Exercicios Calculo (48)

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58 
 
𝑉𝑎𝑙𝑜𝑟𝑒𝑠 𝑚á𝑥𝑖𝑚𝑜𝑠 𝑙𝑜𝑐𝑎𝑖𝑠: {
𝑓 (
𝜋
6
) = 2 sen
𝜋
6
+ cos
𝜋
3
= 2.
1
2
+
1
2
=
3
2
. 
𝑓 (
5𝜋
6
) = 2 sen
5𝜋
6
+ cos
5𝜋
3
= 2.
1
2
+
1
2
=
3
2
.
 
 
𝑉𝑎𝑙𝑜𝑟𝑒𝑠 𝑚í𝑛𝑖𝑚𝑜𝑠 𝑙𝑜𝑐𝑎𝑖𝑠: {
𝑓 (
𝜋
2
) = 2sen
𝜋
2
+ cos𝜋 = 2 + (−1) = 1 
𝑓 (
3𝜋
2
) = 2 sen
3𝜋
2
+ cos3𝜋 = −2 + (−1) = −3
 
 
𝑏) 𝐶𝑎𝑙𝑐𝑢𝑙𝑒 lim
𝑥→0
(3𝑥 + 1)cotg 5𝑥 . 
𝑂𝑏𝑠𝑒𝑟𝑣𝑒 𝑞𝑢𝑒 ,𝑞𝑢𝑎𝑛𝑑𝑜 𝑥 → 0, 𝑡𝑒𝑚𝑜𝑠 3𝑥 + 1 → 1 𝑒 cotg 5𝑥 → ±∞ 𝑎 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑟 𝑠𝑒 
𝑥 → 0+ 𝑜𝑢 𝑠𝑒 𝑥 → 0− ,𝑎𝑠𝑠𝑖𝑚, 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 é 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑑𝑜. 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒 
 
𝑦 = (3𝑥 + 1)cotg5𝑥 
𝐸𝑛𝑡ã𝑜 
 ln𝑦 = ln[(3𝑥 + 1)cotg 5𝑥] = cotg(5𝑥) . ln(3𝑥 + 1) 
𝑒 𝑙𝑜𝑔𝑜,𝑎 𝑅𝑒𝑔𝑟𝑎 𝑑𝑒 𝑙′𝐻ô𝑠𝑝𝑖𝑡𝑎𝑙𝑓𝑜𝑟𝑛𝑒𝑐𝑒 
lim
𝑥→0
ln 𝑦 = lim
𝑥→0
ln(3𝑥 + 1)
tg(5𝑥)
= lim
𝑥→0
3
3𝑥 + 1
5.sec2(5𝑥)
=
3
5
. 
 
𝐶𝑜𝑚𝑜 𝑜 lim
𝑥→0
ln 𝑦 𝑒𝑥𝑖𝑠𝑡𝑒,𝑒𝑛𝑡ã𝑜 𝑜𝑠 𝑙𝑖𝑚𝑖𝑡𝑒𝑠 𝑙𝑎𝑡𝑒𝑟𝑎𝑖𝑠 𝑡𝑎𝑚𝑏é𝑚 𝑒𝑥𝑖𝑠𝑡𝑒𝑚 𝑒 𝑠ã𝑜 𝑖𝑔𝑢𝑎𝑖𝑠. 𝐿𝑜𝑔𝑜, 
𝑢𝑠𝑎𝑛𝑑𝑜 𝑜 𝑓𝑎𝑡𝑜 𝑑𝑒 𝑞𝑢𝑒 𝑦 = 𝑒 ln 𝑦 , 𝑡𝑒𝑚𝑜𝑠: 
 lim
𝑥→0
(3𝑥 + 1)cotg 5𝑥 = lim
𝑥→0
𝑦 = lim
𝑥→0
𝑒 ln 𝑦 = 𝑒
lim
𝑥→0
ln 𝑦
= 𝑒
3
5 . 
 
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟑. 
 
𝑎) 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑝𝑎𝑟𝑎 𝑞𝑢𝑎𝑖𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝛼 𝑎 𝑒𝑞𝑢𝑎çã𝑜 lim
𝑥→∞
(
𝑥 + 𝛼
𝑥 − 𝛼
)
𝑥
= 𝑒, é 𝑣𝑒𝑟𝑑𝑎𝑑𝑒𝑖𝑟𝑎. 
𝑄𝑢𝑎𝑛𝑑𝑜 𝑥 → ∞, 𝑡𝑒𝑚𝑜𝑠 
𝑥 + 𝛼
𝑥 − 𝛼
→ 1, 𝑎𝑠𝑠𝑖𝑚 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 𝑑𝑎𝑑𝑜 é 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑑𝑜. 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒 
𝑦 = (
𝑥 + 𝛼
𝑥 − 𝛼
)
𝑥
 
𝐸𝑛𝑡ã𝑜 
ln 𝑦 = ln [(
𝑥 + 𝛼
𝑥 − 𝛼
)
𝑥
] = 𝑥. ln (
𝑥 + 𝛼
𝑥 − 𝛼
) 
𝑒 𝑙𝑜𝑔𝑜,𝑎 𝑅𝑒𝑔𝑟𝑎 𝑑𝑒 𝑙′𝐻ô𝑠𝑝𝑖𝑡𝑎𝑙 𝑓𝑜𝑟𝑛𝑒𝑐𝑒 
 
lim
𝑥→∞
ln 𝑦 = lim
𝑥→∞
ln (
𝑥 + 𝛼
𝑥 − 𝛼)
1
𝑥
= lim
𝑥→∞
𝑥 − 𝛼
𝑥 + 𝛼 .
−2𝛼
(𝑥 − 𝛼)2
−
1
𝑥2
= lim
𝑥→∞
2𝛼𝑥2
𝑥2 −𝛼2
= lim
𝑥→∞
2𝛼
1 −
𝛼2
𝑥2
= 2𝛼. 
 
𝑈𝑠𝑎𝑛𝑑𝑜 𝑜 𝑓𝑎𝑡𝑜 𝑑𝑒 𝑦 = 𝑒 ln 𝑦 , 𝑡𝑒𝑚𝑜𝑠: 
 
lim
𝑥→∞
(
𝑥 + 𝛼
𝑥 − 𝛼
)
𝑥
= lim
𝑥→∞
𝑦 = lim
𝑥→∞
𝑒 ln 𝑦 = 𝑒
lim
𝑥→∞
ln 𝑦
= 𝑒2𝛼