Chapter one
1-Real number
ℝ = {𝑋: −∞ < 𝑥 < ∞}
If a and b are two real no. then one of the following is true
Some properties of R
1-if a>b then –a<-b
1
1
𝑎
𝑏
2- if a>b then <
3-if a<b ,b<c then a<c
4-if a<b then a+c<b+ c then a<c ∀ real no.
5-if a<b ,c<d then a+c < b+d
6-if a<b ,c any +ve real no. then a.c <b.c
7-if a<b ,c any -ve real no. then a.c >b.c
8-if 0<a<b , 0<c<d then a.c <b.d
9-if a<b<0 , c<d<0 then a.c >b.d
2-Intervals
An interval is asset of real no.s x having one of the following forms
1- Open interval (a,b) all real no.s x a<x<b
(𝑎, 𝑏) = {𝑋: 𝑎 < 𝑥 < 𝑏}
2- closed interval [a,b] all real no.s x a≤x≤b
[𝑎, 𝑏] = {𝑋: 𝑎 ≤ 𝑥 ≤ 𝑏}
1
3- half open from the left or half closed from the right (a,b] , all real no.s
x → a<x<b
(𝑎, 𝑏] = {𝑋: 𝑎 < 𝑥 ≤ 𝑏}
4- half open from the right or half closed from the left [a,b), all real no.s
x → a≤x<b
[𝑎, 𝑏) = {𝑋: 𝑎 ≤ 𝑥 ≤ 𝑏}
Note:
3 − 𝑰𝒏𝒆𝒒𝒖𝒂𝒍𝒊𝒕𝒊𝒆𝒔
Ex: find the set of all the following inequalities:
1−
2
7
> 2 ,𝑥 ≠ 0 ,
𝑥
𝑥∈𝑅
1st case : x>0
7 > 2𝑥 ⇒
7
7
>𝑥⇒𝑥<
2
2
7
Sol. set 1st case=(0, )
2
2st case : x<0
7
7
> 2 ⇒ 7 < 2𝑥 ⇒
<𝑥
𝑥
2
Sol. set 2st case= ∅
2-(x+3)(x+4)>0
Sol:
1st case: (x+3)>0 ʌ (x+4)>0
x>-3
ʌ x>-4
sol. Set 1st case =(-3,∞)
2st case: (x+3)<0
X<-3
ʌ
(x+4)<0
ʌ x<-4
sol. Set 2st case =(-∞,-4)
sol. set=(-3, ∞) U (-∞,-4)=R/[-4,-3]
H.W: find the set of all the following inequalities:
1- 2+3x<5x+8
2- 4<3x-2≤ 10
34-
3
𝑥
𝑥−3
<4, 𝑥 ≠3 , 𝑥 ∈𝑅
𝑥−1
𝑥 2 +𝑥−6
<0
5- Absolute value
The absolute value of real no. x denoted by |𝑥|is define as follows:
𝑥 𝑖𝑓 𝑥 ≥ 0
|𝑥| = {
−𝑥 𝑖𝑓 𝑥 < 0
Properties of absolute value
1-|𝑥| < 𝑎 𝑖𝑓𝑓 − 𝑎 < 𝑥 < 𝑎
2-|𝑥| ≤ 𝑎 𝑖𝑓𝑓 − 𝑎 ≤ 𝑥 ≤ 𝑎
3-|𝑥| > 𝑎 𝑖𝑓𝑓 𝑥 > 𝑎 𝑜𝑟 𝑥 < −𝑎
4-|𝑥| ≥ 𝑎 𝑖𝑓𝑓 𝑥 ≥ 𝑎 𝑜𝑟 𝑥 ≤ −𝑎
Ex: find the set of all the following:
1-|7 − 4𝑥| ≥ 1
Sol:
7 − 4𝑥 ≥ 1
𝑥≤
3
2
Or
7 − 4𝑥 ≤ −1
𝑥≥2
3
3
2
2
Sol. Set:=(∞, ] ∪ [2, ∞)=R/( , 2)
H.W: find the set of all the following:
1-|𝑥 − 4| < 5
2-|2𝑥 − 5| = 7
3-|𝑥 − 2| = |3𝑥 + 4|
4-|2𝑥 − 5| > 3
5-|
3𝑥+8
2𝑥−3
2
|≤4
6-|𝑥 + 2𝑥 − 24| > 0
Theorem:
Let a,b be two no.s then
1-|𝑎| = |−𝑎|
4
2-|𝑎. 𝑏| = |𝑎| . |𝑏|
𝑎
|𝑎|
3-| | = |𝑏|
𝑏
,𝑏 ≠ 0
4-|𝑎 + 𝑏| ≤ |𝑎| + |𝑏|
5-|𝑎 − 𝑏| ≤ |𝑎| + |𝑏|
6-|𝑎 − 𝑏| ≥ |𝑎| − |𝑏|
7-|𝑎 + 𝑏| ≥ |𝑎| − |𝑏|
Proof:
1-|𝑎| = |−𝑎| = √(−𝑎)2 = |−𝑎|
2-|𝑎. 𝑏| = |𝑎| . |𝑏| = √(𝑎𝑏)2 = √𝑎2 𝑏 2 = √𝑎2 . √𝑏 2 = |𝑎||𝑏|
𝑎
|𝑎|
𝑎
𝑎2
𝑏
𝑏2
3-| | = |𝑏| = √( )2 = √
𝑏
=
√𝑎2
√𝑏2
|𝑎|
= |𝑏|
4-|𝑎 + 𝑏| ≤ |𝑎| + |𝑏|
= √(𝑎 + 𝑏)2 = √𝑎2 + 2𝑎𝑏 + 𝑏 2 = √𝑎2 + 2√𝑎2 . √𝑏 2 + 𝑏 2 =
√(√𝑎2 + √𝑏 2 )2 = √𝑎2 + √𝑏 2 = |𝑎| + |𝑏|
5-|𝑎 − 𝑏| ≤ |𝑎| + |𝑏|
|𝑎 − 𝑏| = |𝑎 + (−𝑏)| ≤ |𝑎| + |−𝑏| = |𝑎| + |𝑏|
6-|𝑎 − 𝑏| ≥ |𝑎| − |𝑏|
|𝑎| = |𝑎 + 𝑏 − 𝑏| ≤ |𝑎 − 𝑏| + |𝑏| → |𝑎 − 𝑏| ≥ |𝑎| − |𝑏|
7-|𝑎 + 𝑏| ≥ |𝑎| − |𝑏|
|𝑎| = |𝑎 + 𝑏 − 𝑏| ≤ |𝑎 + 𝑏| + |−𝑏| → |𝑎 + 𝑏| ≥ |𝑎| − |𝑏|
5
Chapter two
1-Relation:
Let a,b≠ ∅ so every set subset of A× 𝐵 is represent relation from A
to B.
2-Functions and its Algebra:
A function from aset A to set B is arule that assigns each element
𝑥 ∈ 𝐴 to only element in B satisfy
𝑓: 𝐴 → 𝐵 𝑖𝑓𝑓 ∀𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 , 𝑓(𝑥) = 𝑦
3-The domain:
The set D is called the domain of f is the set of all x ∈ first component
occurring in the ordered pairs of f , denoted by dom (f)
Dom (f)={𝑥 ∈ 𝐴 ; (𝑥, 𝑦) ∈ 𝑓 , 𝑓𝑟𝑜𝑚 𝑠𝑜𝑚𝑒 𝑦 ∈ 𝐵}
4-The codomain:
The codomain is for all element in two set B .and denoted by 𝑐𝑜𝑑𝑓 such
that
𝑐𝑜𝑑𝑓 = {𝑦: 𝑦 ∈ 𝐵 }
5-Range:
Is the set of second component which assigne by element of first
component and denoted by 𝑅𝑓
𝑅𝑓 = {𝑓(𝑥); 𝑥 ∈ 𝐷𝑓 }
6-Kind of function
The function divided into kind
6
1-Algebraic function
i) polynomial function
𝑎
𝑦 = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + ⋯ + 𝑎𝑛 𝑥 𝑛 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑎0 , [′𝑂𝐿𝑈𝐾𝐽𝐻 , … . . , 𝑎𝑛 ∈ 𝑅
1
ii) Regular function
iii) Radical function
EX: find the domain of the functions:
1- 𝑦 = 𝑥 + 2
2- 𝑦 =
𝑥
𝑥 2 −1
⇒ 𝐷𝑓 = 𝑅
⇒ 𝑥2 − 1 = 0
⇒ 𝑥 2 = 1 ⇒ 𝑥 = ∓1 ⇒ 𝐷𝑓 = 𝑅/{1, −1}
1
1
3- 𝑦 = √2𝑥 − 1 ⇒ 2𝑥 − 1 ≥ 0 ⇒ 2𝑥 ≥ 1 ⇒ 𝑥 ≥ ⇒ 𝐷𝑓 = {𝑥: 𝑥 ≥ }
2
2
4- 𝑦 =
1
√1−𝑥 2
⇒ 1 − 𝑥 2 > 0 ⇒ 1 > 𝑥 2 ⇒ 𝑥 < ∓1 ⇒ 𝐷𝑓 = (−1,1)
H.W: find the domain of the functions:
1- 𝑦 = √𝑥
2- 𝑦 =
1
√𝑋
3- 𝑦 = √𝑥 2 − 9
4- 𝑦 = √1 − 𝑥 2
2-Non-Algebraic function
1-trigonometric function
5678-
Exponential function
Logarithmic function2
Absolute function
Identity function
7-Some operations on functions:
1-(𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥) ; 𝑥 ∈ 𝐷𝑓 ∩ 𝐷𝑔
2- (𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥) ;
7
⇒ 𝑑(𝑓 + 𝑔) = 𝐷𝑓 ∩ 𝐷𝑔
𝑥 ∈ 𝐷𝑓 ∩ 𝐷𝑔 ⇒ 𝑑(𝑓 − 𝑔) = 𝐷𝑓 ∩ 𝐷𝑔
3-(𝑓. 𝑔)(𝑥) = 𝑓(𝑥). 𝑔(𝑥) ; 𝑥 ∈ 𝐷𝑓 ∩ 𝐷𝑔
𝑓
⇒ 𝑑(𝑓. 𝑔) = 𝐷𝑓 ∩ 𝐷𝑔
𝑓(𝑥)
4-( ) (𝑥) =
; 𝑥 ∈ 𝐷𝑓 ∩ 𝐷𝑔 , {𝑥: 𝑔(𝑥) ≠ 0}
𝑔
𝑔(𝑥)
𝑓
⇒ 𝑑 ( ) = 𝐷𝑓 ∩ 𝐷𝑔
𝑔
− {𝑥: 𝑔(𝑥) = 0}
Ex:𝑓(𝑥) = √𝑥 , 𝑔(𝑥) = √1 − 𝑥 find 𝑓 + 𝑔 , 𝑓 − 𝑔 , 𝑓. 𝑔,
𝑓
𝑔
,
𝑔
𝑓
,and
its domain
Sol: 𝐷𝑓 = [0, ∞) , 𝐷𝑔 = (−∞, 1] →
(𝑓 + 𝑔)(𝑥) = √𝑥
𝑑(𝑓 + 𝑔) = 𝐷𝑓 ∩ 𝐷𝑔
= [0,1]
+ √1 − 𝑥
(𝑓 − 𝑔)(𝑥) = √𝑥 − √1 − 𝑥
(𝑓. 𝑔)(𝑥) = √𝑥 . √1 − 𝑥=√𝑥(1 − 𝑥) = √𝑥 − 𝑥 2
𝑓
𝑓(𝑥)
(𝑔) (𝑥) = 𝑔(𝑥) =
𝑔
𝑔(𝑥)
(𝑓 ) (𝑥) = 𝑓(𝑥) =
√𝑥
√1−𝑥
√1−𝑥
√𝑥
=√
=√
𝑥
1−𝑥
1−𝑥
H.W: 𝑓𝑖𝑛𝑑 𝑓 + 𝑔 , 𝑓 − 𝑔 , 𝑓. 𝑔,
𝑥
=
𝑓
𝑔
,
𝑔
𝑓
,and its domain of the following
functions:
1- 𝑓(𝑥) = √𝑥 + 1 , 𝑔(𝑥) = √4 − 𝑥 2
2- 𝑓(𝑥) = √𝑥 2 − 1 , 𝑔(𝑥) = 𝑥 2
3- 𝑓(𝑥) = 𝑥 2 + 1 , 𝑔(𝑥) = √2𝑥 − 3
5-composition of function
If f and g are function such that the range of g is a subset of domain of f
then there is a function fog define as follow:
𝑓𝑜𝑔(𝑥) = 𝑓(𝑔(𝑥))
8
𝑑(𝑓𝑜𝑔) = {𝑥: 𝑔(𝑥) ∈ 𝐷𝑓 , 𝑥 ∈ 𝐷𝑔 }
𝑑(𝑔𝑜𝑓) = {𝑥: 𝑓(𝑥) ∈ 𝐷𝑔 , 𝑥 ∈ 𝐷𝑓 }
In general
𝑓𝑜𝑔 ≠ 𝑔𝑜𝑓
Ex: 𝑓(𝑥) = √𝑥 , 𝑔(𝑥) = 𝑥 2 − 1 find 𝑓𝑜𝑔 , 𝑔𝑜𝑓 , 𝑑𝑓𝑜𝑔 , 𝑑𝑔𝑜𝑓
Sol: 𝑓𝑜𝑔(𝑥) = 𝑓(𝑔(𝑥)) = 𝑓(𝑥 2 − 1 ) = √𝑥 2 − 1
𝑔𝑜𝑓(𝑥) = 𝑔(𝑓(𝑥)) = 𝑔(√𝑥 ) = (√𝑥 )2 − 1 = 𝑥 − 1
𝑑𝑓 = 𝑥 ≥ 0 , 𝑑𝑔 = ℝ
𝑑(𝑓𝑜𝑔) = {𝑥: 𝑔(𝑥) ∈ 𝐷𝑓 , 𝑥 ∈ 𝐷𝑔 } ⇒ 𝑥 ∈ 𝐷𝑔 = ℝ
𝑔(𝑥) ∈ 𝐷𝑓 ⇒ 𝑥 2 − 1 ≥ 0 ⇒ (𝑥 − 1)(𝑥 + 1) ≥ 0
1- X-1≥ 0 ʌ 𝑥 + 1 ≥ 0
X≥ 1 ʌ 𝑥 ≥ −1
2- X-1≤ 0
→ {𝑥: 𝑥 ≥ 1}
ʌ 𝑥+1≤0
X≤ 1 ʌ 𝑥 ≤ −1
→ {𝑥: 𝑥 ≤ −1}
𝑑(𝑓𝑜𝑔) = {𝑥: 𝑥 ≤ −1, 𝑥 ≥ 1} = ℝ\(−1,1)
𝑑(𝑔𝑜𝑓) = {𝑥: 𝑓(𝑥) ∈ 𝐷𝑔 , 𝑥 ∈ 𝐷𝑓 }
= {𝑥: 𝑥 ≥ 0 , 𝑥 ∈ ℝ} = [0, ∞)
H.W: find 𝑓𝑜𝑔 , 𝑔𝑜𝑓 , 𝑑𝑓𝑜𝑔 , 𝑑𝑔𝑜𝑓 of the following functions:
1- 𝑓(𝑥) = √2 − 𝑥 , 𝑔(𝑥) = √𝑥 − 2
2- 𝑓(𝑥) = √𝑥 2 − 1 , 𝑔(𝑥) = 𝑥 2
3- 𝑓(𝑥) = 𝑥 2 + 1 , 𝑔(𝑥) = √2𝑥 − 3
9
4 − 𝑓(𝑥) = |𝑥| , 𝑔(𝑥) = −𝑥
5-𝑓(𝑥) =
𝑥
, 𝑔(𝑥) =
𝑥+2
𝑥−1
𝑥
Chapter three
1- Limits
The limits of f(x) as x approaches a is the no. L if given any∈ there
exists 𝛿 > 0 such that for all x, 0<|𝑥 − 𝑎| < 𝛿implies|𝑓(𝑥) − 𝐿| <∈
Theorem:
𝑙𝑖𝑚
1-if f(x)=c c constant ,then 𝑥→𝑎
𝑓(𝑥) = 𝑐
1-if
i)
ii)
iii)
iv)
𝑙𝑖𝑚
𝑥→𝑎
𝑙𝑖𝑚
and 𝑥→𝑎
𝑔(𝑥) = 𝑀 ,then :
𝑓(𝑥) = 𝐿
𝑙𝑖𝑚
𝑥→𝑎
𝑙𝑖𝑚
𝑥→𝑎
𝑙𝑖𝑚
𝑥→𝑎
𝐾𝑓(𝑥) = 𝐾𝐿 such that K is constant
𝑙𝑖𝑚
𝑙𝑖𝑚
𝑓(𝑥) ∓ 𝑥→𝑎
𝑔(𝑥) = 𝐿 ∓ 𝑀
(𝑓(𝑥) ∓ 𝑔(𝑥)) = 𝑥→𝑎
𝑙𝑖𝑚
𝑙𝑖𝑚
𝑓(𝑥) . 𝑥→𝑎
𝑔(𝑥) = 𝐿 . 𝑀
(𝑓(𝑥) . 𝑔(𝑥)) = 𝑥→𝑎
𝑙𝑖𝑚 𝑓(𝑥)
𝑥→𝑎. 𝑔(𝑥)
=
𝑙𝑖𝑚
𝑥→𝑎
𝑙𝑖𝑚
𝑥→𝑎
𝑓(𝑥)
𝑔(𝑥)
=
𝐿
𝑀
,M≠ 0
𝑛
𝑛
𝑙𝑖𝑚
𝑙𝑖𝑚 𝑛 (𝑥)
V) 𝑥→𝑎
= √ 𝑥→𝑎
𝑓(𝑥) = √𝐿
√𝑓
𝑙𝑖𝑚
𝑙𝑖𝑚
VI) 𝑥→𝑎
(𝑓(𝑥))𝑛 = ( 𝑥→𝑎
𝑓(𝑥))𝑛 = 𝐿𝑛 such that K is constant
𝑙𝑖𝑚
VII) 𝑥→0
.
sin 𝑥
𝑙𝑖𝑚
VIII) 𝑥→0
.
COS 𝑥
𝑥
=1
𝑥
=∞
1
𝑙𝑖𝑚
IX) 𝑥→0
. =∞
𝑥
Limits of polynomials
𝑖𝑓 𝑓(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎0 is any polynomial function
𝑙𝑖𝑚
𝑥→𝑐
𝑓(𝑥) = 𝑓(𝑐) = 𝑎𝑛 𝑐 𝑛 + 𝑎𝑛−1 𝑐 𝑛−1 + ⋯ + 𝑎0
Limits of quotient of poly
10
If f(x) and g(x) are polynomials ,then
𝒍𝒊𝒎 𝒇(𝒙)
𝒙→𝒂. 𝒈(𝒙)
=
𝒇(𝒄)
𝒈(𝒄)
,
Ex: find the limit ,if it exist
1-
3
𝑙𝑖𝑚 𝑥 −1
.
𝑥→1 𝑥−1
2-
𝑙𝑖𝑚
= 𝑥→1
.
𝑙𝑖𝑚 √𝑥+9−3
𝑥→0.
𝑥
(𝑥−1)(𝑥 2 +𝑥+1)
𝑙𝑖𝑚
= 𝑥→0
.
(𝑥−1)
𝑙𝑖𝑚 (𝑥 2
= 𝑥→1
.
+ 𝑥 + 1) = 3 ,
√𝑥+9−3 √𝑥+9+3
.
𝑥
√𝑥+9+3
𝑙𝑖𝑚
= 𝑥→0
.
𝑥+9−9
𝑥(√𝑥+9+3)
=
1
6
,
H.W:
4
2
𝑙𝑖𝑚 𝑥 −2𝑥 −8
1- 𝑥→2
. 2
𝑥 −4
𝑙𝑖𝑚
4- 𝑥→𝑎
.
√𝑥 2 +1−√𝑎2 +1
𝑙𝑖𝑚
7- 𝑥→2
.
𝑥−𝑎
(𝑥 2 −4)
√𝑥−2
3
2
𝑙𝑖𝑚 (1+𝑥)2 −1
- 𝑥→0
.
𝑥
1
𝑙𝑖𝑚
5- 𝑥→0
. (
1
𝑥 𝑥+2
𝑙𝑖𝑚
8- 𝑥→0
.
tan 𝑥
𝑥
1
− )
2
𝑙𝑖𝑚
3- 𝑥→3
.
√3𝑥−3
𝑥−3
𝑙𝑖𝑚
6- 𝑥→0
.
𝑠𝑖𝑛 3𝑥
𝑙𝑖𝑚
9- ℎ→0
.
𝑥
√𝑥+ℎ−√𝑥
ℎ
One -sided limits(right-hand limits and left-hand limits)
Def 1: the limits of the fun. f(x) as x approaches a from the right equals L if
given any ∈> 0 , ∃ 𝑎, 𝛿 > 0 such that for all x,
𝑙𝑖𝑚
𝑎 < 𝑥 < 𝑎 + 𝛿 Implies |𝑓(𝑥) − 𝐿| <∈ denoted by 𝑥→𝑎
+ 𝑓(𝑥) = 𝐿
Def 2: the limits of the fun. f(x) as x approaches a from the left equals L if
given any ∈> 0 , ∃ 𝑎, 𝛿 > 0 such that for all x,
𝑙𝑖𝑚
− 𝑓(𝑥) = 𝐿
𝑎 − 𝛿 < 𝑥 < 𝑎 Implies |𝑓(𝑥) − 𝐿| <∈ denoted by 𝑥→𝑎
Note
A function f(x) has a limit at appoint a iff the right –hand and lefthand limits at exist and are equal
𝑙𝑖𝑚
𝑥→𝑎
11
𝑙𝑖𝑚
𝑙𝑖𝑚
𝑓(𝑥) = 𝐿 ↔ 𝑥→𝑎
+ 𝑓(𝑥) = 𝐿 and 𝑥→𝑎 − 𝑓(𝑥)
Ex: find the limit of the function
𝑥 + 1, 𝑥 ≥ 1
𝑓(𝑥) = { 2
}
𝑥 − 1, 𝑥 < 1
as 1. 𝑥 → 5
2. 𝑥 → −5
𝑙𝑖𝑚
sol: 1. 𝑥→5
𝑓(𝑥) =
𝑙𝑖𝑚
𝑥→5
𝑙𝑖𝑚
2. 𝑥→−5
𝑓(𝑥) =
𝑙𝑖𝑚
b/ 𝑥→1−
𝑓(𝑥) =
∴
𝑙𝑖𝑚
𝑥→1+
𝑓(𝑥) ≠
(𝑥 + 1) = 6
𝑙𝑖𝑚
𝑥→−5
𝑙𝑖𝑚
3. a/ 𝑥→1+
𝑓(𝑥) =
3. 𝑥 → 1
(𝑥 2 − 1) = 24
𝑙𝑖𝑚
𝑥→1+
𝑙𝑖𝑚
𝑥→1−
𝑙𝑖𝑚
𝑥→1+
(𝑥 + 1) = 2
(𝑥 2 − 1) = 0
𝑓(𝑥)
H.w: find the limit of the function
𝑥 2,
𝑥≥0
3
1 − 𝑓(𝑥) = { 𝑥 , − 1 < 𝑥 < 0 }
𝑥+1,
𝑥 ≤ −1
If 1. 𝑥 → 3
2. 𝑥 → −
1
2
3. 𝑥 → −5
4. 𝑥 → 0
5. 𝑥 → −1
𝑥 2 − 1,
𝑥≥1
2 − 𝑓(𝑥) = {𝑥 + 1, 0 ≤ 𝑥 < 1}
𝑥3 − 1 ,
𝑥<0
If 1. 𝑥 → 1
𝑙𝑖𝑚
3- 𝑓(𝑥) = 𝑥→0
.
2. 𝑥 →
1
2
3. 𝑥 → −6
4. 𝑥 → 0
5. 𝑥 → 8
|𝑥|
𝑥
Limits at infinity (limits as 𝒙 → ∞ 𝒐𝒓 𝒙 → −∞ )
Def 1: the limits of the fun. f(x) as x approaches infinity is the no. L
𝑙𝑖𝑚
𝑥→∞ 𝑓(𝑥) = 𝐿 if given any ∈> 0 , ∃ 𝑀 > 0 such that for all x,
𝑀 < 𝑥 Implies |𝑓(𝑥) − 𝐿| <∈
12
Def 2: the limits of the fun. f(x) as x approaches negative infinity is the no. L
𝑙𝑖𝑚
𝑥→−∞ 𝑓(𝑥) = 𝐿 if given any ∈> 0 , ∃ 𝑁 < 0 such that for all x,
𝑥 < 𝑁 Implies |𝑓(𝑥) − 𝐿| <∈
Theorem:
1-
if f(x) =k for any no. k
𝑙𝑖𝑚
𝑥→∞
3- if
𝑙𝑖𝑚
𝑥→∞
i)
ii)
iii)
iv)
v)
vi)
vii)
𝑙𝑖𝑚
𝑥→−∞
𝑘=
𝑙𝑖𝑚
and 𝑥→−∞
𝑔(𝑥) = 𝑀 ,then :
𝑓(𝑥) = 𝐿
𝑙𝑖𝑚
𝑥→∞
𝑙𝑖𝑚
𝑥→∞
𝑘 = 𝑘 such that k is constant
(𝑓(𝑥) ∓ 𝑔(𝑥)) =
(𝑓(𝑥) . 𝑔(𝑥)) =
𝑙𝑖𝑚 𝑓(𝑥)
𝑥→∞. 𝑔(𝑥)
=
𝑙𝑖𝑚
𝑥→∞
𝑙𝑖𝑚
𝑥→∞
𝑓(𝑥)
𝑔(𝑥)
𝑙𝑖𝑚
𝑥→∞ 𝑓(𝑥)
𝑙𝑖𝑚
𝑥→∞
𝐿
=
𝑀
𝑓(𝑥) .
𝑙𝑖𝑚
∓ 𝑥→∞
𝑔(𝑥) = 𝐿 ∓ 𝑀
𝑙𝑖𝑚
𝑥→∞
𝑔(𝑥) = 𝐿 . 𝑀
,M≠ 0
𝑙𝑖𝑚 sin 𝑥
𝑥→∞. 𝑥 = 0
𝑙𝑖𝑚 cos 𝑥
𝑥→∞. 𝑥 = 0
𝑙𝑖𝑚 1
𝑥→∞. 𝑥
=0
Ex: find the limits of:
1.
2.
𝑙𝑖𝑚 𝑥
𝑥→∞. 7𝑥+4
𝑥
𝑙𝑖𝑚
= 𝑥→∞
. 7𝑥𝑥 4
+
𝑥
2
𝑙𝑖𝑚 2𝑥 +2𝑥−3
.
𝑥→∞ 𝑥 2 +4𝑥+4
𝑙𝑖𝑚
= 𝑥→∞
.
𝑥
𝑙𝑖𝑚
= 𝑥→∞
.
3𝑥 4 −𝑥 2
𝑥 4 −1
𝑙𝑖𝑚
2- 𝑥→∞
. (2 +
𝑙𝑖𝑚
3- 𝑥→∞
.
13
sin 𝑥
𝑥
)
3𝑥 2 −5𝑥 2 +𝑥
𝑥4
3
4
4
1+ + 2
𝑥 𝑥
H.W: Ex: find the limits of:
𝑙𝑖𝑚 √
1- 𝑥→∞
.
2
2+ − 2
𝑥 𝑥
1
4
7+
𝑥
=
=
1
7
2 3
𝑙𝑖𝑚
𝑥→∞.(2+𝑥−𝑥2 )
4 4
𝑙𝑖𝑚
𝑥→∞.(1+𝑥+𝑥2 )
=2
Infinite limits
Def 1: the limits of the fun. f(x) as x approaches a , 𝑎 ∈ 𝐷𝑓 such that is the
𝑙𝑖𝑚
infinity 𝑥→𝑎
𝑓(𝑥) = ∞ if given any 𝑀 > 0 , ∃ 𝛿 > 0 such that for all x, S.T
∀ 𝑥 ∈ 𝐷𝑓 , 0 < |𝑥 − 𝑎| < 𝛿 implies f(x)>M
Def 2: the limits of the fun. f(x) as x approaches a , 𝑎 ∈ 𝐷𝑓 such that is the
𝑙𝑖𝑚
-infinity 𝑥→𝑎
𝑓(𝑥) = −∞ if given any 𝑀 < 0 , ∃ 𝛿 > 0 such that for all x,
s.t ∀ 𝑥 ∈ 𝐷𝑓 , 0 < |𝑥 − 𝑎| < 𝛿 implies f(x)<M
Ex: find the limits
𝑙𝑖𝑚
1- 𝑥→2+
.
2.
1
=
𝑥 2 −4
1
0
3
𝑙𝑖𝑚 2𝑥 +2𝑥−1
𝑥→∞. 𝑥 2 −5𝑥+2
=∞
2
1
2+ 2 − 3
𝑙𝑖𝑚
= 𝑥→∞
. 1 𝑥5 𝑥2
− 2+ 3
𝑥 𝑥
2
= =∞
𝑥
0
1
𝑙𝑖𝑚
3- 𝑥→0+
. =∞
𝑥
1
𝑙𝑖𝑚
3- 𝑥→0−
. = −∞
𝑥
H.W: Ex: find the limits
𝑙𝑖𝑚
1- 𝑥→∞
.
𝑥 2 −𝑥
𝑙𝑖𝑚
2- 𝑥→∞
.
𝑥+1
𝑥5
3𝑥 4 +𝑥 3 −𝑥+1
2-continuity
Def1: the fun. y=f(x) is continuous at x=c iff all three of the follow
statements are true:
1- 𝑓 (c) is exists (𝑐 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑜𝑚𝑎𝑖𝑛 𝑜𝑓 𝑓)
𝑙𝑖𝑚
2- 𝑥→𝑐
𝑓(𝑥) is exists (f has a limit as x→ 𝑐)
𝑙𝑖𝑚
3- 𝑥→𝑐 𝑓(𝑥) = 𝑓(𝑐) (the limit equals the fun. value)
sin 𝑥
Ex:𝑓(𝑥) = {
14
,𝑥 ≠ 0
} is f cont. at x=0
1, 𝑥 =0
𝑥
Sol:
1-f(0)=1 is exist
𝑙𝑖𝑚 sin 𝑥
𝑥→𝑎 𝑥
𝑙𝑖𝑚
2- 𝑥→0
𝑓(𝑥) =
∴
𝑙𝑖𝑚
𝑥→0
= 1 𝑖𝑠 𝑒𝑥𝑖𝑠𝑡 𝑎𝑠 𝑥 → 0
𝑓(𝑥) = 𝑓(0) = 1
∴ 𝑓 is cont.
Def2: discontinuity at appoint
Is a fun. F is not cont. at a point c, we say that f is discontinuous at c and
call c appoint of discontinuity of f.
1
, 𝑥 ≠ −1
Ex: 𝑓(𝑥) = { 𝑥+1
} is f cont. at x=-1
1 , 𝑥 = −1
Sol:
1- f(-1)=1
2-
𝑙𝑖𝑚
𝑥→−1
is exist
𝑓(𝑥) =
𝑙𝑖𝑚 1
𝑥→−1 𝑥+1
𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡 𝑎𝑠 𝑥 → −1
∴ 𝑓 is not cont. at x=-1
H.w:
𝑋 2 +𝑋−6
1- 𝑓(𝑥) = {5
4
𝑥
,
2- 𝑓(𝑥) = {
𝑥=2
15
} is f cont. at x=2
𝑥2 − 1 , 𝑥 ≥ 0
} is f cont. at x=0
𝑥
, 𝑥<0
𝑋 2 −9
3-1- 𝑓(𝑥) = {
, 𝑥≠2
, 𝑥 ≠ −3
} is f cont. at x=-3
−6,
= −3
𝑥+3
𝑋 2 −4
4- 𝑓(𝑥) = { 𝑥−2
4,
, 𝑥≠2
} is f cont. at x=2
𝑥=2
|𝑥 + 1| + 2 , 𝑥 ≠ −1
5- 𝑓(𝑥) = {
} is f cont. at x=-1
0,
𝑥 = −1
Chapter 4
The derivative of a function
the derivative of afun. 𝑓 is the fun.𝑓 ´ whose value at is defined by the eq.
𝑙𝑖𝑚
𝑓 ´ (x) = ∆𝑥→0
.
𝑓(𝑥+∆𝑥)−𝑓(𝑥)
∆𝑥
whenever the limit exist
Ex: find the definition of derivative to find the derivative of the fun.
𝑓(𝑥) = 𝑥 2
𝑙𝑖𝑚
Sol: 𝑓 ´ (x) = ∆𝑥→0
.
𝑓(𝑥+∆𝑥)−𝑓(𝑥)
∆𝑥
𝑙𝑖𝑚
= ∆𝑥→0
.
(𝑥+∆𝑥)2 −𝑥 2
∆𝑥
==
2
2
2
𝑙𝑖𝑚 𝑥 −2𝑥∆𝑥+(∆𝑥) −𝑥
.
∆𝑥→0
∆𝑥
𝑙𝑖𝑚
= ∆𝑥→0
.
∆𝑥(2𝑥+∆𝑥)
∆𝑥
= 2x
H.W: find the definition of derivative to find the derivative of the
functions:
1- 𝑓(𝑥) = 2𝑥 − 1
2- 𝑓(𝑥) = (𝑥 + 1)2
4- 𝑓(𝑥) = √𝑥
5- 𝑓(𝑥) =
𝑥
3- 𝑓(𝑥) = 𝑥 2 − 1
6- 𝑓(𝑥) = 𝑥 3
𝑥−9
Some rules of derivative
1- If 𝑦 = 𝑓(𝑥) = 𝑐, where c is constant then
𝑑𝑦
𝑑𝑥
, 𝑓 ´ (x)=0
2- Power rule for positive integer (power of x) 𝑦 = 𝑥 𝑛 if n is positive
integer then
16
𝑑
𝑑𝑥
(𝑥 𝑛 ) = 𝑛𝑥 𝑛−1
3- The sum rule
𝑑
𝑑𝑥
𝑑𝑢
(𝑢 + 𝑣) =
𝑑𝑥
𝑑𝑣
+
= 𝑢´ + 𝑣 ´
𝑑𝑥
4-the constant multiple rule
𝑑
𝑑𝑥
(𝑐𝑓(𝑥)) = 𝑐
𝑑 𝑓(𝑥)
= 𝑐 𝑓 ´ (x)
𝑑𝑥
5-the product rule
𝑑
𝑑𝑥
𝑑𝑣
(𝑢. 𝑣) = 𝑢
+𝑣
𝑑𝑥
𝑑𝑢
𝑑𝑥
= 𝑢´ . 𝑣 ´
6-Positive integer power of adiff. fun. if u is adiff. Fun.n is power of u
then 𝑢𝑛
𝑑
𝑑𝑥
(𝑢𝑛 ) = 𝑛𝑢𝑛−1
𝑑𝑢
𝑑𝑥
7-The quotient rule
𝑑
𝑑𝑥
𝑢
(𝑣 ) =
𝑑𝑢
𝑑𝑣
− 𝑢
𝑑𝑥
𝑑𝑥
𝑣2
𝑣
8-negative integer power of adiff. fun. if u is adiff.
𝑑
𝑑𝑥
(𝑢𝑛 ) = 𝑛𝑢𝑛−1
𝑑𝑢
𝑑𝑥
Ex: find the derivative of the function:
1- 𝑓(𝑥) = 5 ⇒
2- 𝑓(𝑥) = 𝑥 6 ⇒
𝑑
𝑑𝑥
𝑑
𝑓(𝑥) =
𝑑𝑥
3
𝑓(𝑥) =
3- 𝑓(𝑥) = 𝑥 6 ∓ 𝑥 ⇒
4- 𝑓(𝑥) = 3𝑥 5 ⇒
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
(5) = 0
(𝑥 6 ) = 6𝑥 5
𝑑𝑥
𝑑
(𝑥 6 ∓
𝑑𝑥
𝑑
𝑓(𝑥) =
𝑑𝑥
𝑥 3) =
𝑑
𝑑𝑥
(𝑥 6 ) ∓
𝑑
𝑑𝑥
(𝑥 3 ) = 6𝑥 5 ∓ 3𝑥 2
(3𝑥 5 ) = 15𝑥 4
5- 𝑓(𝑥) = (𝑥 2 + 2)(𝑥 3 + 3𝑥 + 1) ⇒
𝑑
𝑑𝑥
𝑓(𝑥) = (𝑥 2 + 2)(3𝑥 2 + 3) + (𝑥 3 + 3𝑥 + 1). 2𝑥
𝑥
6- 𝑓(𝑥) = (𝑥 3 − )6 ⇒
2
17
𝑑
𝑥 5
1
𝑓(𝑥) = 6 (𝑥 3 − ) ( 3𝑥 2 − )
𝑑𝑥
2
2
7- 𝑓(𝑥) =
𝑥 2 −1
𝑥 4 +1
(𝑥 4
⇒
𝑑
𝑥 2 −1
)=
(
𝑑
𝑑
(𝑥 2 −1)−(𝑥 2 −1). (𝑥 4 +1)
𝑑𝑥
𝑑𝑥
(𝑥 4 +1)2
(𝑥 4 +1).
𝑑𝑥 𝑥 4 +1
+ 1). (2𝑥) − [(𝑥 2 − 1). 4𝑥 3 ] 2𝑥 5 + 2𝑥 − [4𝑥 5 − 4𝑥 3 ]
=
=
(𝑥 4 + 1)2
(𝑥 4 + 1)2
2𝑥 5 + 2𝑥 − 4𝑥 5 + 4𝑥 3 ] −2𝑥 5 + 4𝑥 3 + 2𝑥
=
=
(𝑥 4 + 1)2
(𝑥 4 + 1)2
8- 𝑓(𝑥) = (2𝑥 2 − 5𝑥 −2 )−5 ⇒
𝑑
𝑑𝑥
𝑓(𝑥) = −5(2𝑥 2 − 5𝑥 −2 )−6 . (4𝑥 +
Derivative of trigonometric function
1-if 𝑓(𝑥) = sin 𝑥 ⇒ 𝑓´(𝑥) = cos 𝑥
2-if 𝑓(𝑥) = cos 𝑥 ⇒ 𝑓´(𝑥) = −sin 𝑥
3-if 𝑓(𝑥) = tan 𝑥 ⇒ 𝑓´(𝑥) = 𝑠𝑒𝑐 2 𝑥
4-if 𝑓(𝑥) = cot 𝑥 ⇒ 𝑓´(𝑥) = −𝑐𝑠𝑐 2 𝑥
5-if 𝑓(𝑥) = sec 𝑥 ⇒ 𝑓´(𝑥) = sec 𝑥 tan 𝑥
6-if 𝑓(𝑥) = csc 𝑥 ⇒ 𝑓´(𝑥) = −csc 𝑥 𝑐𝑜𝑡𝑥
7-if 𝑓(𝑥) = 𝑒 𝑥 ⇒ 𝑓´(𝑥) = 𝑒 𝑥
8-if 𝑓(𝑥) = ln 𝑥 ⇒ 𝑓´(𝑥) =
1
𝑥
. 𝑓´(𝑥)
Note:
12-
𝑑
𝑑𝑥
𝑑
𝑑𝑥
(sin 𝑢(𝑥)) = cos 𝑢(𝑥). 𝑢´ (𝑥)
(cos 𝑢(𝑥)) = −sin 𝑢(𝑥). 𝑢´ (𝑥)
3−
4-
18
𝑑
𝑑𝑥
𝑑
𝑑𝑥
(tan 𝑢(𝑥)) = 𝑠𝑒𝑐 2 𝑢(𝑥). 𝑢´ (𝑥)
(cot 𝑢(𝑥)) = − 𝑐𝑠𝑐 2 𝑢(𝑥). 𝑢´ (𝑥)
10 𝑥 −3
56-
𝑑
𝑑𝑥
𝑑
𝑑𝑥
(sec 𝑢(𝑥)) = sec 𝑢(𝑥) tan 𝑢(𝑥) . 𝑢´ (𝑥)
(csc 𝑢(𝑥)) = −𝑐𝑠𝑐𝑢(𝑥) cot(𝑢(𝑥)). 𝑢´ (𝑥)
Ex: find the derivative of the function:
1- 𝑓(𝑥) = sin 𝑥 3 ⇒ 𝑓´(𝑥) = cos 𝑥 3 . 3𝑥 2 = 3𝑥 2 cos 𝑥 3
2𝑓(𝑥) = cos(5𝑥 2 + 2)3 + cot(1 − 𝑥 2 ) ⇒
𝑓´(𝑥) =
−𝑠𝑖𝑛(5𝑥 2 + 2)3 . 3(5𝑥 2 + 2)2 . 10 𝑥 − csc(1 − 𝑥 2 ) cot(1 −
𝑥 2 ) . −2𝑥
3- 𝑓(𝑥) = tan(𝑥 2 + 1) ⇒ 𝑓´(𝑥) = 𝑠𝑒𝑐 2 (𝑥 2 + 1). 2𝑥
𝟏
𝟏
𝟏
4- 𝒇(𝒙) = (𝟑 − 𝐬𝐞𝐜(𝟓𝒙))𝟐 ⇒ 𝒇´(𝒙) = 𝟐 (𝟑 − 𝐬𝐞𝐜(𝟓𝒙))−𝟐 . 𝐬𝐞𝐜(𝟓𝒙) 𝐭𝐚𝐧(𝟓𝒙) . 𝟓
5-𝑓(𝑥) = 𝑐𝑠𝑐 2 (2𝑥) + 4𝑥 2 𝑠𝑖𝑛𝑥
𝑓´(𝑥) = 2 csc(2𝑥) . − csc(2𝑥) cot(2𝑥) . 2 + 4(𝑥 2 𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛𝑥. 2𝑥)
𝑓´(𝑥) = −2𝑐𝑠𝑐 2 (2𝑥) cot(2𝑥) . 2 + 4(𝑥 2 𝑐𝑜𝑠𝑥 + 2𝑥 𝑠𝑖𝑛𝑥)
6- = ln(𝑥 2 − 8) ⇒ 𝑓´(𝑥) =
1
𝑥 2 −8
.2𝑥 =
2𝑥
𝑥 2 −8
7- 𝑓(𝑥) = 𝑒 −2𝑥 ⇒ 𝑓´(𝑥) = −2𝑒 −2𝑥
H.W: find the derivative of the function:
𝟏
1- 𝒇(𝒙) = 𝑡𝑎𝑛3 (COS 5𝑥3 )𝟐
3- 𝑓(𝑥) =
19
(𝑠𝑖𝑛√𝑥)3
√𝑥
2- 𝒇(𝒙) = cos(𝑥 4 + 𝟑𝑥 3 + 𝑥 2 )𝟕 + csc(tan𝑥 3 )
4- 𝑓(𝑥) = 𝑥 3 sin(2𝑥 2 + 3)