Exercicios Calculo (177)

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192 
 
lim
𝑥→−1+
𝑓(𝑥) = lim
𝑥→−1+
(𝑥 − 1)3⏞ 
−8
↑
(𝑥 + 1)2⏟ 
↓
0+
= −∞ 𝑒 lim
𝑥→−1−
𝑓(𝑥) = lim
𝑥→−1−
(𝑥 − 1)3⏞ 
−8
↑
(𝑥 + 1)2⏟ 
↓
0+
= −∞ 
 
𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑥 = −1 é 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑎 𝑓𝑢𝑛çã𝑜 𝑓. 
 
𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑖𝑠: 𝐴 𝑟𝑒𝑡𝑎 𝑦 = 𝐿 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑒, 𝑠𝑜𝑚𝑒𝑛𝑡𝑒 𝑠𝑒, 
 
lim
𝑥→∞
𝑓(𝑥) = 𝐿 𝑜𝑢 lim
𝑥→−∞
𝑓(𝑥) = 𝐿. 
 
lim
𝑥→∞
𝑓(𝑥) = lim
𝑥→∞
(𝑥 − 1)3
(𝑥 + 1)2
= lim
𝑥→∞
3(𝑥 − 1)2
2(𝑥 + 1)
= lim
𝑥→∞
6(𝑥 − 1)
2
= ∞. 
 
lim
𝑥→−∞
𝑓(𝑥) = lim
𝑥→−∞
(𝑥 − 1)3
(𝑥 + 1)2
= lim
𝑥→−∞
3(𝑥 − 1)2
2(𝑥 + 1)
= lim
𝑥→−∞
6(𝑥 − 1)
2
= −∞. 
 
𝐶𝑜𝑚𝑜 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒, 𝑒𝑛𝑡ã𝑜 𝑓 𝑛ã𝑜 𝑝𝑜𝑠𝑠𝑢𝑖 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑖𝑠! 
 
𝑂𝑏𝑙í𝑞𝑢𝑎𝑠: 𝐴 𝑟𝑒𝑡𝑎 𝑦 = 𝑎𝑥 + 𝑏 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑜𝑏𝑙í𝑞𝑢𝑎 𝑠𝑒,𝑠𝑜𝑚𝑒𝑛𝑡𝑒 𝑠𝑒, 
 
lim
𝑥→±∞
[𝑓(𝑥)− (𝑎𝑥 + 𝑏)] = 0; 
 
𝑓(𝑥) =
(𝑥 − 1)3
(𝑥 + 1)2
=
𝑥3 −3𝑥2 + 3𝑥 − 1
𝑥2 + 2𝑥 + 1
= (𝑥 − 5) +
12𝑥 + 4
𝑥2 + 2𝑥 + 1
 
 
lim
𝑥→±∞
[𝑓(𝑥) − (𝑥 − 5)] = lim
𝑥→±∞
12𝑥 + 4
𝑥2 +2𝑥 + 1
= lim
𝑥→±∞
12
2𝑥 + 2
= 0. 
 
𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 𝑥 − 5 é 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑜𝑏𝑙í𝑞𝑢𝑎 𝑑𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑎 𝑓𝑢𝑛çã𝑜 𝑓. 
 
(𝑖𝑣)𝑂𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑑𝑒 𝑚á𝑥𝑖𝑚𝑜 𝑒 𝑑𝑒 𝑚í𝑛𝑖𝑚𝑜 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑜𝑠 𝑒 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜𝑠, 𝑠𝑒 𝑒𝑥𝑖𝑠𝑡𝑖𝑟𝑒𝑚: 
 
𝑅𝑒𝑠𝑔𝑎𝑡𝑎𝑛𝑑𝑜 𝑜 𝑒𝑠𝑡𝑢𝑑𝑜 𝑑𝑜 𝑠𝑖𝑛𝑎𝑙 𝑑𝑎 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎: 
 
+ ++(−5) −− −−(−1) ++ ++ + (1) + ++ ++ ++ 𝑓 ′(𝑥) 
 
𝑃𝑒𝑙𝑜 𝑇𝑒𝑠𝑡𝑒 𝑑𝑎 𝑃𝑟𝑖𝑚𝑒𝑖𝑟𝑎 𝐷𝑒𝑟𝑖𝑣𝑎𝑑𝑎, 𝑜 𝑝𝑜𝑛𝑡𝑜 𝑑𝑒 𝑎𝑏𝑠𝑐𝑖𝑠𝑠𝑎 𝑥 = −5 é 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 
𝑑𝑒 𝑚á𝑥𝑖𝑚𝑜 𝑙𝑜𝑐𝑎𝑙. 𝐶𝑜𝑛𝑡𝑢𝑑𝑜, 𝑐𝑜𝑛𝑓𝑜𝑟𝑚𝑒 𝑎𝑛𝑎𝑙𝑖𝑠𝑎𝑑𝑜 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟𝑚𝑒𝑛𝑡𝑒, lim
𝑥→∞
𝑓(𝑥) = ∞ 
𝑒 lim
𝑥→−∞
𝑓(𝑥) = −∞.𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑓 𝑛ã𝑜 𝑝𝑜𝑠𝑠𝑢𝑖 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑚á𝑥𝑖𝑚𝑜𝑠 𝑜𝑢 𝑚í𝑛𝑖𝑚𝑜𝑠 
𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜𝑠. 
𝑓(−5) =
(−5− 1)3
(−5+ 1)2
=
−63
16
=
−(23.33)
16
=
−8× 27
16
= −
27
2