Exercicios Calculo (52)

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62 
 
lim
𝑥→0+
(𝑥𝑒
1
𝑥) = lim
𝑥→0+
𝑒
1
𝑥
1
𝑥
= lim
𝑥→0+
(−
1
𝑥2
) . 𝑒
1
𝑥
−
1
𝑥2
= lim
𝑥→0+
𝑒
1
𝑥 = ∞ 
 
𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 0 é 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓. 
 
𝐴𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑖𝑠: lim
𝑥→∞
𝑓(𝑥) = 𝐿 𝑜𝑢 lim
𝑥→−∞
𝑓(𝑥) = 𝐿 
 
lim
𝑥→∞
𝑓(𝑥) = lim
𝑥→∞
(𝑥. 𝑒
1
𝑥) = ∞ 
 
lim
𝑥→−∞
𝑓(𝑥) = lim
𝑥→−∞
(𝑥. 𝑒
1
𝑥) = −∞ 
 
∗ 𝑂𝑏𝑠. : 𝑆𝑒 𝑥 → ±∞,𝑒𝑛𝑡ã𝑜 
1
𝑥
→ 0 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜,𝑒
1
𝑥 → 1. 
 
𝐿𝑜𝑔𝑜, 𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓 𝑛ã𝑜 𝑝𝑜𝑠𝑠𝑢𝑖 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑖𝑠. 
 
𝐴𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠 𝑂𝑏𝑙í𝑞𝑢𝑎𝑠: 
 
𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 𝑎𝑥 + 𝑏 é 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑜𝑏𝑙í𝑞𝑢𝑎 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑎 𝑓𝑢𝑛çã𝑜 𝑓 𝑠𝑒, 
𝑠𝑜𝑚𝑒𝑛𝑡𝑒 𝑠𝑒, lim
𝑥→±∞
[𝑓(𝑥)− (𝑎𝑥 + 𝑏)] = 0. 
𝑂𝑛𝑑𝑒 𝑎 = lim
𝑥→±∞
𝑓(𝑥)
𝑥
,𝑐𝑜𝑚 𝑎 ≠ 0, 𝑒 𝑏 = lim
𝑥→±∞
[𝑓(𝑥)− 𝑎𝑥]. 
 
𝑎 = lim
𝑥→±∞
𝑓(𝑥)
𝑥
= lim
𝑥→±∞
𝑥. 𝑒
1
𝑥
𝑥
= lim
𝑥→±∞
𝑒
1
𝑥 = 1. 
 
𝑏 = lim
𝑥→±∞
[𝑓(𝑥) − 𝑥] = lim
𝑥→±∞
[𝑥. 𝑒
1
𝑥 − 𝑥] = lim
𝑥→±∞
[
𝑒
1
𝑥
1
𝑥
− 𝑥] = lim
𝑥→±∞
𝑒
1
𝑥 −1
1
𝑥
; 
∗ 𝑈𝑠𝑎𝑛𝑑𝑜 𝑎 𝑟𝑒𝑔𝑟𝑎 𝑑𝑎 𝐿′𝐻ô𝑠𝑝𝑖𝑡𝑎𝑙… 
𝑏 = lim
𝑥→±∞
𝑒
1
𝑥 −1
1
𝑥
= lim
𝑥→±∞
(−
1
𝑥2
)𝑒
1
𝑥
−
1
𝑥2
= lim
𝑥→±∞
𝑒
1
𝑥 = 1. 
𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 𝑥 + 1 é 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑜𝑏𝑙í𝑞𝑢𝑎 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑎 𝑓𝑢𝑛çã𝑜 𝑓. 
 
(𝑖𝑣) 𝑂𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑑𝑒 𝑚á𝑥𝑖𝑚𝑜 𝑒 𝑑𝑒 𝑚í𝑛𝑖𝑚𝑜 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑜𝑠 𝑒 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜𝑠, 𝑠𝑒 𝑒𝑥𝑖𝑠𝑡𝑖𝑟𝑒𝑚; 
 
𝑃𝑒𝑙𝑜 𝑇𝑒𝑠𝑡𝑒 𝑑𝑎 𝑃𝑟𝑖𝑚𝑒𝑖𝑟𝑎 𝐷𝑒𝑟𝑖𝑣𝑎𝑑𝑎, (1, 𝑓(1)) é 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑑𝑒 𝑚í𝑛𝑖𝑚𝑜 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑜. 
𝑃𝑜𝑛𝑡𝑜 𝑑𝑒 𝑚í𝑛𝑖𝑚𝑜: (1,𝑒). 
 
𝐶𝑜𝑚𝑜 lim
𝑥→−∞
𝑓(𝑥) = −∞ 𝑒 lim
𝑥→∞
𝑓(𝑥) = ∞, 𝑒𝑛𝑡ã𝑜 𝑜 𝑝𝑜𝑛𝑡𝑜 (1,𝑒) 𝑛ã𝑜 é 𝑛𝑒𝑚 𝑝𝑜𝑛𝑡𝑜 
𝑑𝑒 𝑚á𝑥𝑖𝑚𝑜 𝑒 𝑛𝑒𝑚 𝑑𝑒 𝑚í𝑛𝑖𝑚𝑜 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜 𝑑𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓. 𝐿𝑜𝑔𝑜,𝑓 𝑛ã𝑜 𝑝𝑜𝑠𝑠𝑢𝑖 
𝑝𝑜𝑛𝑡𝑜𝑠 𝑑𝑒 𝑚á𝑥𝑖𝑚𝑜𝑢 𝑜𝑢 𝑚í𝑛𝑖𝑚𝑜 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜𝑠.