This document contains a series of calculus exercises involving limits as x approaches various values. There are 10 exercises calculating limits of functions as x approaches infinity or specific values. The functions involve polynomials, radicals, and rational expressions. The exercises also determine whether functions are continuous at given values based on their limits.

1. TUGAS CALCULUS
(HAL 8-14)
D
I
S
U
S
U
N
Oleh :
Nama : Cinjy Saylendra Monica
Dita Yoriza
Indah Yanti
Monica Roselina
Prodi : Teknik Elektronika
Kelas : 1E A
Semester : 2 (dua)
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
Kawasan Industri Air Kantung Sungailiat, Bangka 33211
Telp. (0717) 93586, Fax. (0717) 93585
Email : polman@polman-babel.ac.id
Website : www.polman-babel.ac.id
TAHUN AJARAN 2014/2015

2. Latihan 2.2
1. lim
𝑥→∞
(5𝑥 − 7) = ∞
2. lim
𝑥→∞
7
𝑥3
=
7
∞
= 0
3. lim
𝑥→∞
3𝑥 + 95 = ∞ + 95 = ∞
4. lim
𝑥→∞
𝑥3−𝑥2+47𝑥+9
18𝑥3+76𝑥−11
=
1
18
5. lim
𝑥→∞
8
4−𝑥
=
8
∞
= 0
6. lim
𝑥→∞
𝑥−2
𝑥2−5𝑥+6
=
(𝑥−2)
(𝑥−2)(𝑥−3)
=
1
(𝑥−3)
=
1
∞
= 0
7. lim
𝑥→∞
𝑥5+6𝑥−7
5𝑥6+6𝑥2−11
=
𝑥5+6𝑥3−7
𝑥6
5𝑥6+6𝑥2−11
𝑥6
=
1
𝑥
+
6
𝑥3−
7
𝑥6
5+
6
𝑥4−
11
𝑥6
=
1
∞
+
6
∞
+
7
∞
5+
6
∞
−
11
∞
=
0
5
= 0
8. lim
𝑥→∞
7𝑥4+6𝑥2−3𝑥
−3𝑥3−7𝑥+5
=
7𝑥4+6𝑥2−3𝑥
𝑥4
−3𝑥3−7𝑥+5
𝑥4
=
7+
6
𝑥2−
3
𝑥3
−
3
𝑥
−
7
𝑥3+
5
𝑥4
=
7+
6
∞
−
3
∞
−
3
∞
−
7
∞
+
5
∞
=
7
0
= ∞
9. lim
𝑥→∞
2𝑥3+8𝑥−5
−3𝑥2+4
=
2𝑥3+8𝑥−5
𝑥3
−3𝑥2+4
𝑥3
=
2+
8
𝑥2−
5
𝑥3
−3
𝑥
+
4
𝑥3
=
2+
8
∞
−
5
∞
−3
∞
+
4
∞
=
2
0
= ∞
10. lim
𝑥→∞
5
𝑥2−4
=
5
𝑥2
𝑥2−4
𝑥2
=
5
𝑥2
1−
4
𝑥2
=
5
∞
1−
4
∞
=
0
1
= 0

3. Latihan 2.3
1. lim
𝑥→4+
[ 𝑥] + 1 = 4 + 1 = 5
2. lim
𝑥→2−
𝑥2−4
𝑥−2
=
( 𝑥−2)(𝑥+2)
𝑥−2
= 2 + 2 = 4
3. lim
𝑥→8+
4
𝑥−9
=
4
[8−9]
=
4
−1
= 4
4. lim
𝑥→0+
√4𝑥 + 3 = √4(0) + 3 = √3
5. lim
𝑥→5
( 𝑥 − 1) = 5 − 1 = 4
6. lim
𝑥→3−
𝑥2−9
𝑥−3
=
( 𝑥−3)( 𝑥+3)
𝑥−3
= 3 + 3 = −6
7. lim
𝑥→4+
7
𝑥−4
=
7
4−4
=
7
0
= ∞
8. lim
𝑥→4−
𝑥5+𝑥4−8
𝑥+4
=
(−4)5+(−4)4−8
−4+4
=
−1024+256−8
0
= −776
9. lim
𝑥→4+
𝑥2−16
𝑥−4
=
( 𝑥−4)(𝑥+4)
(𝑥−4)
= 4 + 4 = 8
10. lim
𝑥→4−
𝑥2−16
𝑥−4
= ( 𝑥 + 4) = 4−
+ 4 = −8

4. Latihan 3.1
1. 𝑓( 𝑥) = √5𝑥 − 7 𝑑𝑖 𝑥 = 1
𝑓(1) = √−2
lim
𝑥→1
√5𝑥 − 7 = √−2 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢
2. 𝑓( 𝑥) =
𝑥3−8
2−𝑥
𝑑𝑖 𝑥 = 0
𝑓(0) =
−8
2
= −4
lim
𝑥→0
𝑥3−8
2−𝑥
= −4 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢
3. 𝑓( 𝑥) =
4
√2𝑥−3
𝑑𝑖 𝑥 = 1
𝑓(1) =
4
√−1
lim
𝑥→0
4
√2𝑥−3
=
4
√−1
↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢
4. 𝑓( 𝑥) = [ 𝑥] 𝑑𝑖 𝑥 = 3
𝑓(3) = 3
lim
𝑥→0
[ 𝑥] = 3 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢
5. 𝑔( 𝑥) =
𝑥2+6
𝑥−5
𝑑𝑖 𝑥 = 4
𝑓(4) =
10
−1
= −10
lim
𝑥→4
𝑥2+6
𝑥−5
= −10 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢
6. 𝑔( 𝑥) =
√𝑥−5
𝑥−5
𝑑𝑖 𝑥 = 3
𝑔(3) =
√−2
−2
lim
𝑥→3
√𝑥−5
𝑥−5
=
√−2
−2
↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢
7. 𝑔( 𝑥) =
√𝑥−5
𝑥+2
𝑑𝑖 𝑥 = 8

5. 𝑔(8) =
2√2−5
10
lim
𝑥→8
√𝑥−5
𝑥+2
=
2√2−5
10
↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢
8. ℎ( 𝑥) = 5𝑥2
− √ 𝑥 + 7 𝑑𝑖 𝑥 = 5
ℎ(5) = 125 − √5 + 7 = 132 − √5
lim
𝑥→5
5𝑥2
− √ 𝑥 + 7 = 132 − √5 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢
9. 𝑓( 𝑥) =
𝑥−6
𝑥−2
𝑑𝑖 𝑥 = 6
𝑓(6) =
0
4
= 0
lim
𝑥→6
𝑥−6
𝑥−2
= 0 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢
10. ℎ( 𝑥) =
(𝑥−𝑎)2+𝑥−6
𝑥−𝑎+3
𝑑𝑖 𝑥 = 𝑎
ℎ( 𝑎) =
0+𝑎−6
0+3
=
𝑎−6
3
lim
𝑥→𝑎
(𝑥−𝑎)2+𝑥−6
𝑥−𝑎+3
=
𝑎−6
3
↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢