Al Majmaa’h Engineering College
Differential Calculus (Math 105)
Math 105
Appendix 3
Dr. SaMeH Ahmed
Appendix 3– Chapter 3: Math 105
Chapter
3
Limits and Functions
3.1 Limits
(1) To find lim 𝑓(𝑥) then find 𝑓(𝑎) 𝑡𝑜 𝑏𝑒
𝑥→𝑎
(i) If 𝑓(𝑎) ∈ 𝑅
∴ lim 𝑓(𝑥) = 𝑓(𝑎)
𝑥→𝑎
lim
As
𝑥 2 −4
𝑥→𝑎 𝑥+3
𝑓(3) =
9−4
3+3
5
=
6
∴ lim f(x) =
x→3
(ii) If 𝑓(𝑎) = ∞ 𝑜𝑟 = −∞
5
6
(undefined)
∴ lim 𝑓(𝑥) is not exist
𝑥→𝑎
As
lim
3𝑥+2
𝑥→2 𝑥 2 −4
𝑓(2) =
8
4−4
=
8
0
= ∞∉R
Page 2 of 12
Dr SaMeH
Appendix 3– Chapter 3: Math 105
Does not exist
If 𝑥 → 𝑎) 𝑓(𝑎) =
(iii)
0
0
undetermined quantity
(unspecified).
Then try to reduce the zero factors (x-a) by any of:
i. Factorization
ii. Long division by (x-a)
iii. Multiplication by conjugate
Examples:
1) lim 8
= 8
𝑥→2
2) lim 2𝑥
= 2×3=6
𝑥→3
3)
lim 2𝑥 2 − 20
𝑥+3
4) lim
=
2𝑥+7
𝑥→−2
5)
= 3 × (4) − 20 = 28
𝑥→−4
𝐥𝐢𝐦
𝒙𝟐 −𝟒
𝒙→𝟐 𝒙−𝟐
−2+3
1
=3
−4+7
− −−→
substitute by x
The answer will be
=2
0
0
Now try to reduce the zero- factor which is (x – 2) by
factorization as:
= lim
𝑥→2
6)
𝐥𝐢𝐦𝟏
𝒙→
(𝑥−2)(𝑥+2)
(𝑥−2)
𝟐𝒙𝟐 −𝟓𝒙+𝟐
𝟐𝒙−𝟏
=
lim (𝑥 + 2) = 4
𝑥→2
− −−→
substitute by x =1/2
𝟐
The answer will be
0
0
Page 3 of 12
Dr SaMeH
Appendix 3– Chapter 3: Math 105
Now try to reduce the zero- factor which is (2x – 1) by
factorization as:
2𝑥 2 −5𝑥+2
= lim
2𝑥−1
2𝑥→1
= lim (𝑥 − 2)
2𝑥→1
7)
𝐥𝐢𝐦
𝒙𝟑 +𝟖
𝒙→ −𝟐 𝒙+𝟐
=
lim
(2𝑥−1)(𝑥−2)
2𝑥→1
(2𝑥−1)
1
− 2=
2
=
− −−→
−3
2
substitute by x
=-2
The answer will be
0
0
Now try to reduce the zero- factor which is (x + 2) by
(𝑥+2)(𝑥 2 −2𝑥+4)
= lim
(𝑥+2)
𝑥→−2
8)
𝐥𝐢𝐦
−𝟐
𝒙→
= 4 + 4 + 4 = 12
𝟗𝒙𝟐 −𝟒
𝟑𝒙+𝟐
𝟑
(3x−2)(3x+2)
= lim
(3x+2)
3x→−2
9) 𝐥𝐢𝐦
= (- 2) – 2 = - 4
𝒙𝟒 +𝒙
𝒙→ −𝟏
= 𝑙𝑖𝑚
𝑥→−1
𝒙+𝟏
𝑥(𝑥 3 +1)
(𝑥+1)
𝑥(𝑥 + 1)(𝑥 2 − 𝑥 + 1)
= 𝑙𝑖𝑚
𝑥→−1
(𝑥 + 1)
= 𝑙𝑖𝑚 𝑥 × 𝑙𝑖𝑚 (𝑥 2 − 𝑥 + 1)
𝑥→−1
𝑥→−1
Page 4 of 12
Dr SaMeH
Appendix 3– Chapter 3: Math 105
= (-1) (3) = -3
10) 𝐥𝐢𝐦
𝟐𝒙𝟑 +𝟑𝒙𝟐 +𝟒
𝒙𝟑 +𝟖
𝒙→ −𝟐
In this problem, try to reduce the zero factor which is (x + 2) by
using long division by ( x + 2) for numerator on factorization
for denominator.
= 𝑙𝑖𝑚
𝑥→−2
(𝑥 + 2)(2𝑥 2 − 𝑥 + 2)
(𝑥 + 2)(𝑥 2 − 2𝑥 + 4)
2𝑥 2 − 𝑥 + 2
𝑥 + 2|
2𝑥 3 − 3𝑥 2 + 4
2𝑥 3 + 4𝑥 2
−−−−−−−−−−−−
−𝑥 2 + 4
−𝑥 2 − 2𝑥
−−−−−−−−−−−−
2𝑥 + 4
2𝑥 + 4
−−−−−−−−−−−−
0 0
2(−2)2 − (−2) + 2)
12
=
=
=1
(−2)2 − 2(−2) + 4
12
Make sure you understand the above steps and solve the
problem again...
Page 5 of 12
Dr SaMeH
Appendix 3– Chapter 3: Math 105
11) 𝒍𝒊𝒎
𝒙𝟑 +𝟐𝒙−𝟑
𝒙→ 𝟏 𝒙𝟒 −𝒙𝟑 +𝒙−𝟏
Try to reduce the zero factor which is (x -1) by using long
division.
𝑥2 + 𝑥 + 3
𝑥 − 1|
𝑥 3 + 2𝑥 − 3
𝑥3 − 𝑥2
−−−−−−−
𝑥 2 + 2𝑥 − 3
𝑥2 − 𝑥
−−−−−−−
3𝑥 − 3
3𝑥 − 3
−−−−−−−
0 0
Page 6 of 12
Dr SaMeH
Appendix 3– Chapter 3: Math 105
𝑥3 + 1
𝑥 − 1|
𝑥4 − 𝑥3 + 𝑥 − 1
𝑥4 − 𝑥3
−−−−−−−−−−−
0 +0 𝑥−1
𝑥−1
−−−−−−−−−−−
0 0
(𝑥 − 1)(𝑥 2 + 𝑥 + 3)
lim
𝑥→1 (𝑥 − 1)(𝑥 3 + 1)
=
12) 𝒍𝒊𝒎
𝒙𝟐 −(𝒂+𝟑)𝒙 + 𝟑𝒂
𝒙−𝟑
𝒙→ 𝟑
∴
1+1+3
5
=
1+1
2
𝑙𝑖𝑚
∴ 𝑙𝑖𝑚
𝑥→ 3
𝑥→ 3
= 𝟓 find a
𝑥 2 − 3𝑥 + (3𝑎 − 𝑎𝑥)
=5
𝑥−3
𝑥 (𝑥 − 3) − 𝑎(𝑥 − 3)
=5
(𝑥 − 3)
∴ 𝑙𝑖𝑚 (𝑥 − 𝑎) = 5
𝑥→ 3
3−𝑎 =5
∴ 𝑎 = −2
Page 7 of 12
Dr SaMeH
Appendix 3– Chapter 3: Math 105
𝒙𝟐 −(𝟑𝒌+𝟏)𝒙 + 𝟑𝒌
13) 𝒍𝒊𝒎
𝒙−𝟏
𝒙→ 𝟏
∴
= −𝟓 find k
𝑥 2 − 3𝑘𝑥 − 𝑥 + 3𝑘
= −5
𝑥−1
𝑙𝑖𝑚
𝑥→ 1
∴ 𝑙𝑖𝑚
(𝑥 2 − 𝑥 ) − (3𝑘𝑥 − 3𝑘)
= −5
(𝑥 − 1)
∴ 𝑙𝑖𝑚
𝑥(𝑥 − 1) − 3𝑘 (𝑥 − 1)
= −5
(𝑥 − 1)
∴ 𝑙𝑖𝑚
( 𝑥 − 3𝑘) = −5
𝑥→ 1
𝑥→ 1
𝑥→ 1
(1 − 3𝑘) = −5
−3𝑘 = −6
∴𝑘=2
√𝒙+𝟗 − 𝟑
𝒙𝟐 +𝟐𝒙
𝒙→ 𝟎
14) 𝒍𝒊𝒎
Try to reduce the zero factor (x) by multiply by the
conjugate as:
= 𝑙𝑖𝑚
𝑥→ 0
= 𝑙𝑖𝑚
𝑥→ 0
= 𝑙𝑖𝑚
𝑥→ 0
= 𝑙𝑖𝑚
𝑥→ 0
√𝑥 + 9 − 3 √𝑥 + 9 + 3
×
𝑥 2 + 2𝑥
√𝑥 + 9 + 3
𝑥+9−9
𝑥(𝑥 + 2)(√𝑥 + 9 + 3
𝒙
𝒙(𝑥 + 2)(√𝑥 + 9 + 3
1
(𝑥 + 2)(√𝑥 + 9 + 3
Page 8 of 12
Dr SaMeH
Appendix 3– Chapter 3: Math 105
=
1
2(√9+ 3)
=
1
12
H. W.
Find the value of each of the following:
1) lim (−6)
𝑥→1
2) lim 4𝑥
𝑥→0
3) lim (𝑥 3 + 2 )
𝑥→−2
4) lim (4𝑥 − 𝑥 2 )
𝑥→4
5) lim (𝑥 3 − 4𝑥 2 + 𝑥 − 4 )
𝑥→−1
6) 𝑙𝑖𝑚
3𝑥+1
𝑥→−1 𝑥 2 +3
Page 9 of 12
Dr SaMeH
Appendix 3– Chapter 3: Math 105
7) 𝑙𝑖𝑚
𝑥 2 −4
𝑥→2
𝑥−2
𝑥 2 −𝑥−2
8) 𝑙𝑖𝑚
𝑥+1
𝑥→−1
2𝑥 2 −5𝑥+2
9) 𝑙𝑖𝑚1
𝑥→
2𝑥−1
2
10) 𝑙𝑖𝑚
3−𝑥
𝑥→4 𝑥 2 −2𝑥−8
11) 𝑙𝑖𝑚
𝑥→2
12) 𝑙𝑖𝑚
𝑥 2 −4𝑥+4
𝑥 2 +𝑥−6
𝑥 2 +6𝑥+5
𝑥→−1 𝑥 2 −3𝑥−4
13) 𝑙𝑖𝑚
2𝑥 2 −𝑥−15
𝑥→3 3𝑥 2 −10𝑥+3
14) 𝑙𝑖𝑚
𝑥→2
15) 𝑙𝑖𝑚
𝑥 3 −8
𝑥−2
𝑥 3 +27
𝑥→−3 𝑥 2 +𝑥−6
16) 𝑙𝑖𝑚
𝑥→−2
17) 𝑙𝑖𝑚
𝑥→3
18) 𝑙𝑖𝑚
−2
𝑥→
𝑥 3 +8
𝑥+2
𝑥 3 −27
𝑥−3
9𝑥 2 −4
3𝑥+2
3
19) 𝑙𝑖𝑚
𝑥 3 +27
𝑥→−3 𝑥 2 +7𝑥+12
20) 𝑙𝑖𝑚
𝑥→−1
21) 𝑙𝑖𝑚
𝑥 4 +𝑥
𝑥+1
𝑥 3 +2𝑥−3
𝑥→1 𝑥 4 −𝑥 3 +𝑥−1
Page 10 of 12
Dr SaMeH
Appendix 3– Chapter 3: Math 105
2𝑥 3 +2𝑥 2 +4
22) 𝑙𝑖𝑚
𝑥 3 +8
𝑥→−2
𝑥 3 −2𝑥+4
23) 𝑙𝑖𝑚
𝑥→−2
𝑥 2 +2𝑥
√𝑥+9 − 3
𝑥→ 0 𝑥 2 +2𝑥
24) 𝑙𝑖𝑚
√𝑥+7 − 3
𝑥−2
𝑥→ 2
25) 𝑙𝑖𝑚
√𝑥+4 − 2
𝑥
𝑥→ 0
26) 𝑙𝑖𝑚
𝑥−1
27) 𝑙𝑖𝑚
𝑥→ 1 √3𝑥−2−√𝑥
√5𝑥−4 − √𝑥
𝑥−1
𝑥→ 1
28) 𝑙𝑖𝑚
29) If 𝑙𝑖𝑚
𝑥 2 −(3+𝑏)𝑥+3𝑏
𝑥−3
𝑥→3
30) If 𝑙𝑖𝑚
= 8 find b
𝑥 2 −(3𝑎+1)𝑥+3𝑎
𝑥→1
𝑥−1
= −5 find a
Solutions
Problem No.
Solution
Problem No.
Solution
1
(-6)
16
(12)
2
(0)
17
(27)
3
(-6)
18
(-4)
4
(0)
19
(27)
5
(-10)
20
(-3)
Page 11 of 12
Dr SaMeH
Appendix 3– Chapter 3: Math 105
6
(-1/2)
21
(5/2)
7
(4)
22
(1)
8
(-3)
23
(-5)
9
(-3/2)
24
(1/12)
10
(has no limit) 25
(1/16)
11
(0)
26
(1/4)
12
(-4/5)
27
(1)
13
(11/8)
28
(2)
14
(12)
29
b = -5
15
(-27/5)
30
a=2
Page 12 of 12
Dr SaMeH