matematika kalkulus indonesia

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)=
𝑥+2
𝑥−5
a. x= 3.001
Solutions:
f(3.001)=
𝑥+2
𝑥−5
f(x) =
3.001+2
3.001−5
=−
5.001
1.999
= - 2.501
b. x= 2.99
Solutions:
f(2.99)=
𝑥+2
𝑥−5
f(x)=
2.99+2
2.99−5
=−
4.99
2.01
= - 2.482
c. observasi ?
Terlepasdari ituketikax mendekati hasil 3,ketikaf(x) mendekati hasil dari -2.5
2. f(x)=
𝑥−5
4𝑥
a. x= 1.002
f(1.002)=
𝑥−5
4𝑥
f(x)=
1.002−5
4(1.002)
=−
3.998
4.008
= - 0.997
b. x= .993
f(.993)=
𝑥−5
4𝑥
f(x)=
.993−5
4(.993)
=−
4.007
3.972
= - 1.008
c. observasi ?
terlepasdari ituketikax mendekati hasil 1,ketikaf(x) mendekatihasil dari -1.
3. f(x)=
3𝑥
𝑥
2

2. a. x= .001
f(.001)=
3𝑥
𝑥
2
f(x)=
3(.001)
.001
2
=
0.000003
.001
= 0.003
b. x= -.001
f(-.001)=
3𝑥
𝑥
2
f(x)=
3(−.001)
−.001
2
=−
0.000003
.001
= - 0.003
c. observasi ?
terlepasdari ituketikax mendekati hasil 0,berarti f(x) tidakmendekati hasil tetap
latihan1.2
Carilahpersamaandari limitberikutataumenunjukkankeberadaanbebas
1. lim
𝑥→3
𝑥2−4
𝑥+1
Solusi : lim
𝑥→3
𝑥2−4
𝑥+1
=
lim
𝑥→3
𝑥2−4
lim
𝑥→3
𝑥+1
=
5
4
2. lim
𝑥→2
𝑥2−9
𝑥−2
Solusi : lim
𝑥→2
𝑥2−9
𝑥−2
=
−5
0
= ~
3. lim
𝑥→1
√𝑥3 + 7
Solusi : lim
𝑥→1
√𝑥3 + 7 = √8
= 2√2
4. lim
𝑥→𝜋
(5𝑥2 + 9)
Solusi : lim
𝑥→𝜋
( 5𝑥2 + 9) = 5𝜋2 + 9

3. 5. lim
𝑥→0
5−3𝑥
𝑥+11
Solusi : lim
𝑥→0
5−3𝑥
𝑥+11
=
lim
𝑥→0
5−3𝑥
lim
𝑥→0
𝑥+11
=
5
11
6. lim
𝑥→0
9+3𝑥2
𝑥3+11
Solusi : lim
𝑥→0
9+3𝑥2
𝑥3+11
=
lim
𝑥→0
9+3𝑥2
lim
𝑥→0
𝑥3+11
=
9
11
7. lim
𝑥→1
𝑥2−2𝑥+1
𝑥2−1
Solusi: lim
𝑥→1
𝑥2−2𝑥+1
𝑥2−1
= lim
𝑥→1
(𝑥−1)(𝑥−1)
( 𝑥−1)(𝑥+1)
= lim
𝑥→1
𝑥−1
𝑥+1
=
0
2
= 0
8. lim
𝑥→4
6−3𝑥
𝑥2−16
Solusi : lim
𝑥→4
6−3𝑥
𝑥2−16
= lim
𝑥→4
6−3𝑥
(𝑥−4)(𝑥+4)
=
− 6
0
= ~
9. lim
𝑥→−2
√4𝑥3 + 11
Solusi : lim
𝑥→−2
√4𝑥3 + 11= √−32 + 11
= √21
10. lim
𝑥→−6
8−3𝑥
𝑥−6
Solusi : lim
𝑥→−6
8−3𝑥
𝑥−6
=
lim
𝑥→−6
8−3𝑥
lim
𝑥→−6
𝑥−6
= −
26
12
= −
13
6