calculo 1 Exercícios de limite de uma função básico 2018 8304f6cb0a98a83678f0302d579eb4e5

Prévia do material em texto

∈ > 0 𝛿
𝑥 ∈ 𝐷𝑓 .
𝛿 ⇒ | f(x) – L| < ∈
lim
𝑥⟶𝑝
𝑓(𝑥)
𝐷𝑓
lim
𝑥⟶𝑝
𝑓(𝑥) = 𝐿
lim
𝑥⟶𝑝+
𝑓(𝑥) = lim
𝑥⟶𝑝−
𝑓(𝑥) = 𝐿
lim
𝑥⟶1
(4𝑥2 − 7x + 5) = (4. 12 − 7.1 + 5) = 2
lim
𝑥⟶+∞
𝑓(𝑥) = 𝐿 lim
𝑥⟶−∞
𝑓(𝑥) = 𝐿
lim
𝑥⟶+∞
1
𝑥
= 0
lim
𝑥⟶𝑝
1
𝑓(𝑥)
 lim
𝑥⟶𝑝
1
𝑓(𝑥)
 lim
𝑥⟶𝑝+
1
𝑓(𝑥)
 lim
𝑥⟶𝑝+
1
𝑓(𝑥)
 lim
𝑥⟶𝑝−
1
𝑓(𝑥)
 lim
𝑥⟶𝑝−
1
𝑓(𝑥)
 
∞
∞
0
0
1∞ 00 ∞0
lim
𝑥⟶𝑝
𝑓(𝑥) lim
𝑥⟶𝑝
𝑔(𝑥)
lim
𝑥⟶𝑝
[𝑓(𝑥) + 𝑔(𝑥)] = lim
𝑥⟶𝑝
𝑓(𝑥) + lim
𝑥⟶𝑝
𝑔(𝑥)
lim
𝑥⟶𝑝
[𝑓(𝑥) − 𝑔(𝑥)] = lim
𝑥⟶𝑝
𝑓(𝑥) − lim
𝑥⟶𝑝
𝑔(𝑥)
lim
𝑥⟶𝑝
𝑐. 𝑓(𝑥) = 𝑐. lim
𝑥⟶𝑝
𝑓(𝑥)
lim
𝑥⟶𝑝
[𝑓(𝑥) . 𝑔(𝑥)] = lim
𝑥⟶𝑝
𝑓(𝑥) . lim
𝑥⟶𝑝
𝑔(𝑥)
lim
𝑥⟶𝑝
𝑓(𝑥)
𝑔(𝑥)
=
lim
𝑥⟶𝑝
𝑓(𝑥)
lim
𝑥⟶𝑝
𝑔(𝑥)
, 𝑠𝑒 lim
𝑥⟶𝑝
𝑔(𝑥) ≠ 0
lim
𝑥⟶𝑝
[𝑓(𝑥)]𝑛 = [ lim
𝑥⟶𝑝
𝑓(𝑥)]𝑛
lim
𝑥⟶𝑝
𝑐 = 𝑐
lim
𝑥⟶𝑝
𝑥 = 𝑝
lim
𝑥⟶𝑝
𝑥𝑛 = 𝑝𝑛
lim
𝑥⟶𝑝
√𝑥
𝑛
= √𝑝
𝑛
lim
𝑥⟶𝑝
√𝑓(𝑥)
𝑛 = √ lim𝑥⟶𝑝
𝑓(𝑥)𝑛
 
𝑓(𝑥) =
2𝑥2 + 5𝑥 − 3
𝑥2 + 2𝑥 − 3
 
 
 lim
𝑥→1
(4𝑥2 − 7𝑥 + 5) =
 lim
𝑥→−3
𝑥2+2𝑥−3
5−3𝑥
=
 lim
𝑥→2
(
3𝑥2−2𝑥−5
−𝑥2+3𝑥+4
)
3
=
 lim
𝑥→−1
√
2𝑥2+3𝑥−3
5𝑥−4
=
 lim
𝑥→−2
√
3𝑥3−5𝑥2−𝑥+3
4𝑥+3
3
=
 lim
𝑥→2
√2𝑥2+3𝑥+2
6−4𝑥
=
 
 lim
𝑥→1
𝑥2−1
𝑥−1
=
 lim
𝑥→−2
4−𝑥2
2+𝑥
=
 lim
𝑥→1/2
2𝑥2+5𝑥−3
2𝑥2−5𝑥+2
=
 lim
𝑥→1
𝑥3−1
𝑥2−1
=
 lim
𝑥→−2
8+𝑥3
4−𝑥2
=
 lim
𝑥→1
𝑥3−3𝑥2+6𝑥−4
𝑥3−4𝑥2+8𝑥−5
=
 
 lim
𝑥→3
𝑥2−9
𝑥−3
=
 lim
𝑥→−7
49−𝑥2
7+𝑥
=
 lim
𝑥→5
5−𝑥
25−𝑥2
=
 lim
𝑥→0
𝑥2+𝑥
𝑥2−3𝑥
=
 lim
𝑥→0
𝑥3
2𝑥2−𝑥
=
 lim
𝑥→−7
49+14𝑥+𝑥2
7+𝑥
=
 lim
𝑥→3
𝑥2−6𝑥+9
𝑥−3
=
 lim
𝑥→1
𝑥2−4𝑥+3
𝑥−1
=
 lim
𝑥→4
𝑥2−7𝑥+12
𝑥−4
=
 lim
𝑥→1
𝑥−1
𝑥2−3𝑥+2
=
 lim
𝑥→1
𝑥2−2𝑥+1
𝑥−1
=
 lim
𝑥→2
𝑥−2
𝑥2−4
=
 
lim
𝑥→𝑝
2𝑥2+5𝑥−3
𝑥2+2𝑥−3
= ±∞
lim
𝑥→±∞
2𝑥2+5𝑥−3
𝑥2+2𝑥−3
= 2
 lim
𝑥→1
(4.12 − 7.1 + 5) = 4 − 7 + 5 = 𝟐
lim
𝑥→−3
(−3)2+2.(−3)−3
5−3.(−3)
=
9−6−3
5+9
=
0
14
= 𝟎
lim
𝑥→2
(
3.22−2.2−5
−(2)2+3.2+4
)
3
= (
12−4−5
−4+6+4
)
3
= (
3
6
)
3
=
27
216
=
𝟏
𝟖
lim
𝑥→−1
√
2(−1)2+3(−1)−3
5(−1)−4
= √
2−3−3
−5−4
= √
−4
−9
= √
4
9
=
𝟐
𝟑
lim
𝑥→−2
√
3(−2)3−5(−2)2−(−2)+3
4(−2)+3
3
= √
−24−20+2+3
−8+3
3
= √
−39
−5
3
= √
𝟑𝟗
𝟓
𝟑
lim
𝑥→2
√2.22+3.2+2
6−4.2
=
√8+6+2
−2
=
4
−2
= −𝟐
 
 lim
𝑥→1
𝑥2−1
𝑥−1
= lim
𝑥→1
(𝑥−1)(𝑥+1)
𝑥−1
= lim
𝑥→1
(𝑥 + 1) = 1 + 1 = 𝟐
 lim
𝑥→−2
4−𝑥2
2+𝑥
= lim
𝑥→−2
(2−𝑥)(2+𝑥)
2+𝑥
= lim
𝑥→−2
(2 − 𝑥) = 2 − (−2) = 𝟒
 lim
𝑥→1/2
2𝑥2+5𝑥−3
2𝑥2−5𝑥+2
= lim
𝑥→1/2
(2𝑥−1)(𝑥+3)
(2𝑥−1)(𝑥−2)
=
(1/2+3)
(1/2−2)
=
7/2
−3/2
= −
𝟕
𝟑
 lim
𝑥→1
𝑥3−1
𝑥2−1
= lim
𝑥→1
(𝑥−1)(𝑥2+𝑥+1)
(𝑥−1)(𝑥+1)
=
(12+1+1)
(1+1)
=
𝟑
𝟐
 lim
𝑥→−2
8+𝑥3
4−𝑥2
= lim
𝑥→−2
(𝑥+2)(𝑥2−2𝑥+4)
(2+𝑥)(2−𝑥)
=
((−2)2−2.(−2)+4)
(2−(−2))
=
12
4
= 𝟑
 lim
𝑥→1
𝑥3−3𝑥2+6𝑥−4
𝑥3−4𝑥2+8𝑥−5
= lim
𝑥→1
(𝑥−1)(𝑥2−2𝑥+4)
(𝑥−1)(𝑥2−3𝑥+5)
=
(12−2.1+4)
(12−3.1+5)
=
3
3
= 𝟏
 
 lim
𝑥→3
𝑥2−9
𝑥−3
= lim
𝑥→3
(𝑥−3)(𝑥+3)
𝑥−3
= lim
𝑥→3
(𝑥 + 3) = 6
 lim
𝑥→−7
49−𝑥2
7+𝑥
= lim
𝑥→−7
(7+𝑥)(7−𝑥)
7+𝑥
= lim
𝑥→−7
(7 − 𝑥) = 14
 lim
𝑥→5
5−𝑥
25−𝑥2
= lim
𝑥→5
5−𝑥
(5+𝑥)(5−𝑥)
= lim
𝑥→5
1
(5+𝑥)
= 1/10 
 lim
𝑥→0
𝑥2+𝑥
𝑥2−3𝑥
= lim
𝑥→0
𝑥(𝑥+1)
𝑥(𝑥−3)
= −1/3 
 lim
𝑥→0
𝑥3
2𝑥2−𝑥
= lim
𝑥→0
𝑥.𝑥2
𝑥(2𝑥−1)
= 0
 lim
𝑥→−7
49+14𝑥+𝑥2
7+𝑥
= lim
𝑥→−7
(7+𝑥)2
7+𝑥
= 0
 lim
𝑥→3
𝑥2−6𝑥+9
𝑥−3
= lim
𝑥→3
(𝑥−3)2
𝑥−3
= 0
 lim
𝑥→1
𝑥2−4𝑥+3
𝑥−1
= lim
𝑥→1
(𝑥−1)(𝑥−3)
𝑥−1
= −2
 lim
𝑥→4
𝑥2−7𝑥+12
𝑥−4
= lim
𝑥→4
(𝑥−3)(𝑥−4)
𝑥−4
= 1
 lim
𝑥→1
𝑥−1
𝑥2−3𝑥+2
= lim
𝑥→1
𝑥−1
(𝑥−1)(𝑥−2)
= −1
 lim
𝑥→1
𝑥2−2𝑥+1
𝑥−1
= lim
𝑥→1
(𝑥−1)2
𝑥−1
= 0
 lim
𝑥→2
𝑥−2
𝑥2−4
= lim
𝑥→2
𝑥−2
(𝑥−2)(𝑥+2)
=
1
4