1. 1
Group : 1 ( Page 1-8 Kalkulus)
Name :
1. Azhari Rahman
2. MuhammadPachroni Suryana
3. Yudiansyah
Class: 1EA
Exercise 1.1
Compute the value of f(x) whenx hasthe indicatedvaluesgivenin(a) and(b).For(c),make an
observationbasedonyourresultsin(a) and(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. Observation?
It appearsthan whenx isclose to 3 invalue, thenf(x) isclose to -2.5in value.
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. Observation?
It appearsthan whenx isclose to 1 invalue,thenf(x) isclose to -1in value.

2. 2
3. f(x)=
3𝑥
𝑥
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. Observation?
It appearsthan whenx isclose to 0 invalue,f(x) isnotclose toanyfixednumberinvalue.
Exercise 1.2
Findthe followinglimitsorindicate nonexistence.
1. lim
𝑥→3
𝑥2−4
𝑥+1
Solutions: 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
Solutions: lim
𝑥→2
𝑥2−9
𝑥−2
=
−5
0
= ~
3. lim
𝑥→1
√𝑥3 + 7
Solutions: lim
𝑥→1
√𝑥3 + 7 = √8
= 2√2
4. lim
𝑥→𝜋
(5𝑥2 + 9)
Solutions: lim
𝑥→𝜋
(5𝑥2 + 9) = 5𝜋2 + 9

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

4. 4
Exercise 2.1
Evalute the followinglimits.
1. lim
𝑥→3
𝑥−3
𝑥2+𝑥−12
Solutions: lim
𝑥→3
𝑥−3
𝑥2+𝑥−12
= lim
𝑥→3
𝑥−3
( 𝑥−3)(𝑥+4)
= lim
𝑥→3
1
(𝑥+4)
=
1
7
2. lim
ℎ→0
(𝑥+ℎ)2−𝑥2
ℎ
Solutions: lim
ℎ→0
(𝑥+ℎ)2−𝑥2
ℎ
= lim
ℎ→0
𝑥2+2ℎ𝑥+ℎ2−𝑥2
ℎ
= lim
ℎ→0
2ℎ𝑥+ℎ2
ℎ
= lim
ℎ→0
ℎ(2𝑥+ℎ)
ℎ
= lim
ℎ→0
2𝑥 + ℎ
= 2x
3. lim
𝑥→4
𝑥3−64
𝑥2−16
Solutions: lim
𝑥→4
𝑥3−64
𝑥2−16
= lim
𝑥→4
𝑥3−64
𝑥2−16
= lim
𝑥→4
(𝑥−4)(𝑥2+4𝑥+16)
( 𝑥−4)(𝑥+4)
= lim
𝑥→4
𝑥2+4𝑥+16
𝑥+4
=
48
8
= 6
4. If f(x) = 5x+8, find lim
ℎ→0
𝑓( 𝑥+ℎ)−𝑓(𝑥)
ℎ
Solutions: lim
ℎ→0
𝑓( 𝑥+ℎ)−𝑓(𝑥)
ℎ
= lim
ℎ→0
(5( 𝑥+ℎ)+8)−(5x+8)
ℎ
= lim
ℎ→0
(5𝑥+5ℎ+8)−(5x+8)
ℎ
= lim
ℎ→0
5ℎ
ℎ

5. 5
= ~
5. lim
𝑥→−3
5𝑥+7
𝑥2−3
Solutions: lim
𝑥→−3
5𝑥+7
𝑥2−3
=
lim
𝑥→3
5𝑥+7
lim
𝑥→3
𝑥2−3
= −
8
6
= -
4
3
6. lim
𝑥→25
√ 𝑥−5
𝑥−25
Solutions: lim
𝑥→25
√ 𝑥−5
𝑥−25
= lim
𝑥→25
(√ 𝑥−5)
(𝑥−25)
(√ 𝑥+5)
(√ 𝑥+5)
= lim
𝑥→25
𝑥−25
𝑥√ 𝑥+ 5𝑥−25√ 𝑥−125
=
0
0
= ~
7. If g(x) =𝑥2 , find lim
𝑥→2
𝑔( 𝑥)−𝑔(2)
𝑥−2
Solutions: lim
𝑥→2
𝑔( 𝑥)−𝑔(2)
𝑥−2
= lim
𝑥→2
𝑥2−4
𝑥−2
= lim
𝑥→2
(𝑥+2)(𝑥−2)
𝑥−2
= lim
𝑥→2
𝑥 + 2
= 4
8. lim
𝑥→0
2𝑥2− 4𝑥
𝑥
Solutions: lim
𝑥→0
2𝑥2− 4𝑥
𝑥
= lim
𝑥→0
(2𝑥− 4)𝑥
𝑥
= lim
𝑥→0
(2𝑥 − 4)
= -4
9. lim
𝑟→0
√ 𝑥+𝑟−√ 𝑥
𝑟
Solutions:lim
𝑟→0
√ 𝑥+𝑟−√ 𝑥
𝑟
= lim
𝑟→0
(√ 𝑥+𝑟−√𝑥)
𝑟
(√ 𝑥+𝑟+√𝑥)
√ 𝑥+𝑟+√ 𝑥
= lim
𝑟→0
𝑥+𝑟−𝑥
𝑟(√ 𝑥+𝑟+√ 𝑥)
=
0
0
= ~
10. lim
𝑥→4
𝑥3+6
𝑥−4

6. 6
Solutions: lim
𝑥→4
𝑥3+6
𝑥−4
= lim
𝑥→4
𝑥3+6
𝑥−4
𝑥+4
𝑥+4
=
𝑥4+4𝑥3+6𝑥+24
(𝑥−4)(𝑥+4)
=
70
0
= ~