𝑝
lim
→
𝑥 + 𝑝 = 2𝑝
𝑥 𝑝
lim
→
𝑠𝑒𝑛(𝑥2−𝑝2)
𝑥
2
−𝑝2
● se > 0𝑥
2
− 𝑝2
𝑠𝑒𝑛 (𝑥
2
− 𝑝2 ) ≤ 𝑥
2
− 𝑝2 ≤ 𝑡𝑔 (𝑥
2
− 𝑝2 ) ⇒ 1 ≤ 𝑥
2
−𝑝2
𝑠𝑒𝑛 (𝑥
2
−𝑝2 )
≤ 1
𝑐𝑜𝑠 (𝑥
2
−𝑝2 )
⇒ 1 ≥ 𝑠𝑒𝑛 (𝑥
2
−𝑝2 )
𝑥
2
−𝑝2
≥ 𝑐𝑜𝑠 (𝑥
2
− 𝑝2 )
Já que 1=1 e cos ( ) = 1
𝑥 𝑝
lim
→ 𝑥 𝑝
lim
→
𝑥
2
− 𝑝2
Pelo Teorema do Confronto,
𝑥 𝑝+
lim
→
 ( 𝑠𝑒𝑛 (𝑥
2
−𝑝2 )
𝑥
2
−𝑝2
) = 1
● se < 0𝑥
2
− 𝑝2
𝑠𝑒𝑛 (𝑥
2
− 𝑝2 ) ≥ 𝑥
2
− 𝑝2 ≥ 𝑡𝑔 (𝑥
2
− 𝑝2 ) ⇒ 1 ≥ 𝑥
2
−𝑝2
𝑠𝑒𝑛 (𝑥
2
−𝑝2 )
≥ 1
𝑐𝑜𝑠 (𝑥
2
−𝑝2 )
⇒ 1 ≤ 𝑠𝑒𝑛 (𝑥
2
−𝑝2 )
𝑥
2
−𝑝2
≤ 𝑐𝑜𝑠 (𝑥
2
− 𝑝2 )
Já que 1=1 e cos ( ) = 1
𝑥 𝑝
lim
→ 𝑥 𝑝
lim
→
𝑥
2
− 𝑝2
Pelo Teorema do Confronto,
𝑥 𝑝−
lim
→
 ( 𝑠𝑒𝑛 (𝑥
2
−𝑝2 )
𝑥
2
−𝑝2
) = 1
Então, já que
𝑥 𝑝+
lim
→
 ( 𝑠𝑒𝑛 (𝑥
2
−𝑝2 )
𝑥
2
−𝑝2
) =
𝑥 𝑝−
lim
→
 ( 𝑠𝑒𝑛 (𝑥
2
−𝑝2 )
𝑥
2
−𝑝2
) = 1,
𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 
𝑥 𝑝
lim
→
 ( 𝑠𝑒𝑛 (𝑥
2
−𝑝2 )
𝑥
2
−𝑝2
) = 1
𝑆𝑒 
𝑥 𝑎
lim
→
 𝑓(𝑥) 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚, 𝑒𝑛𝑡ã𝑜 
𝑥 𝑎
lim
→
 ( 𝑓(𝑥) * 𝑔(𝑥))=
𝑥 𝑎
lim
→
 𝑓(𝑥) *
𝑥 𝑎
lim
→
 𝑔(𝑥) 
18
= =
𝑥 𝑝
lim
→
𝑠𝑒𝑛(𝑥2−𝑝2)
𝑥 −𝑝 𝑥 𝑝
lim
→
𝑠𝑒𝑛(𝑥2−𝑝2)
𝑥
2
−𝑝2
* (𝑥 + 𝑝)
𝑥 𝑝
lim
→
𝑠𝑒𝑛(𝑥2−𝑝2)
𝑥
2
−𝑝2
*
𝑥 𝑝
lim
→
 (𝑥 + 𝑝) = 1 * 2𝑝 = 2𝑝
Logo
𝑥 𝑝
lim
→
𝑠𝑒𝑛(𝑥2−𝑝2)
𝑥 −𝑝
= 2𝑝
_____________________________________________________________________
n)
𝑥 0
lim
→
𝑠𝑒𝑛(𝑥2− 1𝑥 )−𝑠𝑒𝑛(
1
𝑥 )
𝑥
𝑥 0
lim
→
𝑠𝑒𝑛(𝑥2− 1𝑥 )−𝑠𝑒𝑛(
1
𝑥 )
𝑥
=
𝑥 0
lim
→
𝑠𝑒𝑛(𝑥2)*𝑐𝑜𝑠( 1𝑥 )−𝑠𝑒𝑛(
1
𝑥 )*𝑐𝑜𝑠(𝑥
2)−𝑠𝑒𝑛( 1𝑥 )
𝑥
=
𝑥 0
lim
→
𝑠𝑒𝑛(𝑥2)*𝑐𝑜𝑠( 1𝑥 )+𝑠𝑒𝑛(
1
𝑥 )*(𝑐𝑜𝑠(𝑥
2)−1)
𝑥
𝑥 0
lim
→
𝑠𝑒𝑛(𝑥2)*𝑐𝑜𝑠( 1𝑥 )+𝑠𝑒𝑛(
1
𝑥 )*(𝑐𝑜𝑠(𝑥
2)−1)
𝑥
=
𝑥 0
lim
→
𝑠𝑒𝑛(𝑥2)
𝑥 * 𝑐𝑜𝑠(
1
𝑥 ) + 𝑠𝑒𝑛(
1
𝑥 ) *
(𝑐𝑜𝑠(𝑥2)−1)
𝑥
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥
2)
𝑥 ) 
● 𝑆𝑒 𝑥 > 0
𝑠𝑒𝑛 (𝑥2) ≤ 𝑥
2
≤ 𝑡𝑔 (𝑥2) ⇒ 1 ≤ 𝑥
2
𝑠𝑒𝑛 (𝑥2)
≤ 1
𝑐𝑜𝑠 (𝑥2)
⇒ 𝑥 ≥ 𝑠𝑒𝑛 (𝑥
2)
𝑥 ≥ 𝑥 * 𝑐𝑜𝑠 (𝑥
2)
Já que x=0 e = 0
𝑥 0
lim
→ 𝑥 0
lim
→
𝑥 * 𝑐𝑜𝑠(𝑥2)
Pelo Teorema do Confronto,
𝑥 0+
lim
→
 ( 𝑠𝑒𝑛 (𝑥
2)
𝑥 ) = 0
● se x < 0
𝑠𝑒𝑛 (𝑥2) ≥ 𝑥2 ≥ 𝑡𝑔 (𝑥2) ⇒ 1 ≥ 𝑥
2
𝑠𝑒𝑛 (𝑥2)
≥ 1
𝑐𝑜𝑠 (𝑥2)
⇒ 𝑥 ≤ 𝑠𝑒𝑛 (𝑥
2)
𝑥 ≤ 𝑥 * 𝑐𝑜𝑠 (𝑥
2)
Já que x=0 e = 0
𝑥 0
lim
→ 𝑥 0
lim
→
𝑥 * 𝑐𝑜𝑠(𝑥2)
19
Pelo Teorema do Confronto,
𝑥 0−
lim
→
 ( 𝑠𝑒𝑛 (𝑥
2)
𝑥 ) = 0
Então, já que
𝑥 0+
lim
→
 ( 𝑠𝑒𝑛 (𝑥
2)
𝑥 ) =
𝑥 0−
lim
→
 ( 𝑠𝑒𝑛 (𝑥
2)
𝑥 ) = 0,
𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥
2)
𝑥 ) = 0
𝑃𝑒𝑙𝑜 𝐶𝑜𝑟𝑜𝑙á𝑟𝑖𝑜 𝑑𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝐶𝑜𝑛𝑓𝑟𝑜𝑛𝑡𝑜
𝐽á 𝑞𝑢𝑒 
𝑥 0
lim
→
𝑠𝑒𝑛 (𝑥2)
𝑥 = 0 𝑒 𝑐𝑜𝑠 (1/𝑥) é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑙𝑖𝑚𝑖𝑡𝑎𝑑𝑎 
𝑥 0
lim
→
𝑠𝑒𝑛 (𝑥2)
𝑥 * 𝑐𝑜𝑠 (1/𝑥) = 0
𝑥 0
lim
→
(𝑐𝑜𝑠(𝑥2)−1)
𝑥 =
𝑥 0
lim
→
(𝑐𝑜𝑠2(𝑥2)−1)
𝑥*(𝑐𝑜𝑠(𝑥2)+1)
=
𝑥 0
lim
→
(𝑠𝑒𝑛(𝑥2))
𝑥 *
1
(𝑐𝑜𝑠(𝑥2)+1)
𝑥 0
lim
→
1
(𝑐𝑜𝑠(𝑥2)+1)
= 1/2
𝑆𝑒 
𝑥 𝑎
lim
→
 𝑓(𝑥) 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚, 𝑒𝑛𝑡ã𝑜 
𝑥 𝑎
lim
→
 ( 𝑓(𝑥) * 𝑔(𝑥)) =
𝑥 𝑎
lim
→
 𝑓(𝑥) *
𝑥 𝑎
lim
→
 𝑔(𝑥) 
𝑥 0
lim
→
(𝑐𝑜𝑠(𝑥2)−1)
𝑥 =
𝑥 0
lim
→
(𝑠𝑒𝑛(𝑥2))
𝑥 *
1
(𝑐𝑜𝑠(𝑥2)+1)
=
𝑥 0
lim
→
(𝑠𝑒𝑛(𝑥2))
𝑥 *
𝑥 0
lim
→
1
(𝑐𝑜𝑠(𝑥2)+1)
= 0 * (1/2) = 0
𝑃𝑒𝑙𝑜 𝐶𝑜𝑟𝑜𝑙á𝑟𝑖𝑜 𝑑𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝐶𝑜𝑛𝑓𝑟𝑜𝑛𝑡𝑜
𝐽á 𝑞𝑢𝑒 
𝑥 0
lim
→
(𝑐𝑜𝑠(𝑥2)−1)
𝑥 = 0 𝑒 𝑠𝑒𝑛 (1/𝑥) é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑙𝑖𝑚𝑖𝑡𝑎𝑑𝑎 
𝑥 0
lim
→
𝑠𝑒𝑛( 1𝑥 ) *
(𝑐𝑜𝑠(𝑥2)−1)
𝑥 = 0
𝑆𝑒 
𝑥 𝑎
lim
→
 𝑓(𝑥) 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚, 𝑒𝑛𝑡ã𝑜 
𝑥 𝑎
lim
→
 ( 𝑓(𝑥) + 𝑔(𝑥)) =
𝑥 𝑎
lim
→
 𝑓(𝑥) +
𝑥 𝑎
lim
→
 𝑔(𝑥) 
𝑥 0
lim
→
𝑠𝑒𝑛(𝑥2− 1𝑥 )−𝑠𝑒𝑛(
1
𝑥 )
𝑥
=
𝑥 0
lim
→
𝑠𝑒𝑛(𝑥2)
𝑥 * 𝑐𝑜𝑠(
1
𝑥 ) + 𝑠𝑒𝑛(
1
𝑥 ) *
(𝑐𝑜𝑠(𝑥2)−1)
𝑥 =
𝑥 0
lim
→
𝑠𝑒𝑛(𝑥2)
𝑥 * 𝑐𝑜𝑠(
1
𝑥 ) +
𝑥 0
lim
→
𝑠𝑒𝑛( 1𝑥 ) *
(𝑐𝑜𝑠(𝑥2)−1)
𝑥
𝑥 0
lim
→
𝑠𝑒𝑛(𝑥2)
𝑥 * 𝑐𝑜𝑠(
1
𝑥 ) +
𝑥 0
lim
→
𝑠𝑒𝑛( 1𝑥 ) *
(𝑐𝑜𝑠(𝑥2)−1)
𝑥 = 0 + 0 = 0
20
Logo
𝑥 0
lim
→
𝑠𝑒𝑛(𝑥2− 1𝑥 )−𝑠𝑒𝑛(
1
𝑥 )
𝑥
= 0
_____________________________________________________________________
21
o)
𝑥 0
lim
→
𝑥+𝑠𝑒𝑛(𝑥)
𝑥2−𝑠𝑒𝑛(𝑥)
𝑥 0
lim
→
𝑥+𝑠𝑒𝑛(𝑥)
𝑥2−𝑠𝑒𝑛(𝑥)
=
𝑥 0
lim
→
𝑥+𝑠𝑒𝑛(𝑥)
𝑥2−𝑠𝑒𝑛(𝑥)
* 𝑥𝑥 =
𝑥 0
lim
→
1+ 𝑠𝑒𝑛(𝑥)𝑥
𝑥− 𝑠𝑒𝑛(𝑥)𝑥
𝑥 0
lim
→
 (− 𝑠𝑒𝑛 (𝑥)𝑥 ) 
● se x > 0
𝑠𝑒𝑛 (𝑥) ≤ 𝑥 ≤ 𝑡𝑔 (𝑥) ⇒ 1 ≤ 𝑥𝑠𝑒𝑛 (𝑥) ≤
1
𝑐𝑜𝑠 (𝑥) ⇒ − 1 ≤ −
𝑠𝑒𝑛 (𝑥)
𝑥 ≤ − 𝑐𝑜𝑠 (𝑥)
Já que -1=-1 e cos (x) = -1
𝑥 0
lim
→ 𝑥 0
lim
→
−
Pelo Teorema do Confronto,
𝑥 0+
lim
→
 (− 𝑠𝑒𝑛 (𝑥)𝑥 ) =− 1
● se x < 0
𝑠𝑒𝑛 (𝑥) ≥ 𝑥 ≥ 𝑡𝑔 (𝑥) ⇒ 1 ≥ 𝑥𝑠𝑒𝑛 (𝑥) ≥
1
𝑐𝑜𝑠 (𝑥) ⇒ − 1 ≥ −
𝑠𝑒𝑛 (𝑥)
𝑥 ≥ − 𝑐𝑜𝑠 (𝑥)
Já que -1=-1 e -cos (x) = -1
𝑥 0
lim
→ 𝑥 0
lim
→
Pelo Teorema do Confronto,
𝑥 0−
lim
→
 (− 𝑠𝑒𝑛 (𝑥)𝑥 ) =− 1
Então, já que
𝑥 0+
lim
→
 (− 𝑠𝑒𝑛 (𝑥)𝑥 ) =
𝑥 0−
lim
→
 (− 𝑠𝑒𝑛 (𝑥)𝑥 ) =− 1,
𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 
𝑥 0
lim
→
 (− 𝑠𝑒𝑛 (𝑥)𝑥 ) =− 1
𝑆𝑒 
𝑥 𝑎
lim
→
 𝑓(𝑥) 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚, 𝑒𝑛𝑡ã𝑜 
𝑥 𝑎
lim
→
 ( 𝑓(𝑥) + 𝑔(𝑥)) =
𝑥 𝑎
lim
→
 𝑓(𝑥) +
𝑥 𝑎
lim
→
 𝑔(𝑥) 
𝑥 0
lim
→
1 + 𝑠𝑒𝑛(𝑥)𝑥 =
𝑥 0
lim
→
1 −
𝑥 0
lim
→
 − 𝑠𝑒𝑛(𝑥)𝑥 = 1 + 1 = 2
𝑥 0
lim
→
𝑥 − 𝑠𝑒𝑛(𝑥)𝑥 =
𝑥 0
lim
→
𝑥 +
𝑥 0
lim
→
 − 𝑠𝑒𝑛(𝑥)𝑥 = 0 − 1 = − 1
22
𝑆𝑒 
𝑥 𝑎
lim
→
 𝑓(𝑥) 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) ≠ 0, 𝑒𝑛𝑡ã𝑜 
𝑥 𝑎
lim
→
 ( 𝑓(𝑥) / 𝑔(𝑥)) =
𝑥 𝑎
lim
→
 𝑓(𝑥) /
𝑥 𝑎
lim
→
 𝑔(𝑥) 
𝑥 0
lim
→
𝑥−𝑠𝑒𝑛(𝑥)
𝑥2−𝑠𝑒𝑛(𝑥)
=
𝑥 0
lim
→
1− 𝑠𝑒𝑛(𝑥)𝑥
𝑥− 𝑠𝑒𝑛(𝑥)𝑥
=
𝑥 0
lim
→
1 − 𝑠𝑒𝑛(𝑥)𝑥 /
𝑥 0
lim
→
𝑥 − 𝑠𝑒𝑛(𝑥)𝑥 = −
2
1 =− 2
Logo
𝑥 0
lim
→
𝑥−𝑠𝑒𝑛(𝑥)
𝑥2−𝑠𝑒𝑛(𝑥)
=− 2
_____________________________________________________________________
23
p)
𝑥 0
lim
→
𝑥−𝑡𝑔(𝑥)
𝑥+ 𝑡𝑔(𝑥)
𝑥 0
lim
→
𝑥−𝑡𝑔(𝑥)
𝑥+ 𝑡𝑔(𝑥) =
𝑥 0
lim
→
𝑥* 𝑐𝑜𝑠 (𝑥)−𝑠𝑒𝑛(𝑥)
𝑥* 𝑐𝑜𝑠 (𝑥)+𝑠𝑒𝑛(𝑥) =
𝑥 0
lim
→
𝑥* 𝑐𝑜𝑠 (𝑥)−𝑠𝑒𝑛(𝑥)
𝑥* 𝑐𝑜𝑠 (𝑥)+𝑠𝑒𝑛(𝑥) *
𝑥
𝑥 =
𝑥 0
lim
→
𝑐𝑜𝑠(𝑥)− 𝑠𝑒𝑛(𝑥)𝑥
𝑐𝑜𝑠 (𝑥)+ 𝑠𝑒𝑛(𝑥)𝑥
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) 
● se x > 0
𝑠𝑒𝑛 (𝑥) ≤ 𝑥 ≤ 𝑡𝑔 (𝑥) ⇒ 1 ≤ 𝑥𝑠𝑒𝑛 (𝑥) ≤
1
𝑐𝑜𝑠 (𝑥) ⇒ 1 ≥
𝑠𝑒𝑛 (𝑥)
𝑥 ≥ 𝑐𝑜𝑠 (𝑥)
Já que 1=1 e cos (x) = 1
𝑥 0
lim
→ 𝑥 0
lim
→
Pelo Teorema do Confronto,
𝑥 0+
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) = 1
● se x < 0
𝑠𝑒𝑛 (𝑥) ≥ 𝑥 ≥ 𝑡𝑔 (𝑥) ⇒ 1 ≥ 𝑥𝑠𝑒𝑛 (𝑥) ≥
1
𝑐𝑜𝑠 (𝑥) ⇒ 1 ≤
𝑠𝑒𝑛 (𝑥)
𝑥 ≤ 𝑐𝑜𝑠 (𝑥)
Já que 1=1 e cos (x) = 1
𝑥 0
lim
→ 𝑥 0
lim
→
Pelo Teorema do Confronto,
𝑥 0−
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) = 1
Então, já que
𝑥 0+
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) =
𝑥 0−
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) = 1,
𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 
𝑥 0
lim
→
 ( 𝑠𝑒𝑛 (𝑥)𝑥 ) = 1
𝑆𝑒 
𝑥 𝑎
lim
→
 𝑓(𝑥) 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚, 𝑒𝑛𝑡ã𝑜 
𝑥 𝑎
lim
→
 ( 𝑓(𝑥) + 𝑔(𝑥)) =
𝑥 𝑎
lim
→
 𝑓(𝑥) +
𝑥 𝑎
lim
→
 𝑔(𝑥) 
𝑥 0
lim
→
𝑐𝑜𝑠(𝑥) − 𝑠𝑒𝑛(𝑥)𝑥 =
𝑥 0
lim
→
𝑐𝑜𝑠(𝑥) +
𝑥 0
lim
→
 − 𝑠𝑒𝑛(𝑥)𝑥 =
𝑥 0
lim
→
𝑐𝑜𝑠(𝑥) −
𝑥 0
lim
→
 𝑠𝑒𝑛(𝑥)𝑥 = 1 − 1 = 0
24
𝑥 0
lim
→
𝑐𝑜𝑠(𝑥) + 𝑠𝑒𝑛(𝑥)𝑥 =
𝑥 0
lim
→
𝑐𝑜𝑠(𝑥) +
𝑥 0
lim
→
 𝑠𝑒𝑛(𝑥)𝑥 = 1 + 1 = 2
𝑆𝑒 
𝑥 𝑎
lim
→
 𝑓(𝑥) 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) ≠ 0, 𝑒𝑛𝑡ã𝑜 
𝑥 𝑎
lim
→
 ( 𝑓(𝑥) / 𝑔(𝑥)) =
𝑥 𝑎
lim
→
 𝑓(𝑥) /
𝑥 𝑎
lim
→
 𝑔(𝑥) 
𝑥 0
lim
→
𝑥−𝑡𝑔(𝑥)
𝑥+ 𝑡𝑔(𝑥) =
𝑥 0
lim
→
𝑐𝑜𝑠(𝑥)− 𝑠𝑒𝑛(𝑥)𝑥
𝑐𝑜𝑠(𝑥)+ 𝑠𝑒𝑛(𝑥)𝑥
=
𝑥 0
lim
→
𝑐𝑜𝑠(𝑥) − 𝑠𝑒𝑛(𝑥)𝑥 /
𝑥 0
lim
→
𝑐𝑜𝑠(𝑥) + 𝑠𝑒𝑛(𝑥)𝑥 = 
0
2 = 0
Logo
𝑥 0
lim
→
𝑥−𝑡𝑔(𝑥)
𝑥+ 𝑡𝑔(𝑥) = 0
_____________________________________________________________________
25
q)
𝑥 1
lim
→
𝑠𝑒𝑛(π𝑥)
𝑥−1
𝑆𝑒𝑛𝑑𝑜 𝑎 = 𝑥 − 1 ⇒ 𝑥 = 𝑎 + 1
𝑥 1
lim
→
𝑠𝑒𝑛(π𝑥)
𝑥−1 =
𝑎 0
lim
→
𝑠𝑒𝑛(π𝑎+π)
𝑎
𝑠𝑒𝑛(π𝑎 + π) = 𝑠𝑒𝑛(π𝑎) * 𝑐𝑜𝑠(π) + 𝑠𝑒𝑛(π) * 𝑐𝑜𝑠(π𝑎) =− 𝑠𝑒𝑛(π𝑎)
𝑎 0
lim
→
𝑠𝑒𝑛(π𝑎+π)
𝑎 =
𝑎 0
lim
→
−𝑠𝑒𝑛(π𝑎)
𝑎 =
𝑎 0
lim
→
−𝑠𝑒𝑛(π𝑎)
𝑎 *
π
π =
𝑎 0
lim
→
−π*𝑠𝑒𝑛(π𝑎)
π𝑎
𝑎 0
lim
→
 (− 𝑠𝑒𝑛 (π𝑎)π𝑎 ) 
● se a > 0
𝑠𝑒𝑛 (π𝑎) ≤ π𝑎 ≤ 𝑡𝑔 (π𝑎) ⇒ 1 ≤ π𝑎𝑠𝑒𝑛 (π𝑎) ≤
1
𝑐𝑜𝑠 (π𝑎) ⇒ − 1 ≤ −
𝑠𝑒𝑛 (π𝑎)
π𝑎 ≤ − 𝑐𝑜𝑠 (π𝑎)
Já que -1=-1 e cos ( ) = -1
𝑎 0
lim
→ 𝑎 0
lim
→
− π𝑎
Pelo Teorema do Confronto,
𝑎 0+
lim
→
 (− 𝑠𝑒𝑛 (π𝑎)π𝑎 ) =− 1
● se a < 0
𝑠𝑒𝑛 (π𝑎) ≥ π𝑎 ≥ 𝑡𝑔 (π𝑎) ⇒ 1 ≥ π𝑎𝑠𝑒𝑛 (π𝑎) ≥
1
𝑐𝑜𝑠 (π𝑎) ⇒ − 1 ≥ −
𝑠𝑒𝑛 (π𝑎)
π𝑎 ≥ − 𝑐𝑜𝑠 (π𝑎)
Já que -1=-1 e -cos ( ) = -1
𝑎 0
lim
→ 𝑎 0
lim
→
π𝑎
Pelo Teorema do Confronto,
𝑎 0−
lim
→
 (− 𝑠𝑒𝑛 (π𝑎)π𝑎 ) =− 1
Então, já que
𝑎 0+
lim
→
 (− 𝑠𝑒𝑛 (π𝑎)π𝑎 ) =
𝑎 0−
lim
→
 (− 𝑠𝑒𝑛 (π𝑎)π𝑎 ) =− 1,
𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 
π𝑎 0
lim
→
 (− 𝑠𝑒𝑛 (π𝑎)π𝑎 ) =− 1
26
𝑎 0
lim
→
 π = π
𝑆𝑒 
𝑥 𝑎
lim
→
 𝑓(𝑥) 𝑒 
𝑥 𝑎
lim
→
 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚, 𝑒𝑛𝑡ã𝑜 
𝑥 𝑎
lim
→
 ( 𝑓(𝑥) * 𝑔(𝑥)) =
𝑥 𝑎
lim
→
 𝑓(𝑥) *
𝑥 𝑎
lim
→
 𝑔(𝑥) 
= =
𝑥 1
lim
→
𝑠𝑒𝑛(π𝑥)
𝑥−1
𝑎 0
lim→
−π*𝑠𝑒𝑛(π𝑎)
π𝑎
𝑎 0
lim
→
 π *
𝑎 0
lim
→
−𝑠𝑒𝑛(π𝑎)
π𝑎 =− π
Logo
𝑥 1
lim
→
𝑠𝑒𝑛(π𝑥)
𝑥−1 =− π
_____________________________________________________________________
27