tugas kelompok 1 mtk tentang limit hal : 1-8jawaban mtk
TRANSCRIPT
Materi limit
Question
Latihan 1
1. F(x)=x+2x−5
a. X=3,001b. X=2,99c. Observation?
2. F(=x)= x−54 x
a. x=1,002b. x=0,993c. observation?
3. f(x)=3x2
xa. x=.001b. x= -.001c. observation?
Answer :
Latihan 1
1. F(x)=x+2x−5
a. X=3,001
maka f(x)=3,001+23,001−5
=5,001
−1,999 = -2,5017
b. X=2,99
maka f(x)=2,99+22,99−5
=4,99
−2,01 =-2,482
Observation It appears that when x is close to 3 in value, then f(x) is close to -2 in value.
1
Materi limit
2. F(=x)= x−54 x
a. x=1,002
maka f(x)=1,002−54 (1,002)
=-−3,9984,008
=0,9975b. 0,993
maka f(x)=0,993−54 (0,993)
=−4,0073,972
=1,008
Observation: It appears that when x is close to 1 in value, then f(x) is close to 1 in value.
3. f(x)=3x2
xa. x=.0,001
maka f(x)= 3(0.001)2
0,001
=0,0000030,001
=0.0015b. x= -0,001
maka f(x)= 3(−0,001)2
−0,001
=0,000003−0,001
=-0,0015 observation: It appears that when x is close to 0 in value, f(x) is not close to
any fixed number in value.
2
Materi limit
Question
Latihan 1.2
1.limX→3
x2−4
x+1=
2.limX→2
x2−9
x−2
3. limX→1
√ x3+7 =
4. limX→π
¿(5x2+9) =
5.limX→0
5−3 x
x+11 =
6.limX→0
9+3 x2
x3+11 =
7. limX→1
x2−2x+1
x2−1 =
8.limX→0
6−3 x
x2−16 =
9. limX→−2
√4 x3+11 =
3
Materi limit
10. .limX→1
8−3 x
x−6=
Answer :1. lim
x→3
x2−4x+1
=limx→3
x2−4
limx→3
x+1=54
2. limx→2
x2−9x−2
=limx→2
x2−9
limx→2
x−2=−5
3. limx→1
¿√x3+7=2√24. limx→π
(5 x2+9 )=¿585. lim
x→0
5−3 xx+11
=limx→05−3 x
limx→0
x+11= 511
6. limx→0
9−3 x2
x3+11=limx→0
9−3 x2
limx→0
x3+11= 911
7. limx→1
x2−2 x+1x2−1
=limx→1
x2−2x+1
limx→1
x3+11=12
8. limx→4
6−3 xx2−16
=limx→46−3 x
limx→4
x2−16= 6−3 x
(x−4)(x+4)
9. limx→−2
√4 x3+11=√4(−8)+11=√−21
4
Materi limit
10. limx→-6
8−3 xx−6
=limx→-6
8−3x
limx→-6
x−6= 26
−12
Latihan 2.1
1.limX→3
x−3
x2+x−12=
2. limh→o¿¿=
3.limX→4
x3−64
x2−16=
4. IF F(x)=5(x+b), find limX→ 4¿
5.limX→−3
5x+7
x2−3=
5
Materi limit
6.limX→25
√x−5x−25
= . limX→25
(√ x−5)
(√ x−5)(√ x+5)
7. IF g ( x )= x2 , FindlimX→2
g ( x )−9 (2 )
x−2=¿
8. limX→ 0¿ 2x
2−4 xx
=
9.limr →0
√ x+r−√xr
=
10.limx→4
x3+6
x−4=
Answer
Latihan 2.1
1.limX→3
x−3
x2+x−12=limX→3
x−3
( x−3 ) ( x+4 )
= limX→3
1
x+4
6
Materi limit
=13+4
=17
2. limh→o¿¿=
limh→o
−x3−xh4
h
=limh→o
¿-xh3
=-x.03
=0
3.limX→4
x3−64
x2−16=limX→4
x3−64
x2−16= limX→4
( x−4 ) (x2+16 )−(16 x+4 x2 )
(x−4 ) ( x+4 )
=limX→4
(x2+16 )−(16 x+4 x2)
(x+4)
=(42+16 )−¿¿
=(16+16 )−(64−64 )
8
=32−1288
= -968
= -12
4. IF F(x)=5(x+b), find limh→0
f (x+h )−f (x)
h
find limh→0
f (x+h )−f (x)
h= find
limh→0
(5 ( x+h )+8)−(5 x+8)
h
7
Materi limit
=limh→0
5 x+5h+8−5x−8
h
=limh→0
5h
h
=limh→ 0
5
=5
5.limX→−3
5x+7
x2−3=5(−3)+7(−3)2−3
=−15+79−3
=−86
=−43
6.limX→25
√ x−5x−25
= . limX→25
(√ x−5)
(√ x−5)(√ x+5)
= limX→25
1
(√ x+5)
=1
(√25+5)
=1
(5+5)
=110
7. IF g ( x )= x2 , FindlimX→2
g ( x )−9 (2 )
x−2
limX→2
g ( x )−9 (2 )
x−2 = limX→2
2x2−(2)2
x−2
= limX→2
x2−4
x−2
=limX→2
(x−2)(x+2)
(x−2)
8
Materi limit
=limX→ 2
¿(x+2)
= 4
8. limX→ 0
¿ 2x2−4 xx
= limX→ 0
¿2x-4
= 2.0 – 4= -4
9.limr →0
√ x+r−√ xr
= limr →0
√ x+r−√xr
× √ x+r+√x√ x+r+√x
limr→0
( x+r )−x
r (√ x+r+√x )
limr→0
r
r (√ x+r+√x )
limr→0
1
√x+r+√x
1
√x+0+√x
1
√x+√ x
1
2√x
9
Materi limit
10.limx→4
x3+6
x−4= 4
3+64−4
=640
= ∞
10