Exercicios Calculo (128)

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142 
 
𝐶𝑜𝑚𝑜 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 𝑓𝑒𝑐ℎ𝑎𝑑𝑜 [−100,0] 𝑒 0 é 𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑒𝑛𝑡𝑟𝑒 𝑓(−100) 
𝑒 𝑓(0),𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝑉𝑎𝑙𝑜𝑟 𝐼𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖á𝑟𝑖𝑜,𝑒𝑥𝑖𝑠𝑡𝑒 𝑎𝑙𝑔𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑐 ∈ (−100,0) 
𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓(𝑐) = 0. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑓 𝑝𝑜𝑠𝑠𝑢𝑖 𝑢𝑚𝑎 𝑟𝑎í𝑧 𝑟𝑒𝑎𝑙. 
 
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟒. 
 
𝑎) 𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑒 lim
𝑥→∞
𝑓(𝑥) , 𝑠𝑒 𝑝𝑎𝑟𝑎 𝑡𝑜𝑑𝑜 𝑥 > 1,
10.2𝑥 − 21
2𝑥+1
 1 
𝑔(𝑥) 1 𝑒 lim
𝑥→∞
𝑔(𝑥) = lim
𝑥→∞
ℎ(𝑥) = 5, 𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 
𝑑𝑜 𝐶𝑜𝑛𝑓𝑟𝑜𝑛𝑡𝑜, lim
𝑥→∞
𝑓(𝑥) = 5. 
 
𝑏)𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑒 𝑎𝑠 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑖𝑠 𝑑𝑒 𝑓(𝑥) = (√𝑥2 +𝑎𝑥 −√𝑥2 + 𝑏𝑥). 
 
𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 𝐿 é 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑎 𝑓𝑢𝑛çã𝑜 𝑓 𝑠𝑒, 
𝑠𝑜𝑚𝑒𝑛𝑡𝑒 𝑠𝑒, 
lim
𝑥→∞
𝑓(𝑥) = 𝐿 𝑜𝑢 lim
𝑥→−∞
𝑓(𝑥) = 𝐿 
 
lim
𝑥→∞
𝑓(𝑥) = lim
𝑥→∞
(√𝑥2 +𝑎𝑥 −√𝑥2 + 𝑏𝑥) 
 = lim
𝑥→∞
[(√𝑥2 + 𝑎𝑥 −√𝑥2 +𝑏𝑥) ∙
√𝑥2 + 𝑎𝑥 + √𝑥2+ 𝑏𝑥
√𝑥2 + 𝑎𝑥 + √𝑥2+ 𝑏𝑥
] 
 = lim
𝑥→∞
𝑥2 + 𝑎𝑥 − (𝑥2 + 𝑏𝑥)
√𝑥2 +𝑎𝑥 +√𝑥2 +𝑏𝑥
 
 = lim
𝑥→∞
𝑥(𝑎 − 𝑏)
√𝑥2 +𝑎𝑥 +√𝑥2 +𝑏𝑥
 
 = lim
𝑥→∞
𝑥(𝑎 − 𝑏)
√𝑥2 (1 +
𝑎
𝑥) +
√𝑥2 (1 +
𝑏
𝑥)
 ; √𝑥2 = |𝑥|. 𝑆𝑒 𝑥 → ∞, 𝑒𝑛𝑡ã𝑜 |𝑥|
= 𝑥. 
 = lim
𝑥→∞
𝑥(𝑎 − 𝑏)
𝑥√1+
𝑎
𝑥
+ 𝑥√1 +
𝑏
𝑥
 
 = lim
𝑥→∞
𝑎 − 𝑏
√1+
𝑎
𝑥 +
√1 +
𝑏
𝑥