1. Materi limit
PoliteknikManufakturNegeri BangkaBelitung 1
Question
Latihan1
1. F(x)=
𝑥+2
𝑥−5
a. X=3,001
b. X=2,99
c. Observation?
2. F(=x)=
𝑥−5
4𝑥
a. x=1,002
b. x=0,993
c. observation?
3. f(x)=
3𝑥2
𝑥
a. x=.001
b. x= -.001
c. observation?
Answer:
Latihan1
1. F(x)=
𝑥+2
𝑥−5
a. X=3,001
maka f(x)=
3,001+2
3,001−5
=
5,001
−1,999
= -2,5017
b. X=2,99
maka f(x)=
2,99+2
2,99−5
=
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.

2. Materi limit
PoliteknikManufakturNegeri BangkaBelitung 2
2. F(=x)=
𝑥−5
4𝑥
a. x=1,002
maka f(x)=
1,002−5
4(1,002)
=-
−3,998
4,008
=0,9975
b. 0,993
maka f(x)=
0,993−5
4(0,993)
=
−4,007
3,972
=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
𝑥
a. x=.0,001
maka f(x)=
3(0.001)
2
0,001
=
0,000003
0,001
=0.0015
b. 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.

3. Materi limit
PoliteknikManufakturNegeri BangkaBelitung 3
Question
Latihan1.2
1. lim
𝑋→3
𝑥2
−4
𝑥+1
=
2. lim
𝑋→2
𝑥2
−9
𝑥−2
3. lim
𝑋→1
√𝑥3 + 7 =
4. lim
𝑋→𝜋
(5𝑥2
+ 9) =
5. lim
𝑋→0
5−3𝑥
𝑥+11
=
6. lim
𝑋→0
9+3𝑥2
𝑥3 +11
=
7. lim
𝑋→1
𝑥2
−2𝑥+1
𝑥2−1
=
8. lim
𝑋→0
6−3𝑥
𝑥2−16
=
9. lim
𝑋→−2
√4𝑥3 + 11 =
10. . lim
𝑋→1
8−3𝑥
𝑥−6
=

4. Materi limit
PoliteknikManufakturNegeri BangkaBelitung 4
Answer :
1. 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
=
lim
𝑥→2
𝑥2−9
lim
𝑥→2
𝑥−2
= −5
3. lim
𝑥→1
= √ 𝑥3 + 7 = 2√2
4. lim
𝑥→𝜋
(5𝑥2
+ 9) =58
5. 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
=
lim
𝑥→0
9−3𝑥2
lim
𝑥→0
𝑥3+11
=
9
11
7. lim
𝑥→1
𝑥2−2𝑥+1
𝑥2−1
=
lim
𝑥→1
𝑥2−2𝑥+1
lim
𝑥→1
𝑥3+11
= 12
8. lim
𝑥→4
6−3𝑥
𝑥2−16
=
lim
𝑥→4
6−3𝑥
lim
𝑥→4
𝑥2−16
=
6−3𝑥
(𝑥−4)(𝑥+4)
9. lim
𝑥→−2
√4𝑥3 + 11 = √4(−8) + 11 = √−21
10. lim
𝑥→-6
8−3𝑥
𝑥−6
=
lim
𝑥→-6
8−3𝑥
lim
𝑥→-6
𝑥−6
=
26
−12

5. Materi limit
PoliteknikManufakturNegeri BangkaBelitung 5
Latihan2.1
1. lim
𝑋→3
𝑥−3
𝑥2+𝑥−12
=
2. lim
ℎ→𝑜
(𝑥+ℎ)2
−𝑥2
ℎ
=
3. lim
𝑋→4
𝑥3
−64
𝑥2−16
=
4. IF F(x)=5(x+b), find lim
𝑋→4
5. lim
𝑋→−3
5𝑥+7
𝑥2−3
=
6. lim
𝑋→25
√ 𝑥−5
𝑥−25
= . lim
𝑋→25
(√ 𝑥−5)
(√ 𝑥−5)(√ 𝑥+5)
7. 𝐼𝐹 𝑔( 𝑥)= 𝑥2
, 𝐹𝑖𝑛𝑑 lim
𝑋→2
𝑔( 𝑥)−9(2)
𝑥−2
=
8. lim
𝑋→0
2𝑥2
−4𝑥
𝑥
=
9. lim
𝑟→0
√ 𝑥+𝑟−√ 𝑥
𝑟
=
10.lim
𝑥→4
𝑥3
+6
𝑥−4
=

6. Materi limit
PoliteknikManufakturNegeri BangkaBelitung 6
Answer
Latihan2.1
1. lim
𝑋→3
𝑥−3
𝑥2+𝑥−12
=lim
𝑋→3
𝑥−3
( 𝑥−3)( 𝑥+4)
= lim
𝑋→3
1
𝑥+4
=
1
3+4
=
1
7
2. lim
ℎ→𝑜
(𝑥+ℎ)2
−𝑥2
ℎ
=lim
ℎ→𝑜
−𝑥3
−𝑥ℎ4
ℎ
=lim
ℎ→𝑜
-𝑥ℎ3
=-x. 03
=0
3. lim
𝑋→4
𝑥3
−64
𝑥2−16
=lim
𝑋→4
𝑥3
−64
𝑥2−16
= lim
𝑋→4
( 𝑥−4)( 𝑥2
+16)−(16𝑥+4𝑥2)
( 𝑥−4)( 𝑥+4)
=lim
𝑋→4
( 𝑥2
+16)−(16𝑥+4𝑥2
)
(𝑥+4)
=
(42
+16)−(16.4+(4)2
4+4
=
(16+16)−(64−64)
8
=
32−128
8
= -
96
8
= -12

7. Materi limit
PoliteknikManufakturNegeri BangkaBelitung 7
4. IF F(x)=5(x+b), find lim
ℎ→0
𝑓( 𝑥+ℎ)−𝑓(𝑥)
ℎ
find lim
ℎ→0
𝑓( 𝑥+ℎ)−𝑓(𝑥)
ℎ
= find lim
ℎ→0
(5( 𝑥+ℎ)+8)−(5𝑥+8)
ℎ
=lim
ℎ→0
5𝑥+5ℎ+8−5𝑥−8
ℎ
=lim
ℎ→0
5ℎ
ℎ
=lim
ℎ→0
5
=5
5. lim
𝑋→−3
5𝑥+7
𝑥2−3
=
5(−3)+7
(−3)2−3
=
−15+7
9−3
=
−8
6
=
−4
3
6. lim
𝑋→25
√ 𝑥−5
𝑥−25
= . lim
𝑋→25
(√ 𝑥−5)
(√ 𝑥−5)(√ 𝑥+5)
= lim
𝑋→25
1
(√ 𝑥+5)
=
1
(√25+5)
=
1
(5+5)
=
1
10
7. 𝐼𝐹 𝑔( 𝑥)= 𝑥2
, 𝐹𝑖𝑛𝑑 lim
𝑋→2
𝑔( 𝑥)−9(2)
𝑥−2
lim
𝑋→2
𝑔( 𝑥)−9(2)
𝑥−2
= lim
𝑋→2
2𝑥2
−(2)2
𝑥−2
= lim
𝑋→2
𝑥2
−4
𝑥−2
=lim
𝑋→2
(𝑥−2)(𝑥+2)
(𝑥−2)

8. Materi limit
PoliteknikManufakturNegeri BangkaBelitung 8
=lim
𝑋→2
(x+2)
= 4
8. lim
𝑋→0
2𝑥2
−4𝑥
𝑥
= lim
𝑋→0
2x-4
= 2.0 – 4
= -4
9. 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)
lim
𝑟→0
𝑟
𝑟(√ 𝑥+𝑟+√ 𝑥)
lim
𝑟→0
1
√ 𝑥+𝑟+√ 𝑥
1
√ 𝑥+0+√ 𝑥
1
√ 𝑥+√ 𝑥
1
2√ 𝑥
10. lim
𝑥→4
𝑥3
+6
𝑥−4
=
43
+6
4−4
=
64
0
= ∞

9. Materi limit
PoliteknikManufakturNegeri BangkaBelitung 9