Exercicios Calculo (44)

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54 
 
− −− −− −− −− −− −− −− −− −− −− −− −− −− −− −− −− −− − 
𝑄𝑢𝑒𝑠𝑡ã𝑜 1𝑏 (𝐼𝑛𝑓𝑜𝑟𝑚𝑎çã𝑜 𝐴𝑑𝑖𝑐𝑖𝑜𝑛𝑎𝑙) 
− −− −− −− −− −− −− −− −− −− −− −− −− −− −− −− −− −− − 
 
∗ 𝑀𝑜𝑠𝑡𝑟𝑎𝑛𝑑𝑜 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 𝑓𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑐𝑖𝑎𝑙 𝑢𝑠𝑎𝑛𝑑𝑜 𝑎 𝑓𝑢𝑛çã𝑜 𝑓(𝑥) = 𝑒𝑥 . 
 
 𝐷𝑎𝑑𝑜 𝑓(𝑥) = 𝑒𝑥 , 𝑡𝑒𝑚𝑜𝑠 𝑓 ′(𝑥) = 𝑒𝑥 𝑒,𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑓(0) = 𝑓 ′(0) = 𝑒0 = 1. 
 
𝑃𝑒𝑙𝑎 𝑑𝑒𝑓𝑖𝑛𝑖çã𝑜 𝑑𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑑𝑒 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜,𝑡𝑒𝑚𝑜𝑠: 
𝑓 ′(0) = lim
ℎ→0
𝑓(0 + ℎ) − 𝑓(0)
ℎ
= lim
ℎ→0
𝑒ℎ −1
ℎ
= 1 
𝐹𝑎ç𝑎𝑚𝑜𝑠 𝑎 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖çã𝑜 𝑥 = 𝑒ℎ −1. 
∗ 𝑂𝑏𝑠. : 𝑆𝑒 ℎ → 0, 𝑒𝑛𝑡ã𝑜 𝑥 → 0. 
𝑥 = 𝑒ℎ − 1⟹ 𝑒ℎ = 1 + 𝑥 ∴ ℎ = ln(1 + 𝑥) 
lim
ℎ→0
𝑒ℎ −1
ℎ
= lim
𝑥→0
𝑥
ln(1 + 𝑥)
= lim
𝑥→0
1
1
𝑥
. ln(1 + 𝑥)
= lim
𝑥→0
1
ln(1 + 𝑥)1 𝑥⁄
= 1 
𝐶𝑜𝑚𝑜 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 𝑒𝑥𝑖𝑠𝑡𝑒,𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑑𝑖𝑧𝑒𝑟 𝑞𝑢𝑒 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 𝑑𝑜 𝑞𝑢𝑜𝑐𝑖𝑒𝑛𝑡𝑒 é 𝑜 𝑞𝑢𝑜𝑐𝑖𝑒𝑛𝑡𝑒 𝑑𝑜𝑠 
𝑙𝑖𝑚𝑖𝑡𝑒𝑠 𝑒,𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 
 lim
𝑥→0
1
ln(1 + 𝑥)1 𝑥⁄
=
lim
𝑥→0
1
lim
𝑥→0
[ln(1 + 𝑥)1 𝑥⁄ ]
= 1 
𝐸𝑛𝑡ã𝑜 … 
lim
𝑥→0
[ln(1 + 𝑥)1 𝑥⁄ ] = 1 
𝑆𝑒𝑗𝑎 𝑔(𝑥) = ln 𝑥 𝑒 ℎ(𝑥) = (1 + 𝑥)1 𝑥⁄ 𝑒 𝑞𝑢𝑒 lim
𝑥→0
𝑔(ℎ(𝑥)) = 1. 𝐿𝑜𝑔𝑜, 𝑐𝑜𝑚𝑜 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 
𝑑𝑎 𝑓𝑢𝑛çã𝑜 𝑐𝑜𝑚𝑝𝑜𝑠𝑡𝑎 𝑒𝑥𝑖𝑠𝑡𝑒, 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑑𝑖𝑧𝑒𝑟 𝑞𝑢𝑒 
 
lim
𝑥→0
𝑔(ℎ(𝑥)) = 𝑔 (lim
𝑥→0
ℎ(𝑥)) = 1 
ln [lim
𝑥→0
(1 + 𝑥)1 𝑥⁄ ] = 1 
𝑒
ln[lim
𝑥→0
(1+𝑥)1 𝑥⁄ ]
= 𝑒1 
lim
𝑥→0
(1 + 𝑥)1 𝑥⁄ = 𝑒.