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Rosalys Herrera 20.348.002
1. Lim 𝑦 → −1 (
𝑦2−2𝑦+2
𝑦−4
+ 1)
=(
(−1)2−2(−1)+2
−1−4
+ 1)
= (
1+2+2
−5
+ 1)
= (
5
−5
)
= (−1 + 1)
= 0
2. lim 𝑥 ⟶ 9
1
√ 𝑥
−
1
3
𝑥−9
=
1
√9
−
1
3
9−9
=
1
√9
−
1
3
0
=
1
3
−
1
3
0
=
0
0
= +∞
3. lim 𝑇 ⟶ 2 (
𝑇2−6𝑇+8
𝑇2−5𝑇+6
)
𝑇−2
√𝑇−√2
= (
22−6(2)+8
22−5(2)+6
)
2−2
√2−√2
= (
4−12+8
4+10+6
)
0
0
= (0
0
)+∞
= (+∞)+∞
= +∞
4. lim 𝑘 ⟶ 6
2−√𝐾−2
𝐾2−36
=
2−√6−2
62−36
=
2−2
36−36
= +∞
5. lim 𝑥 ⟶
𝜋
4
𝑠𝑒𝑛 𝑥−cos 𝑥
1−tan 𝑥
=
𝑠𝑒𝑛(
𝜋
4
)−cos(
𝜋
4
)
1−tan(
𝜋
4
)
=
0,01370−0,9999
0,9862
= -1
6. lim ℎ ⟶ 0
(ℎ+3)2−9
ℎ
=
(0+3)2−9
0
= +∞
7. lim 𝑥 ⟶ ∞
6𝑥
√𝑥2+1
=
6(∞)
√∞2+1
=
∞
∞
= ∞

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Ejercicios limites

  • 1. Rosalys Herrera 20.348.002 1. Lim 𝑦 → −1 ( 𝑦2−2𝑦+2 𝑦−4 + 1) =( (−1)2−2(−1)+2 −1−4 + 1) = ( 1+2+2 −5 + 1) = ( 5 −5 ) = (−1 + 1) = 0 2. lim 𝑥 ⟶ 9 1 √ 𝑥 − 1 3 𝑥−9 = 1 √9 − 1 3 9−9 = 1 √9 − 1 3 0 = 1 3 − 1 3 0 = 0 0 = +∞
  • 2. 3. lim 𝑇 ⟶ 2 ( 𝑇2−6𝑇+8 𝑇2−5𝑇+6 ) 𝑇−2 √𝑇−√2 = ( 22−6(2)+8 22−5(2)+6 ) 2−2 √2−√2 = ( 4−12+8 4+10+6 ) 0 0 = (0 0 )+∞ = (+∞)+∞ = +∞ 4. lim 𝑘 ⟶ 6 2−√𝐾−2 𝐾2−36 = 2−√6−2 62−36 = 2−2 36−36 = +∞ 5. lim 𝑥 ⟶ 𝜋 4 𝑠𝑒𝑛 𝑥−cos 𝑥 1−tan 𝑥 = 𝑠𝑒𝑛( 𝜋 4 )−cos( 𝜋 4 ) 1−tan( 𝜋 4 ) = 0,01370−0,9999 0,9862 = -1 6. lim ℎ ⟶ 0 (ℎ+3)2−9 ℎ = (0+3)2−9 0 = +∞ 7. lim 𝑥 ⟶ ∞ 6𝑥 √𝑥2+1 = 6(∞) √∞2+1 = ∞ ∞ = ∞