Exercicios Calculo (2)

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7 
 
−𝐸𝑚 𝑥 = 0, 𝑡𝑒𝑚𝑜𝑠: 
 
𝑓(0) = 2 − 02 = 2 − 0 = 2. 
lim
𝑥→0+
𝑓(𝑥) = lim
𝑥→0+
(2 − 𝑥2) = lim
𝑥→0+
2 − lim
𝑥→0+
𝑥2 = 2 − 02 = 2 − 0 = 2. 
lim
𝑥→0−
𝑓(𝑥) = lim
𝑥→0−
sen𝑥 = sen0 = 0. 
 
∗ 𝐶𝑜𝑚𝑜 lim
𝑥→0−
𝑓(𝑥) ≠ lim
𝑥→0+
𝑓(𝑥) , 𝑒𝑛𝑡ã𝑜 lim
𝑥→0
𝑓(𝑥) 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑓 é 
𝑑𝑒𝑠𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 0. 𝐴𝑙é𝑚 𝑑𝑖𝑠𝑠𝑜, 𝑐𝑜𝑚𝑜 𝑜𝑠 𝑙𝑖𝑚𝑖𝑡𝑒𝑠 𝑙𝑎𝑡𝑒𝑟𝑎𝑖𝑠 𝑒𝑥𝑖𝑠𝑡𝑒𝑚, 𝑖𝑚𝑝𝑙𝑖𝑐𝑎 
𝑑𝑖𝑧𝑒𝑟 𝑞𝑢𝑒 𝑎 𝑑𝑒𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑑𝑎𝑑𝑒 𝑑𝑒 𝑓 𝑒𝑚 0 é 𝑑𝑜 𝑡𝑖𝑝𝑜 𝑠𝑎𝑙𝑡𝑜. 
 
−𝐸𝑚 𝑥 = √2,𝑡𝑒𝑚𝑜𝑠: 
 
𝑓(√2) = 2− (√2)
2
= 2− 2 = 0. 
lim
𝑥→√2
−
𝑓(𝑥) = lim
𝑥→√2
−
(2 − 𝑥2) = lim
𝑥→√2
−
2 − lim
𝑥→√2
−
𝑥2 = 2 − (√2)
2
= 2− 2 = 0. 
lim
𝑥→√2
+
𝑓(𝑥) = lim
𝑥→√2
+
arctg(
𝑥 −√2
2
) = arctg(
√2 − √2
2
) = arctg 0 = 0. 
 
∗ 𝐶𝑜𝑚𝑜 lim
𝑥→√2
+
𝑓(𝑥) = lim
𝑥→√2
−
𝑓(𝑥) , 𝑒𝑛𝑡ã𝑜 lim
𝑥→√2
𝑓(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒 𝑒 lim
𝑥→√2
𝑓(𝑥) = 𝑓(√2). 
𝐿𝑜𝑔𝑜, 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 √2. 
 
𝐶𝑜𝑚 𝑖𝑠𝑠𝑜, 𝑗𝑢𝑛𝑡𝑎𝑚𝑒𝑛𝑡𝑒 𝑐𝑜𝑚 𝑎 𝑎𝑛á𝑙𝑖𝑠𝑒 𝑓𝑒𝑖𝑡𝑎 𝑖𝑛𝑖𝑐𝑖𝑎𝑙𝑚𝑒𝑛𝑡𝑒, 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑚𝑜𝑠 𝑞𝑢𝑒 𝑓 é 
𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 (−∞,0) ∪ (0,∞),𝑜𝑢 𝑎𝑖𝑛𝑑𝑎, 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 ℝ∗. 
 
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟑. 
 
𝑎)𝑆𝑒𝑗𝑎 𝑓(𝑥) = {
1− 𝑥
1− √𝑥
3
, 𝑠𝑒 𝑥 ≠ 1
𝑐, 𝑠𝑒 𝑥 = 1 
. 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑐 𝑝𝑎𝑟𝑎 𝑞𝑢𝑒 𝑎 𝑓𝑢𝑛çã𝑜 
𝑠𝑒𝑗𝑎 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑡𝑜𝑑𝑜𝑠 𝑜𝑠 𝑟𝑒𝑎𝑖𝑠. 
 
𝑆𝑒𝑗𝑎 𝑔(𝑥) =
1 − 𝑥
1 − √𝑥
3
. 𝑇𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝐷(𝑔) = {𝑥 ∈ ℝ ∶ 𝑥 ≠ 1}. 𝐶𝑜𝑚𝑜 𝑔 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 
𝑟𝑎𝑐𝑖𝑜𝑛𝑎𝑙, 𝑒𝑛𝑡ã𝑜 𝑔 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑠𝑒𝑢 𝑑𝑜𝑚í𝑛𝑖𝑜. 𝐿𝑜𝑔𝑜, 𝑔 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 (−∞,1) ∪ (1,∞). 
𝐼𝑠𝑡𝑜 𝑖𝑚𝑝𝑙𝑖𝑐𝑎 𝑑𝑖𝑧𝑒𝑟 𝑞𝑢𝑒 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 (−∞,1) ∪ (1,∞). 
 
𝑃𝑎𝑟𝑎 𝑞𝑢𝑒 𝑓 𝑠𝑒𝑗𝑎 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑡𝑜𝑑𝑜𝑠 𝑜𝑠 𝑟𝑒𝑎𝑖𝑠 é 𝑐𝑜𝑛𝑑𝑖çã𝑜 𝑛𝑒𝑐𝑒𝑠𝑠á𝑟𝑖𝑎 𝑒 𝑠𝑢𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 
𝑞𝑢𝑒 𝑓 𝑠𝑒𝑗𝑎 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 1. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑝𝑒𝑙𝑎 𝑑𝑒𝑓𝑖𝑛𝑖çã𝑜 𝑑𝑒 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑑𝑎𝑑𝑒 𝑑𝑒 
𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑒𝑚 𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜,𝑡𝑒𝑚𝑜𝑠: 
lim
𝑥→1
𝑓(𝑥) = 𝑓(1) = 𝑐 
𝑐 = lim
𝑥→1
𝑓(𝑥) = lim
𝑥→1
1 − 𝑥
1− √𝑥
3
 
 = lim
𝑥→1
[
1 − 𝑥
1 − √𝑥
3
.
1 + √𝑥
3 + √𝑥2
3
1 + √𝑥
3
+ √𝑥2
3
]