More Related Content
Similar to Tugas 3 Matematika kalkulus (20)
More from yudiansyah1996 (7)
Tugas 3 Matematika kalkulus
- 1. Group : 1 ( Page 1-8 Kalkulus)
Name :
1. Azhari Rahman
2. MuhammadPachroni Suryana
3. Yudiansyah
Latihan1.1
Hitunglahhasil dari f(X),jikax menyatakannilai adanb. untukc, buatlahsebuahobservasi dari hasil a
dan b.
1. f(x)=
π₯+2
π₯β5
a. x= 3.001
Solutions:
f(3.001)=
π₯+2
π₯β5
f(x) =
3.001+2
3.001β5
=β
5.001
1.999
= - 2.501
b. x= 2.99
Solutions:
f(2.99)=
π₯+2
π₯β5
f(x)=
2.99+2
2.99β5
=β
4.99
2.01
= - 2.482
c. observasi ?
Terlepasdari ituketikax mendekati hasil 3,ketikaf(x) mendekati hasil dari -2.5
2. f(x)=
π₯β5
4π₯
a. x= 1.002
f(1.002)=
π₯β5
4π₯
f(x)=
1.002β5
4(1.002)
=β
3.998
4.008
= - 0.997
b. x= .993
f(.993)=
π₯β5
4π₯
f(x)=
.993β5
4(.993)
=β
4.007
3.972
= - 1.008
c. observasi ?
terlepasdari ituketikax mendekati hasil 1,ketikaf(x) mendekatihasil dari -1.
3. f(x)=
3π₯
π₯
2
- 2. a. x= .001
f(.001)=
3π₯
π₯
2
f(x)=
3(.001)
.001
2
=
0.000003
.001
= 0.003
b. x= -.001
f(-.001)=
3π₯
π₯
2
f(x)=
3(β.001)
β.001
2
=β
0.000003
.001
= - 0.003
c. observasi ?
terlepasdari ituketikax mendekati hasil 0,berarti f(x) tidakmendekati hasil tetap
latihan1.2
Carilahpersamaandari limitberikutataumenunjukkankeberadaanbebas
1. lim
π₯β3
π₯2β4
π₯+1
Solusi : lim
π₯β3
π₯2β4
π₯+1
=
lim
π₯β3
π₯2β4
lim
π₯β3
π₯+1
=
5
4
2. lim
π₯β2
π₯2β9
π₯β2
Solusi : lim
π₯β2
π₯2β9
π₯β2
=
β5
0
= ~
3. lim
π₯β1
βπ₯3 + 7
Solusi : lim
π₯β1
βπ₯3 + 7 = β8
= 2β2
4. lim
π₯βπ
(5π₯2 + 9)
Solusi : lim
π₯βπ
( 5π₯2 + 9) = 5π2 + 9
- 3. 5. lim
π₯β0
5β3π₯
π₯+11
Solusi : lim
π₯β0
5β3π₯
π₯+11
=
lim
π₯β0
5β3π₯
lim
π₯β0
π₯+11
=
5
11
6. lim
π₯β0
9+3π₯2
π₯3+11
Solusi : lim
π₯β0
9+3π₯2
π₯3+11
=
lim
π₯β0
9+3π₯2
lim
π₯β0
π₯3+11
=
9
11
7. lim
π₯β1
π₯2β2π₯+1
π₯2β1
Solusi: lim
π₯β1
π₯2β2π₯+1
π₯2β1
= lim
π₯β1
(π₯β1)(π₯β1)
( π₯β1)(π₯+1)
= lim
π₯β1
π₯β1
π₯+1
=
0
2
= 0
8. lim
π₯β4
6β3π₯
π₯2β16
Solusi : lim
π₯β4
6β3π₯
π₯2β16
= lim
π₯β4
6β3π₯
(π₯β4)(π₯+4)
=
β 6
0
= ~
9. lim
π₯ββ2
β4π₯3 + 11
Solusi : lim
π₯ββ2
β4π₯3 + 11= ββ32 + 11
= β21
10. lim
π₯ββ6
8β3π₯
π₯β6
Solusi : lim
π₯ββ6
8β3π₯
π₯β6
=
lim
π₯ββ6
8β3π₯
lim
π₯ββ6
π₯β6
= β
26
12
= β
13
6
- 4. Latihan2.1
Tentukanlahlimitberikut:
1. lim
π₯β3
π₯β3
π₯2+π₯β12
Solusi : lim
π₯β3
π₯β3
π₯2+π₯β12
= lim
π₯β3
π₯β3
( π₯β3)(π₯+4)
= lim
π₯β3
1
(π₯+4)
=
1
7
2. lim
ββ0
(π₯+β)2βπ₯2
β
Solusi : lim
ββ0
(π₯+β)2βπ₯2
β
= lim
ββ0
π₯2+2βπ₯+β2βπ₯2
β
= lim
ββ0
2βπ₯+β2
β
= lim
ββ0
β(2π₯+β)
β
= lim
ββ0
2π₯ + β
= 2x
3. lim
π₯β4
π₯3β64
π₯2β16
Solusi : lim
π₯β4
π₯3β64
π₯2β16
= lim
π₯β4
π₯3β64
π₯2β16
= lim
π₯β4
(π₯β4)(π₯2+4π₯+16)
( π₯β4)(π₯+4)
= lim
π₯β4
π₯2+4π₯+16
π₯+4
=
48
8
= 6
4. If f(x) = 5x+8, find lim
ββ0
π( π₯+β)βπ(π₯)
β
Solusi : lim
ββ0
π( π₯+β)βπ(π₯)
β
= lim
ββ0
(5( π₯+β)+8)β(5x+8)
β
= lim
ββ0
(5π₯+5β+8)β(5x+8)
β
- 5. = lim
ββ0
5β
β
= ~
5. lim
π₯ββ3
5π₯+7
π₯2β3
Solusi : lim
π₯ββ3
5π₯+7
π₯2β3
=
lim
π₯β3
5π₯+7
lim
π₯β3
π₯2β3
= β
8
6
= -
4
3
6. lim
π₯β25
β π₯β5
π₯β25
Solusi : lim
π₯β25
β π₯β5
π₯β25
= lim
π₯β25
(β π₯β5)
(π₯β25)
(β π₯+5)
(β π₯+5)
= lim
π₯β25
π₯β25
π₯β π₯+ 5π₯β25β π₯β125
=
0
0
= ~
7. If g(x) =π₯2 , find lim
π₯β2
π( π₯)βπ(2)
π₯β2
Solusi : lim
π₯β2
π( π₯)βπ(2)
π₯β2
= lim
π₯β2
π₯2β4
π₯β2
= lim
π₯β2
(π₯+2)(π₯β2)
π₯β2
= lim
π₯β2
π₯ + 2
= 4
8. lim
π₯β0
2π₯2β 4π₯
π₯
Solusi : lim
π₯β0
2π₯2β 4π₯
π₯
= lim
π₯β0
(2π₯β 4)π₯
π₯
= lim
π₯β0
(2π₯ β 4)
= -4
9. lim
πβ0
β π₯+πββ π₯
π
Solusi :lim
πβ0
β π₯+πββ π₯
π
= lim
πβ0
(β π₯+πββπ₯)
π
(β π₯+π+βπ₯)
β π₯+π+β π₯
= lim
πβ0
π₯+πβπ₯
π(β π₯+π+β π₯)
=
0
0
= ~