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𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟏.
𝑎) 𝐶𝑎𝑙𝑐𝑢𝑙𝑒 lim
𝑥→0
√1+ 𝑒𝜋𝑥
3 −1
𝑥
.
lim
𝑥→0
√1+ 𝑒𝜋𝑥
3 − 1
𝑥
= lim
𝑥→0
[
√1+ 𝑒𝜋𝑥
3 −1
𝑥
.
√(1+ 𝑒𝜋𝑥)23 + √1+ 𝑒𝜋𝑥
3 +1
√(1+ 𝑒𝜋𝑥)23 + √1+ 𝑒𝜋𝑥
3 +1
]
= lim
𝑥→0
1 + 𝑒𝜋𝑥 − 1
𝑥. [√(1 + 𝑒𝜋𝑥)23 + √1+ 𝑒𝜋𝑥
3 + 1]
= lim
𝑥→0
𝑒𝜋𝑥
𝑥. [√(1 + 𝑒𝜋𝑥)23 + √1+ 𝑒𝜋𝑥
3 + 1]
;
𝑆𝑒 𝑥 → 0,𝑒𝑛𝑡ã𝑜
𝑥 ≠ 0.
= lim
𝑥→0
𝑒𝜋
√(1+ 𝑒𝜋𝑥)23 + √1+ 𝑒𝜋𝑥
3 + 1
=
lim
𝑥→0
𝑒𝜋
√[lim
𝑥→0
(1 + 𝑒𝜋𝑥)]
23
+ √lim
𝑥→0
(1 + 𝑒𝜋𝑥)3 + lim
𝑥→0
1
=
𝑒𝜋
√12
3
+ √1
3 + 1
=
𝑒𝜋
3
.
𝑏) 𝐶𝑎𝑙𝑐𝑢𝑙𝑒 lim
𝑥→−1+
𝑥 + 1
|𝑥2 − 2𝑥 − 3|
.
∗ 𝑥2 − 2𝑥 − 3 = (𝑥 + 1)(𝑥 − 3). |𝑥2 −2𝑥 − 3| = {
𝑥2 − 2𝑥 − 3, 𝑠𝑒 𝑥 ≤ −1 𝑜𝑢 𝑥 ≥ 3
−(𝑥2− 2𝑥 − 3), 𝑠𝑒 − 1 −1 𝑒,𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, |𝑥2 − 2𝑥 − 3| = −(𝑥2 − 2𝑥 − 3).
lim
𝑥→−1+
𝑥 + 1
|𝑥2 −2𝑥 − 3|
= lim
𝑥→−1+
𝑥 + 1
−(𝑥2 − 2𝑥 − 3)
= lim
𝑥→−1+
𝑥 + 1
−(𝑥 + 1)(𝑥 − 3)
;
𝑆𝑒 𝑥 → −1, 𝑒𝑛𝑡ã𝑜 𝑥 ≠ −1.
𝐿𝑜𝑔𝑜, (𝑥 + 1) ≠ 0
= lim
𝑥→−1+
1
−(𝑥 − 3)
=
lim
𝑥→−1+
1
− lim
𝑥→−1+
𝑥 + lim
𝑥→−1+
3
=
1
−(−1)+ 3
=
1
4
.
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟐.
𝑎) 𝐶𝑎𝑙𝑐𝑢𝑙𝑒 lim
𝑥→∞
⟦𝑥2 +𝜋⟧
⟦𝑥2⟧
.
𝑆𝑎𝑏𝑒-𝑠𝑒 𝑞𝑢𝑒 𝑥 − 1 ≤ ⟦𝑥⟧ ≤ 𝑥. 𝐿𝑜𝑔𝑜,𝑡𝑒𝑚𝑜𝑠
𝑥2 + 𝜋 − 1 ≤ ⟦𝑥2 + 𝜋⟧ ≤ 𝑥2 +𝜋 𝑒
𝑥2 −1 ≤ ⟦𝑥2⟧ ≤ 𝑥2