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42)
4−x2
lim (
n
√x−1−1 ) √x−1−1
) (m
x→2 3→√x2 −5
Solución lim (
n
n
n
(4−x2 ) (3+√x2 −5
√x−1−1 ( √(x−1)n−1 + √(x−1)n−2 +⋯+1 )
)(m
x→2 (3−√x2 −5) (3+√x2 −5)
m
m
√x−1−1 ( √(x−1)m−1 + √(x−1)m−2 +⋯+1)
(2−x)(2+x) (3+√x2 −5
(x−2)( n√(x−1)n−1 +⋯+1 )
(2−x)(2+x)
(x−2)( m√(x−1)m−1 +⋯+1)
lim ( x→2
)(
lim((3 + √x 2 − 5) ( x→2
n
n
m
m
)
)
( √(x−1)n−1 + √(x−1)n−2 +⋯+1 )
)
( √(x−1)m−1 + √(x−1)m−2 +⋯+1)
m
6n
3
43)
(3−√3 sgn (x3 +6)−2x )( √28√x2 +13−196 )
lim
x→−6
3
3
x
√x+6 ( 2− √4+√|x−2|−4[|4|]
solución
Sgn (x 3 + 6)= si x 3 + 6 > 0 → x > √6 3 0 , si x 3 + 6 = 0 → x =√6 3 -1, si x 3 + 6 < 0 → x < √6
[|4|] → x<-6
3
x
𝐱 𝟒
<
−𝟔 𝟒
x
−6
x
[|4|] < [| 4 |] → [|4|]= -2 3
lim
(3−√3(−1)−2x )( √28√x2 +13−196 ) 3
l1 =
3
√x+6 ( 2− √4+√2−x+8
x→−6
lim
(3−√3(−1)−2x )(3+√3(−1)−2x )
x→−6
lim
x→−6
(x+6)(3+√3(−1)−2x) −2( x+6 )
x→−6
2
1
6
3
= =
(x+6)(3+√3(−1)−2x)
= lim
−2( x+6 ) (x+6)(3+√3(−1)−2x)
3
2
3
3
(2− √4+√10−x)(4+ √4+√1−x+ √(4+√10−x) )
l2 = lim
x→−6
2
3
3
(x+6)(4+ √4+√1−x+ √(4+√10−x) ) (4−√10−x )(4+√10−x)
lim
X+6
= lim (x+6)(FR)(4+√10−x)
x→−6 (x+6)(FR)(4+√10−x)
x→−6
1
lim
x→−6
1
2
3
3
(x+6)(4+ √4+√1−x+ √(4+√10−x) )(4+√10−x)
1
= 12.8 = 96
3
( √28(√x2 +13−7))
l3= lim
√x+6
x→−6 3
lim √
3
x→−6
28(√x2 +13−7)(√x2 +13+7) (x+6)(√x2 +13+7)
x→−6 3
2
28(√x +13−7) = lim √ x+6
3
3
28(X+6)(X−6)
3
28(−12)
= lim √ = √ (x+6)(√x2 +13+7) x→−6
14
=
3
√−24 = −2√3
1 3 1 96
l1
L= l2 l3=
3
3
(−2√3) = -64 √3
3
44) lim
√3x2 +x+4+√x2 +5x+10−6x2 3
√√x+3+6+√x+8−5x2
x→1
solución 3
lim
( √3x2 +x+4 −2)+(√x2 +5x+10−4)−(6x2 −6) 3
( √√x+3+6−2)+(√x+8−3)−(5x2 −5)
x→1
3
l1 = lim
3
3
( √3x2 +x+4 −2)( √(3x2 +x+4)2 +2 √3x2 +x+4+4) 3
3
(x−1)( √(3x2 +x+4)2 +2 √3x2 +x+4+4)
x→1
3x2 +x−4
lim
3
3
x→1 (x−1)( √(3x2 +x+4)2 +2 √3x2 +x+4+4)
lim
(3x+4) 3
3
x→1 ( √(3x2 +x+4)2 +2 √3x2 +x+4+4)
l2= lim
= lim
=7/2 x2 +5x−6
(√x2 +5x+10−4)(√x2 +5x+10+4)
x→1
lim
(x−1)(x+6)
x→1 (x−1)(√x2 +5x−10+4)
l3= lim
=lim
(x−1)(√x2 +5x−10+4)
6x2 −6
x→1 x−1
= lim
x→1
x→1 (x−1)(√x2 +5x−10+4)
= lim
x+6
x→1 (√x2 +5x−10+4)
6(x−1)(x+1) x−1
(3x+4)(x−1) 3
= 12
3
x→1 (x−1)( √(3x2 +x+4)2 +2 √3x2 +x+4+4)
= 7/8
3
3
( √3x2 +x+4 −2)( √(√x+3+6)2 +2 3√√x+3+6+4)
l4= lim
3
(x−1)( √(√x+3+6)2 +2 3√√x+3+6+4)
x→1
√x+3−2
lim
3
2
x→1 (x−1)( √ (√x+3+6) +2 √√x+3+6+4) X−1
lim (x−1)FR(
l5= lim
(√x+8−3)(√x+8+3) (X−1)(√x+8+3)
x→1
l6
lim
=
5(X−1)(X+1) X−1
x→1
1
= lim
√x+3+2)
x→1
(√x+3−2)(√x+3+2) = x→1 (x−1)FR(√x+3+2)
=lim
3
= 1/(12)(4)= 1/48
x→1 FR(√x+3+2)
X−1
= lim (X−1)(√x+8+3)= 1/6 x→1
= 10
L= l1+ l2− l3 / l4+ l5− l6 = 506/371 3
45) lim
√5x2 +7 + √x4 +9−8
3
x→2 √√x2 +5+9x+6+√x+2−5
solución 3
lim
( √5x2 +7−3)+(√x4 +9−5) 3
x→2 ( √√x2 +5+9x+6−3)(√x+2−2) 3
l1 = lim
2
3
3
( √5x2 +7−3)( √5x2 +7 +3 √5x2 +7+9)
x→2
l2= lim
(x−2)(
(√x4 +9−5)(√x4 +9+5)
x→2
lim
2 3 √5x2 +7 +3 √5x2 +7+9)
3
x→1
(x−2)(√x4 +9+5) (x2 +4)(x+2) (√x4 +9+5)
= lim
x→2
lim
(x−2)(√x4 +9+5)
x→1
= 32/10
x→2
( √√x2 +5+9x+6−3).FR 2
3
(x−2)( √((√x2 +5−3)+9(x+1)) +⋯+9 (√x2 +5−3)+9(x−2) 3
5(x+2)(x−2) (x−2)FR
(x2 +4)(x−2)(x+2)
= lim
3
l3= lim
x→1
2
(x−2)( √((√x2 +5−3)+9(x+1)) +⋯+9
= 20/27
(√x2 +5−3)(√x2 +5+3)
M= lim
x→2
(x−2)(x+2)
= lim
(x−2)(FR)(√x2 +5+3) 9(x−2)
S = lim
= 9/27 = 1/3
2
3
x→2
= 4/(27)(6) = 4/281
x→2 (x−2)(FR)(√x2 +5+3)
(x−2)( √((√x2 +5−3)+9(x+1)) +⋯+9
l3= m + s= 58/6(27) = 29/81 l4 = lim
(√x+2−2)(√x+2+2)
x−2
= lim
(x−2)(√x+2+2)
x→2
x→2 (x−2)(√x+2+2)
=¼
L= l1+ l2 /l3+ l4 46) lim
x→−3
√ 3√−9x+1−2 3
2− √x+11
solución 3 3 (√ √−9x+1−2)(√ √−9x+1+2)
l1= lim
x→−3
3 (x+3)(√ √−9x+1+2)
3 3 (√ √−9x+1−2)(√ √−9x+1+2)
lim
x→−3
x→−3
3
3 (x+3)(√ √−9x+1+2)
3
3 3 3 (x+3)(√ √−9x+1+2)( √(−9x)2 +3 √−9x+9)
−9(x+3) 3
√3
3
=
(x+3)( √−9x+1+2)( √(−9x)2 +3 √−9x+9) 3
l2= lim
( √−9x−3)
( √−9x−3)( √(−9x)2 +3 √−9x+9)
lim
x→−3
x→−3
3 (x+3)(√ √−9x+1+2) 3
lim
3
= lim
x→−3
3
3
(2− √x+11)(4+2 √x+11+ √(x+11)2 ) 3
(x+3)(4+2 3√x+11+ √(x+11)2 )
lim
−(x+3) 3
3
x→−3 (x+3)(4+2 √x+11+ √(x+11)2 )
L= l1 /l2= 1
= −1/12
4 4(27)
3
3
√(x2 +1)2 −2 √2x2 +2+ 3√4
47) lim
(x−1)2
x→1
solución 3
3
( √(x2 +1)− √2)
lim
(x−1)(
x→1
3
3 √4)
)2 =
3
3
3 3 ( √(x2 +1)−√2)( √(x2 +1)2 + √2. √x2 +1+ √4 3
3
(x−1)2 (( √(x2 +1)2 + 3√2( √x2 +1)+ 3√4)
x→1
x2 −2
lim ( 2
3
= lim (
(x−1)2
x→1
(
2
3 √(x2 +1)2 + 3√2( √x2 +1)+ 3√4)
3
)2
1
)2 = lim (
(
x→1
3 √(x2 +1)2 + 3√2( √x2 +1)+ 3√4)
3
)2
3
√4 9
9
13
11
√x−1+4 √x−1− √x−1+4
48) lim
x→0
3
5
7
√x−1−5 √x−1+ √x−1−3
solución 9
11
13
5
7
( √x−1+1)+(4 √x−1+4)−( √x−1+1)
lim
x→0
3
( √x−1+1)−(5 √x−1+5)+( √x−1−1) 9
( √x−1+1)FR
l1= lim
x→0 x(
= lim
1 9 9 √(x−1)8 − √(χ−1)7 + √(x−1)6 …+1
9
x→0 x(
x
9 9 √(x−1)8 − √(χ−1)7 + √(x−1)6 …+1)
11
l2= lim
(4( √x−1+1))FR 11
4x
= lim
11
x→0 x( √(x−1)10 − √(χ−1)9 +⋯+1)
11
( √x−1+1)FR 13
13
x→0 x( √(x−1)12 − √(χ−1)11 +⋯+1)
x
= lim
13
l4= lim
x→0 x(
3
3 √(x−1)2 − √(χ−1)3 +1
= lim
x
x→0 x(
3
3 √(x−1)2 − √(χ−1)3 +1)
5
l5= lim
(5( √x−1+1))FR 5
5
x→0 x( √(x−1)4 − √(χ−1)3 +⋯+1)
5
5
= lim
7
7
( √x−1−1)FR 7
7
x→0 x( √(x−1)6 − √(χ−1)5 +⋯+1)
= 1/3
5x
= lim
x→0 x( √(x−1)4 − √(χ−1)3 +⋯+1)
7
l6= lim
= 1/13
13
x→0 x( √(x−1)12 − √(χ−1)11 +⋯+1)
3
( √x−1+1)FR
= 4/11
11
x→0 x( √(x−1)10 − √(χ−1)9 +⋯+1)
13
l3= lim
L=l1+ l2− l3 / l4− l5+ l6 =-3584/4719
= 1/9
9
x
x→0 x( √(x−1)6 − √(χ−1)5 +⋯+1)
=1 = 1/7
49) lim
x→0
5
6
18
25
√x+1−3 √x+6+2 √x+1+ √x+1−2
solución 5
6
18
25
( √x+1−1)−(3 √x+6−3)
lim
x→0 ( √x+1−1)+( √x+1−1) 5
( √x+1−1)FR
l1= lim
x→0 x(
5
x
= lim
5
= lim
6
5 √(x+1)4 − √(χ+1)3 +⋯+1)
x→0 x(
6
3( √x+6−1)FR
l2= lim
6
6
x→0 x( √(x−1)5 − √(χ−1)4 +⋯+1)
=1/5
5 √(x+1)4 − √(χ+1)3 +⋯+1)
3x 6
x→0 x( √(x+1)5 − √(χ+1)4 +⋯+1)
=½
18
( √x+1−1)(FR) x(FR) x→0
l3= lim
= 1/18
25
( √x+1−1)FR (X)FR x→0
l4= lim l
= 1/25
l
L= l1− l2 = -3(45) /43= -135/43 3+ 4
50) lim
√x3 +3√x−3x−1 3
3
x→1 x+3 √x→ √x2 −1
solución lim
(√x−1)3 3
x→1 ( √x−1)3 3
lim
3
3
( (√x−1)(√x+1)( √x2 + √x+1))3 3
x→1 (( √x−1)(√x+1)(
3
= lim 3
√x2 + 3√x+1))
x→1
3
{ (x−1)( √x2 + √x+1)}3 {(x−1)(√x+1)}3
33
= 23 = 27/8
5
51)
x
√2+ √8 −2 5
√5x−10 lim (2√5− x) √ x→20
x2 −400
(
)
solución √5x−10 ) 2√5−√x
l1= lim ( x→20
(√5x−10)(√5x+10)(2√5+√x)
lim ((2√5− x→20
)
√x)(√5x+10)(2√5+√x)
5(2√5+√x)
5(4√5)
lim (−(√5x+10))= x→20
5
−20
x
x
5
√2+ √8 −2 .√2+ √8 +2 5
l2= lim
x→20
5
5 8x 5
(x−20)(x+20).√2+ √ +2
( 5 8x √ 5
lim
x→20
)
−2 5 8x 5
(x−20)(x+20).√2+ √ +2
(
)
5 8x √ −2(FR) 5
lim
x→20
5 8x 5
5
(x−20)(x+20).(√2+ √ +2)( 5√(8x/5)4 −2 √(8χ/5)3 +⋯+16)
(
)
8(x−20)
lim
x→20
5
8x 5 5 5(x−20)(x+20).(√2+ √ +2)( √(8x/5)4 −2 √(8x/5)3 +⋯+16) 5
( 8
lim (5(40)(4)(8/2) x→20
L=l1. l2 = −
√5 400
)=1/400
)
4−x2
lim (
n
√x−1−1 ) √x−1−1
) (m
x→2 3→√x2 −5
Solución lim (
n
n
n
(4−x2 ) (3+√x2 −5
√x−1−1 ( √(x−1)n−1 + √(x−1)n−2 +⋯+1 )
)(m
x→2 (3−√x2 −5) (3+√x2 −5)
m
m
√x−1−1 ( √(x−1)m−1 + √(x−1)m−2 +⋯+1)
(2−x)(2+x) (3+√x2 −5
(x−2)( n√(x−1)n−1 +⋯+1 )
(2−x)(2+x)
(x−2)( m√(x−1)m−1 +⋯+1)
lim ( x→2
)(
lim((3 + √x 2 − 5) ( x→2
n
n
m
m
)
)
( √(x−1)n−1 + √(x−1)n−2 +⋯+1 )
)
( √(x−1)m−1 + √(x−1)m−2 +⋯+1)
m
6n
3
43)
(3−√3 sgn (x3 +6)−2x )( √28√x2 +13−196 )
lim
x→−6
3
3
x
√x+6 ( 2− √4+√|x−2|−4[|4|]
solución
Sgn (x 3 + 6)= si x 3 + 6 > 0 → x > √6 3 0 , si x 3 + 6 = 0 → x =√6 3 -1, si x 3 + 6 < 0 → x < √6
[|4|] → x<-6
3
x
𝐱 𝟒
<
−𝟔 𝟒
x
−6
x
[|4|] < [| 4 |] → [|4|]= -2 3
lim
(3−√3(−1)−2x )( √28√x2 +13−196 ) 3
l1 =
3
√x+6 ( 2− √4+√2−x+8
x→−6
lim
(3−√3(−1)−2x )(3+√3(−1)−2x )
x→−6
lim
x→−6
(x+6)(3+√3(−1)−2x) −2( x+6 )
x→−6
2
1
6
3
= =
(x+6)(3+√3(−1)−2x)
= lim
−2( x+6 ) (x+6)(3+√3(−1)−2x)
3
2
3
3
(2− √4+√10−x)(4+ √4+√1−x+ √(4+√10−x) )
l2 = lim
x→−6
2
3
3
(x+6)(4+ √4+√1−x+ √(4+√10−x) ) (4−√10−x )(4+√10−x)
lim
X+6
= lim (x+6)(FR)(4+√10−x)
x→−6 (x+6)(FR)(4+√10−x)
x→−6
1
lim
x→−6
1
2
3
3
(x+6)(4+ √4+√1−x+ √(4+√10−x) )(4+√10−x)
1
= 12.8 = 96
3
( √28(√x2 +13−7))
l3= lim
√x+6
x→−6 3
lim √
3
x→−6
28(√x2 +13−7)(√x2 +13+7) (x+6)(√x2 +13+7)
x→−6 3
2
28(√x +13−7) = lim √ x+6
3
3
28(X+6)(X−6)
3
28(−12)
= lim √ = √ (x+6)(√x2 +13+7) x→−6
14
=
3
√−24 = −2√3
1 3 1 96
l1
L= l2 l3=
3
3
(−2√3) = -64 √3
3
44) lim
√3x2 +x+4+√x2 +5x+10−6x2 3
√√x+3+6+√x+8−5x2
x→1
solución 3
lim
( √3x2 +x+4 −2)+(√x2 +5x+10−4)−(6x2 −6) 3
( √√x+3+6−2)+(√x+8−3)−(5x2 −5)
x→1
3
l1 = lim
3
3
( √3x2 +x+4 −2)( √(3x2 +x+4)2 +2 √3x2 +x+4+4) 3
3
(x−1)( √(3x2 +x+4)2 +2 √3x2 +x+4+4)
x→1
3x2 +x−4
lim
3
3
x→1 (x−1)( √(3x2 +x+4)2 +2 √3x2 +x+4+4)
lim
(3x+4) 3
3
x→1 ( √(3x2 +x+4)2 +2 √3x2 +x+4+4)
l2= lim
= lim
=7/2 x2 +5x−6
(√x2 +5x+10−4)(√x2 +5x+10+4)
x→1
lim
(x−1)(x+6)
x→1 (x−1)(√x2 +5x−10+4)
l3= lim
=lim
(x−1)(√x2 +5x−10+4)
6x2 −6
x→1 x−1
= lim
x→1
x→1 (x−1)(√x2 +5x−10+4)
= lim
x+6
x→1 (√x2 +5x−10+4)
6(x−1)(x+1) x−1
(3x+4)(x−1) 3
= 12
3
x→1 (x−1)( √(3x2 +x+4)2 +2 √3x2 +x+4+4)
= 7/8
3
3
( √3x2 +x+4 −2)( √(√x+3+6)2 +2 3√√x+3+6+4)
l4= lim
3
(x−1)( √(√x+3+6)2 +2 3√√x+3+6+4)
x→1
√x+3−2
lim
3
2
x→1 (x−1)( √ (√x+3+6) +2 √√x+3+6+4) X−1
lim (x−1)FR(
l5= lim
(√x+8−3)(√x+8+3) (X−1)(√x+8+3)
x→1
l6
lim
=
5(X−1)(X+1) X−1
x→1
1
= lim
√x+3+2)
x→1
(√x+3−2)(√x+3+2) = x→1 (x−1)FR(√x+3+2)
=lim
3
= 1/(12)(4)= 1/48
x→1 FR(√x+3+2)
X−1
= lim (X−1)(√x+8+3)= 1/6 x→1
= 10
L= l1+ l2− l3 / l4+ l5− l6 = 506/371 3
45) lim
√5x2 +7 + √x4 +9−8
3
x→2 √√x2 +5+9x+6+√x+2−5
solución 3
lim
( √5x2 +7−3)+(√x4 +9−5) 3
x→2 ( √√x2 +5+9x+6−3)(√x+2−2) 3
l1 = lim
2
3
3
( √5x2 +7−3)( √5x2 +7 +3 √5x2 +7+9)
x→2
l2= lim
(x−2)(
(√x4 +9−5)(√x4 +9+5)
x→2
lim
2 3 √5x2 +7 +3 √5x2 +7+9)
3
x→1
(x−2)(√x4 +9+5) (x2 +4)(x+2) (√x4 +9+5)
= lim
x→2
lim
(x−2)(√x4 +9+5)
x→1
= 32/10
x→2
( √√x2 +5+9x+6−3).FR 2
3
(x−2)( √((√x2 +5−3)+9(x+1)) +⋯+9 (√x2 +5−3)+9(x−2) 3
5(x+2)(x−2) (x−2)FR
(x2 +4)(x−2)(x+2)
= lim
3
l3= lim
x→1
2
(x−2)( √((√x2 +5−3)+9(x+1)) +⋯+9
= 20/27
(√x2 +5−3)(√x2 +5+3)
M= lim
x→2
(x−2)(x+2)
= lim
(x−2)(FR)(√x2 +5+3) 9(x−2)
S = lim
= 9/27 = 1/3
2
3
x→2
= 4/(27)(6) = 4/281
x→2 (x−2)(FR)(√x2 +5+3)
(x−2)( √((√x2 +5−3)+9(x+1)) +⋯+9
l3= m + s= 58/6(27) = 29/81 l4 = lim
(√x+2−2)(√x+2+2)
x−2
= lim
(x−2)(√x+2+2)
x→2
x→2 (x−2)(√x+2+2)
=¼
L= l1+ l2 /l3+ l4 46) lim
x→−3
√ 3√−9x+1−2 3
2− √x+11
solución 3 3 (√ √−9x+1−2)(√ √−9x+1+2)
l1= lim
x→−3
3 (x+3)(√ √−9x+1+2)
3 3 (√ √−9x+1−2)(√ √−9x+1+2)
lim
x→−3
x→−3
3
3 (x+3)(√ √−9x+1+2)
3
3 3 3 (x+3)(√ √−9x+1+2)( √(−9x)2 +3 √−9x+9)
−9(x+3) 3
√3
3
=
(x+3)( √−9x+1+2)( √(−9x)2 +3 √−9x+9) 3
l2= lim
( √−9x−3)
( √−9x−3)( √(−9x)2 +3 √−9x+9)
lim
x→−3
x→−3
3 (x+3)(√ √−9x+1+2) 3
lim
3
= lim
x→−3
3
3
(2− √x+11)(4+2 √x+11+ √(x+11)2 ) 3
(x+3)(4+2 3√x+11+ √(x+11)2 )
lim
−(x+3) 3
3
x→−3 (x+3)(4+2 √x+11+ √(x+11)2 )
L= l1 /l2= 1
= −1/12
4 4(27)
3
3
√(x2 +1)2 −2 √2x2 +2+ 3√4
47) lim
(x−1)2
x→1
solución 3
3
( √(x2 +1)− √2)
lim
(x−1)(
x→1
3
3 √4)
)2 =
3
3
3 3 ( √(x2 +1)−√2)( √(x2 +1)2 + √2. √x2 +1+ √4 3
3
(x−1)2 (( √(x2 +1)2 + 3√2( √x2 +1)+ 3√4)
x→1
x2 −2
lim ( 2
3
= lim (
(x−1)2
x→1
(
2
3 √(x2 +1)2 + 3√2( √x2 +1)+ 3√4)
3
)2
1
)2 = lim (
(
x→1
3 √(x2 +1)2 + 3√2( √x2 +1)+ 3√4)
3
)2
3
√4 9
9
13
11
√x−1+4 √x−1− √x−1+4
48) lim
x→0
3
5
7
√x−1−5 √x−1+ √x−1−3
solución 9
11
13
5
7
( √x−1+1)+(4 √x−1+4)−( √x−1+1)
lim
x→0
3
( √x−1+1)−(5 √x−1+5)+( √x−1−1) 9
( √x−1+1)FR
l1= lim
x→0 x(
= lim
1 9 9 √(x−1)8 − √(χ−1)7 + √(x−1)6 …+1
9
x→0 x(
x
9 9 √(x−1)8 − √(χ−1)7 + √(x−1)6 …+1)
11
l2= lim
(4( √x−1+1))FR 11
4x
= lim
11
x→0 x( √(x−1)10 − √(χ−1)9 +⋯+1)
11
( √x−1+1)FR 13
13
x→0 x( √(x−1)12 − √(χ−1)11 +⋯+1)
x
= lim
13
l4= lim
x→0 x(
3
3 √(x−1)2 − √(χ−1)3 +1
= lim
x
x→0 x(
3
3 √(x−1)2 − √(χ−1)3 +1)
5
l5= lim
(5( √x−1+1))FR 5
5
x→0 x( √(x−1)4 − √(χ−1)3 +⋯+1)
5
5
= lim
7
7
( √x−1−1)FR 7
7
x→0 x( √(x−1)6 − √(χ−1)5 +⋯+1)
= 1/3
5x
= lim
x→0 x( √(x−1)4 − √(χ−1)3 +⋯+1)
7
l6= lim
= 1/13
13
x→0 x( √(x−1)12 − √(χ−1)11 +⋯+1)
3
( √x−1+1)FR
= 4/11
11
x→0 x( √(x−1)10 − √(χ−1)9 +⋯+1)
13
l3= lim
L=l1+ l2− l3 / l4− l5+ l6 =-3584/4719
= 1/9
9
x
x→0 x( √(x−1)6 − √(χ−1)5 +⋯+1)
=1 = 1/7
49) lim
x→0
5
6
18
25
√x+1−3 √x+6+2 √x+1+ √x+1−2
solución 5
6
18
25
( √x+1−1)−(3 √x+6−3)
lim
x→0 ( √x+1−1)+( √x+1−1) 5
( √x+1−1)FR
l1= lim
x→0 x(
5
x
= lim
5
= lim
6
5 √(x+1)4 − √(χ+1)3 +⋯+1)
x→0 x(
6
3( √x+6−1)FR
l2= lim
6
6
x→0 x( √(x−1)5 − √(χ−1)4 +⋯+1)
=1/5
5 √(x+1)4 − √(χ+1)3 +⋯+1)
3x 6
x→0 x( √(x+1)5 − √(χ+1)4 +⋯+1)
=½
18
( √x+1−1)(FR) x(FR) x→0
l3= lim
= 1/18
25
( √x+1−1)FR (X)FR x→0
l4= lim l
= 1/25
l
L= l1− l2 = -3(45) /43= -135/43 3+ 4
50) lim
√x3 +3√x−3x−1 3
3
x→1 x+3 √x→ √x2 −1
solución lim
(√x−1)3 3
x→1 ( √x−1)3 3
lim
3
3
( (√x−1)(√x+1)( √x2 + √x+1))3 3
x→1 (( √x−1)(√x+1)(
3
= lim 3
√x2 + 3√x+1))
x→1
3
{ (x−1)( √x2 + √x+1)}3 {(x−1)(√x+1)}3
33
= 23 = 27/8
5
51)
x
√2+ √8 −2 5
√5x−10 lim (2√5− x) √ x→20
x2 −400
(
)
solución √5x−10 ) 2√5−√x
l1= lim ( x→20
(√5x−10)(√5x+10)(2√5+√x)
lim ((2√5− x→20
)
√x)(√5x+10)(2√5+√x)
5(2√5+√x)
5(4√5)
lim (−(√5x+10))= x→20
5
−20
x
x
5
√2+ √8 −2 .√2+ √8 +2 5
l2= lim
x→20
5
5 8x 5
(x−20)(x+20).√2+ √ +2
( 5 8x √ 5
lim
x→20
)
−2 5 8x 5
(x−20)(x+20).√2+ √ +2
(
)
5 8x √ −2(FR) 5
lim
x→20
5 8x 5
5
(x−20)(x+20).(√2+ √ +2)( 5√(8x/5)4 −2 √(8χ/5)3 +⋯+16)
(
)
8(x−20)
lim
x→20
5
8x 5 5 5(x−20)(x+20).(√2+ √ +2)( √(8x/5)4 −2 √(8x/5)3 +⋯+16) 5
( 8
lim (5(40)(4)(8/2) x→20
L=l1. l2 = −
√5 400
)=1/400
)
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