Exercicios Calculo (119)

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133 
 
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟐. 
 
𝑎)𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒, 𝑠𝑒 𝑒𝑥𝑖𝑠𝑡𝑖𝑟, lim
𝑥→0
⟦
𝑥2 −1
𝑥2 +1
⟧. 
 
𝑆𝑎𝑏𝑒-𝑠𝑒 𝑞𝑢𝑒 (𝑥2 −1) ≥ −1 𝑒 (𝑥2 +1) ≥ 1,∀𝑥 ∈ ℝ 𝑒 𝑞𝑢𝑒 𝑎𝑚𝑏𝑎𝑠 𝑠ã𝑜 𝑓𝑢𝑛çõ𝑒𝑠 
𝑝𝑎𝑟𝑒𝑠 𝑒, 𝑐𝑜𝑛𝑠𝑒𝑞𝑢𝑒𝑛𝑡𝑒𝑚𝑒𝑛𝑡𝑒 ,𝑜 𝑞𝑢𝑜𝑐𝑖𝑒𝑛𝑡𝑒 𝑡𝑎𝑚𝑏é𝑚 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑝𝑎𝑟.𝐷𝑒𝑠𝑠𝑎 
𝑓𝑜𝑟𝑚𝑎, 𝑡𝑒𝑚𝑜𝑠: 
𝑥2 −1
𝑥2 +1
≥ −1, ∀𝑥 ∈ ℝ 
 
𝑆𝑒𝑛𝑑𝑜 𝑓(𝑥) =
𝑥2 −1
𝑥2 +1
 𝑒 𝑞𝑢𝑒 𝑓(0) = −1, 𝑡𝑒𝑚𝑜𝑠 𝑎 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑎𝑛á𝑙𝑖𝑠𝑒: 
 
𝑠𝑒 𝑥 → 0+ , 𝑒𝑛𝑡ã𝑜 𝑓(𝑥) → −1+ 𝑒,𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, lim
𝑥→0+
⟦𝑓(𝑥)⟧ = −1 𝑒, 
 
𝑠𝑒 𝑥 → 0− , 𝑒𝑛𝑡ã𝑜 𝑓(𝑥) → −1+ 𝑒,𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, lim
𝑥→0−
⟦𝑓(𝑥)⟧ = −1. 
 
𝐶𝑜𝑚𝑜 𝑜𝑠 𝑙𝑖𝑚𝑖𝑡𝑒𝑠 𝑙𝑎𝑡𝑒𝑟𝑎𝑖𝑠 𝑒𝑚 0 𝑒𝑥𝑖𝑠𝑡𝑒𝑚 𝑒 𝑠ã𝑜 𝑖𝑔𝑢𝑎𝑖𝑠, 𝑒𝑛𝑡ã𝑜 lim
𝑥→0
⟦𝑓(𝑥)⟧ 𝑒𝑥𝑖𝑠𝑡𝑒 𝑒 
lim
𝑥→0
⟦𝑓(𝑥)⟧ = −1.𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 
 lim
𝑥→0
⟦
𝑥2 −1
𝑥2 +1
⟧= −1. 
 
𝑏)𝐶𝑎𝑙𝑐𝑢𝑙𝑒 lim
𝑥→−∞
2𝑥 cos𝑥. 
 
∀𝑥 ∈ ℝ, 𝑡𝑒𝑚𝑜𝑠: 
−1 ≤ cos𝑥 ≤ 1 
 
𝐶𝑜𝑚𝑜 2𝑥 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑐𝑖𝑎𝑙 𝑐𝑟𝑒𝑠𝑐𝑒𝑛𝑡𝑒, 𝑒𝑛𝑡ã𝑜 … 
 
−2𝑥⏟
𝑓(𝑥)
≤ 2𝑥 cos𝑥⏟ 
𝑔(𝑥)
≤ 2𝑥⏟
ℎ(𝑥)
 
 
𝑂𝑏𝑠. : 𝑆𝑒 𝑥 → −∞, 𝑒𝑛𝑡ã𝑜 2𝑥 → 0. 𝐿𝑜𝑔𝑜, 
 
lim
𝑥→−∞
𝑓(𝑥) = lim
𝑥→−∞
(−2𝑥) = 0 𝑒 lim
𝑥→−∞
ℎ(𝑥) = lim
𝑥→−∞
(2𝑥) = 0 
 
𝑆𝑒 𝑓(𝑥) ≤ 𝑔(𝑥) ≤ ℎ(𝑥) 𝑞𝑢𝑎𝑛𝑑𝑜 𝑥 → −∞ 𝑒 lim
𝑥→−∞
𝑓(𝑥) = lim
𝑥→−∞
ℎ(𝑥) = 0, 𝑝𝑒𝑙𝑜 
𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝐶𝑜𝑛𝑓𝑟𝑜𝑛𝑡𝑜, lim
𝑥→−∞
𝑔(𝑥) = 0. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 
 
lim
𝑥→−∞
2𝑥 cos𝑥 = 0.