The Basic Knowledge for Calculus

8/23/2019 The Basic Knowledge for Calculus

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LESSON ONE

THE BASIC KNOWLEDGE FOR CALCULUS

1. An important tautology for using in a proof

1.1 p 1.2 1.3 p 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11

1.12

1.13 1.14 1.15 2 REAL NUMBERS

Real numbers consist of counting numbers, integers, rational numbers and irrational numbers.

Co-ordinate lines or real lines amount to the straight line lying below. An each of number of real

Numbers correspond to only one point on the real line

The point replacing a digit zero is called an origin. All points lying on the right-hand replace

positive

Real numbers, All points lying on the left-hand replace negative real numbers, Definition1.1 Let a,b There exist the numbers 0,1 such that

1. a 2. 1b 3. a 4. b b ,when b where

The numbers 0,1,

are called the identities for

RespectivelyTheorem 1.1 In the real numbers system, The following expressions are to be true.

1. There is the unique identity for addition and multiplication.

2. There is the unique inverse of each of a number for addition and multiplication.


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3. If ab then , when a,b Proof Leave the proof as exercises.

3 An inequality

Since, = and the trichotomy law as follows.1. 0 2. If a,b then ab and a 3. If x then x or x=0 or THE ONLY ONE CASE.

We can make the following definition.

Definition1.2 Let a,b

, We say that

1. a is greater than b which be written by a if and only if and2. a is less than b which be written by a if and only if b From this definition and the trichotomy law, we can make the following theorems.

Theorem1.2 Let a,b,c and d are real numbers. The following expressions are to be true.

1. If a then ab 2. If a then ab 3. If a then ab 4. If ab then 5. If a

then

6. If ab

then

7. If a then a 8.If a then 9. a if and only if a b 10. a if and only if ac , when c 11. a if and only if ac ,when c 12. If a then a 13. If a then a 14. If a then ac , when a,b,c,d 15. If ab then 16. If a then ac , when a,b,c,d 17. If a then , when a,b,c,d 18. If a

then

, when a,b,c,d

Proof we prove , and and leave the remaining points as exercises.14. Since, a bd 15. Since, ab 0


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18 Since a

, where a,b,c,d

Definition1.3 Let a , b and c are real numbers. It follows that.1. a 2. a 3. a 4. a

Example1.1 Show that if a

and d

then

1. a 2. a Solution we want to show the only point 2 and leave the point 1 as exercise.

Since, a and EXERCISE 1.1

1. Let a show that there exists a number k such that a and exemplify.2. Let a, b, c, d, e and f are positive real number, where a and d Show that 1. ad

2.

3. From the point 2, RESTRICT both inequalities when a, b, c, d, e and f are negative real

numbers.

4. Prove that

,

5. Let a,b , Prove that1. If a then there exists an c that make b 2. If there is c that make then a

6. Let a, b where a , Prove that.


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1. a b and 2. a 7. Prove that if a

then

8. Prove that if a then b Each man can approach to lifes highest success by tries and tries itself continuously.4. INTERVALS

Interval amount to SETS being subsets of SET of the real number possessing elements are all

Real numbers according to the restricted condition

Definition1.4 Let a, b , Where 1. Open interval

amount to

|

2. Closed interval amount to | 3. Half-open interval amount to | 4. Half-open interval amount to |

REMARK: ALL are called bounded INTERVALS.

Definition1.5 Let a 1. Infinite interval amount to | 2. Infinite interval

amount to

|

3. Infinite interval amount to | 4. Infinite interval amount to |

REMARK: ALL are called bounded INTERVALS. W e use the symbol and to refer topositive

And negative infinity and do not denote real numbers, they merely enable us to describe

Unbounded conditions more concisely.

5. ABSOLUTE VALUE

Definition1.6 Let a, b , the absolute value of a is written by ||And defined by ||

Theorem1.3 Let a, b , It follows that.1. || 2. || ||


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3. || || 4. || ||||5.

| | | |6.

||||

7. || || 8. ||9. | | || || 10. || || | |

11. || || | |Proof we only prove ,, points and exempt the remaining properties as exercises.

2. According to the definition || Therefore

||

And || Therefore ||

From the both cases, we conclude || || 4. We apply the definition as follows. ||

Consider the following expression.

|| |||| ||||

And || |||| |||| From the both cases, we can conclude that || |||| ,

9. we apply by the point as follows. || || || || So Implies that || || || || And then

| |

Therefore | | || || 6. ABSOLUTE VALUE AND DISTANCE

Definition1.7 When a and b are real numbers corresponding to points A and B on a real line

respectively and d is the distance between A and B. It will get.


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d Theorem1.4 If d is the distance between A and B then d | |Proof It is so easy.

Theorem1.5 Let k and a , it will get that.| | Proof It is so easy.

Theorem1.6 Let k and a , It will get that.

| |

Proof It is obviously.

Theorem1.7 Let k and a , It will get that.| | Proof It is obviously.

EXERCISE 1.2

1. Let a and a where a,b Find

that make x

lie together in the interval

2. Find same the one. When let a,b 3. Defined | | ( ) Let are intervals. Show that | | possesses properties same the theorem 1.3

Exempt a point 6 c,d Although it will be little success but, if it is arisen from the doing correctly by you, then

It is proudly and it is called that is the success that can be eaten.

7. MAXIMUM AND MINIMUM

Definition1.8 Let a and b are any real numbers.

Maximum of a and b are written by Max , Denoted by Max Minimum of a and b are written by Min , Denoted by Min


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REMARK: 1. Max 2. Min

3. Max 4. Min Theorem1.8 Let a, b

1. Max Min 2. Min Max 3. Max Min

Remark: 1. Max Max 2. Min Min Proof It is left as exercises.

Theorem1.9 the following expressions are to be true.

1. Max Max2. Min Min3. Max Min{ }4. Min Max{ }

Proof It is exempted as exercises. Hint: consider the six cases as follows.

a

EXERCISE 1.3

LET a ,b ,c and d are real numbers. Prove that.

1. Max Min 2. Min Max3. Max|| || || || || 4. Min|| || || || | |5. Max{ } Min{ }6. Min{ } Max{}7. Max

{ } Min

{ }

8. Min{ } Max{ }The giver gave and the receiver received but the giver possesses the success more

Than the receiver

8. THE AXIOM OF COMFLETENESS

Let S , We see that 5, 5.2, 6, 100 are greater than all elements of S.


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Similar these numbers are less than all elements of S.Definition1.9 Let S

and S

1. A real number a is called that is an upper bound of S if and only if a In this case, we say that a set S is bounded above.2. A real number b is called that is a lower bound of S if and only if b

In this case, we say that a set S is bounded below.

EXAMPLE1.2 Find lower bound and upper bound of each of the following sets 1. A B 3. C

4.

Solution 1.since x , So are a lower bound and an upper bound respectively.2. Since, x if let k is a lower bound. It follows that.

k but k , so k Thus k means that B has not a lower bound. There exists a contradiction.But B has an upper bound Such as 10

3. Similar C has a lower bound such as

but it has not an upper bound.

4. Similar D has not both a lower bound and an upper bound.

Definition1.10 Let S and S A set S is bounded if and only if a set S are bounded above and bounded below.

It follows that there exist a and b such that b Definition1.11 Let S and S C is called that be the least upper bound of S

if and only If 1. C is an upper bound of S 2. If k is any upper bound of S then C Or, if k

, then there exists an

such that

and we write

Similar D is called that be the greatest lower bound of S if and only if

1. D is a lower bound of S 2. If k is any lower bound of S then K Or, if k , then there exists an such that and we write

A set S is called that is bounded set when has both bounded above and bounded

below


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EXAMPLE1.3 Find the L.u.b and G.l.b of each of the following sets. 1. A

2. B

3. C | 4. D Solution 1.Since, , So

That is A which make o, are the G.l.b, the L.u.b respectively.2. Since, 0 So 0 That is 0 , Therefore B which make o,1 are the G.l.b,L.u.b

respectively.

3. Since, 0 , , So 0 ,That is 1 Therefore C which make 1, 2 are the G.l.b,L.u.b respectively.

4. Since, 2 , , So 0 But

, That is, , We can see that

is the G.l.b but there is no L.u.b, The uniqueness of L.u.b and G.l.b

Theorem1.10 Let . It obtain that1. If there is , then has a unique

2. If there is , then has a unique Proof Exempt as exercises by supposing that there are the two, and then find the contradiction.

The axiom of completeness

Definition1.12

Let A set S which is bounded above has a least upper bound. OrIf S is a set having an upper bound, then S possesses the least upper bound.

Now, If S is a set having an upper bound, then there exists an C such that l.u.b Example1.4 Let , we find that


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Therefore a set S is bounded set.

Important notice: 1. . We find that 2.

Theorem1.11 L et And if |

Proof According to definition 1.11 And 2.if

Since And 2.If ( ) Therefore

Theorem1.12 Let , if S is bounded below, then S has a greatest lower boundProof Since S

, and S is bounded below

So, there exists a real numbers such that And since which make a set is bounded above.And since is an upper bound of, thus according to the axiom of completenessSet have to possess the least upper bound

We choose is L.u.b now finally, according to the theorem above (), or Upper bound and lower bound for function

Definition1.13 Let f is real valued function having domain D if and | , then we say that f is a bounded function ona set S if and only if T is a bounded set.

REMARK: A set S is bounded set if and only if S are bounded above and bounded below

If and only if there exist any elements being both upper bound and lower bound


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Example1.5 Let , | We find that

and T is bounded set.

Therefore f is bounded function on And if let | Consider That is, and T is bounded set.Therefore f is bounded function on But if let | Consider

That is, and T is bounded set.Therefore f is bounded function on But if | ; So, is unbounded set.Therefore f is unbounded function on

Definition1.14 if f is a function with domain D and A D, then the image of A under f isDenoted by

, where

|

Definition1.15 a real valued function f with domain D is bounded function on If and only if is bounded set, where | Example1.6 Let f , find the image of A, where A Solution Since |

So, is the image of under fAccording to definition1.14 Let S= , |

Hence,

( ) ( )

( ) And if there is a number c which and , It follows that ( )


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Example1.7 Let f , Find

, since

Thus Since Thus Since Thus | ( ) From 1-3 we see that f is bounded function on Since are bounded sets.

EXERCISE 1.4

Find the least upper bound and the greatest lower bound of each of the following sets.

If possesses. 1. A 2. B 3. C 4. D 5. E 6. F 7. G 8. H | | |

If everybody recall to only mistakes itself and discover some good of other persons

Moreover, have to improve itself and congratulate with other person, then the society

Will has only common happiness

9. PRINCIPLE OF MATHEMATICAL INDUCTION.All sets being subsets of NATURAL number have to possess an element having least value.

Definition1.16 Let S and S , It follows that there exists xS which x , S.

Lemma1.13 Let S it will get that S if and only if


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1. 1S and2. If n

S then

Proof Let S see that are to be true. Suppose that are to be true. And SSo, , by wellordering principle have to exists m SWhich m S and since m-1 thus m-1 SAccording to 1S means that m1

Hence, , and from choosing m, we find that SAnd according to

will get that m

which is impossible.

Since arise a contradiction with respect to assumption m Therefore Theorem1.14

Let P instead of an expression concerning n positive integer. If P is to be true. And If P is to be true, then P is to be true, Then will obtain that P

is to be true,

Proof Let P instead of the expression gratifying the condition And let S | According to will get that 1S and according to will get that if kS implies that , then by the lemma 1.10 will obtain that S is the set of positive integer.

Therefore P is to be true for all n EXAMPLE1.8 Prove that Solution Let P

:

We have to show that P is to be true for all n . P is to be true, as . If accept P is to be true then have to show that P is to be true.Suppose that is to be true.Consider ( )


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We can see that is to be true.That is, P

is to be true.

Therefore, , Theorem1.15 Let P instead of an expression concerning n integer and a . P is to be true. . For k integer which k If P is to be true, then P is to be true.It follows that P is to be true foe all n integer that n .

Proof Since, n so n If let m

so will obtain m

And will obtain P and P are the same expressions.So in the proof that the expression P is to be true for all n integers which n ,It is the enough to prove that the expression P is to be true, Let Q instead of P Now, it will obtain that Q instead of P which is P

Q instead of P Q

instead of P

P

Consider, . Q is to be true, as P is to be true according to the given 1.. If Qis to be true will get p is to be true, where k It follows that k+a-1 and when P is to be true.According to the given 2 will get that P is to be true.

That is, Q is to be true, according to 1,2 can conclude thatQ is to be true, , hence P is to be true, .Therefore P

is to be true for all n

that n

Example1.9 Prove that Solution Let P: 1. P is to be true as 2. Suppose that P is to be true will get that

Consider


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( )

So, P is to be true.Therefore EXERCISE 1.5

1. Let P , Prove that P+1

2. Prove that 3. Prove that n

4. Prove that 5. Prove that 1+3+6+10+15+21+28++ , 6. Prove that 7. Prove that 1+2+ 8. Prove that ( ) () 9. Prove that 10. Prove that 2.6.10.14.18.22 The accuracy often be bored or kept down so deep, as our society are material societyAlmost man can not evade. How to lie without hardness and pain.

10. FUNCTIONS

If f is a function then f is a relation but if f is a relation then f is a function Definition1.17 A relation f that each of element in its domain is used only one to be called

A function and if f is a function from A into B then it is written by

f : AB, defined by y That is, if then Definition1.18 a function f is the function from A into B, if and only if , written by f: ABDefinition1.19 a function f is the one-to-one function from A into B if and only if


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1. f is the function from A into B and

2. Each of elements in its range if is used will be used only one, and written by f:

A BTest, if then Definition1.20 a function f is the function from A onto B, if and only if

1. f is the function from A into B

2. And written by Test, let there exists such that Definition1.21 a function f is the one-to-one correspondence function from

A onto B if and only if is f the one-to-one function with And written by

Example1.10 let , find the domain and range of f andShow that 1. f is a function from

2. f is a one-to-one function from 3. f is a function from

Solution Since,

, now, since

and

1. Let if we have then ,Thus, , where ,

2. Let , if we have Then , thus therefore

3. Let , we have to choose Finally, we substitute as follows , therefore 11. ALGEBRA OF FUNCTION ANDCOMPOSITE FUNCTION

Definition1.21 Let f and g are functions having domain and range are subset of

Real number, then The Sum, The Difference, The Product and The Quotient

Of f and g are written by and and defined by


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1. 2.

3. 4. EXAMPLE1.11 Let and , Find

And its domain and range

Solution since domain and range of f and g are respectively.So,

, and it follows that

1. 2. 3. 4. ,

Demonstration of each of ranges above

1.

4. 2, 3 have to utilize an application of differentiation.Defintion1.22 The composite function of g and f is written by

And defined by ( ), and { }EXAMPLE1.12 Let , and find Solution as

so, we have

1. ( ) ( ) { } And, then [ ] [ ]

Therefore [ ]


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2 () ( )

{ }

{ } | And, then

Therefore 3. ( ) ( )

{ }

And, then [ ] [ ]Therefore [ ]4. () ( ) +1 { }

And, then Therefore

12. THE INVERSE FUNCTIONS

Definition1.23 when the inverse of a function f is the function, it is called the inverse function of

f

And written by which And defined by y , Therefore

EXAMPLE1.13 Find the inverse function of the following function.

1. 2. 3. 4. | | Solution 1 As Defined by Considering,

So, Defined by , substitute y by x


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2. As , Defined by Considering,

by Quadratic formula, we have So, Defined by , substitute y by x

3. As Defined by Considering, by Quadratic formula, we have

, substitute y by x

So, , Defined by , respectively4. As Defined by | |

Considering, | | , substitute y by x

So,

,

Defined by

respectively

Theorem1.16 A function f possesses the inverse function if and only f is one to one function.

Proof when the inverse of f is a function, we have to show that f is one to one function.Let ; Therefore f is one to one function. When f is one to one function, we have to show that f possesses the inverseIs a function, Let

Therefore f possesses the inverse is a function.

13. INCREASING AND DECREASING FUNCTION.

Definition1.24 let f is real valued function of real variables, and A , it follows that1. f is a increasing function on A, if and only if


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If 2. f is a decreasing function on A, if and only if

If 3. f is a non-increasing function on A, if and only ifIf ( )

4. f is a non-decreasing function on A, if and only if

If ( ) Remark: A may be equal to such as f is decreasing function on its entire domain.EXAMPLE1.14 1 let

show that

f is a increasing on and a decreasing function 2 let show that q is a non-decreasing function on

Solution 1 As , ,where Hence, Therefore f is a increasing function on And, if , where Hence,

Therefore f is a decreasing function on 2 As

If Hence, If Hence,

, so, if

,

Therefore q is a non-decreasing function on its domain.

Theorem1.17 let f possesses the inverse function, it will obtain that

1. ifis increasing on its domain, and then is increasing on its domain.2. ifis decreasing on its domain, and then is decreasing on its domain.

Proof we prove the first part and leave the proof of the second part as an exercise.


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Let , then there exist such that

, and since

is increasing, hence

Holds precisely, when , therefore Which implies that is increasing.14. PERIODIC FUNCTIONS

We call functions that its graphs have recurrent character that is the periodic functions

Definition1.25 if g is a periodic function and p is a length of an interval such that

Its graph is recurred, then will obtain Definition1.26 if g is no constant function, then will be the periodic function there exists

p such that = Definition1.27 the primitive period or fundamental period amount to the shortest lengthSuch that its graph is recurrent character

Definition1.28 the periodic function g will has a period is p if and only if p is

The primitive period such that Amplitude and Asymptote

Definition1.29 let g is a periodic function which has maximum value and minimum

Value, g will has an amplitude is k if and only if ||Definition1.30 let is restricted equation

1 vertical asymptote, rearrange into and thenConsider a value of x such that is meaningless.Such as ; is a vertical asymptote

2 Horizontal asymptote; rearrange into and thenConsider a value of y such that is meaningless.Such as ; is a horizontal asymptote.

3 Oblique asymptote amounts to asymptote inclining and angle with x- axis

Substitute y by into and rearrange term intoA pattern , solve equations


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Will obtain m and b such that is oblique asymptoteTheorem1.18

Let is a function having a period is p and Then a function g has a period is ||Proof

|| is a period of g if and only if || is the least positive number such that || Let m is them such that Since So, we have And then f has the period is p such that

So, it follows that p , we will obtain ||Test, let g , hence || ||.

And then f has a period is p, now we have

( )

That is || According to will get || Therefore g has a period is

|| 15. MONOTONIC AND STRICTLY MONOTONIC FUNCTIONS

Definition1.31 a function f is monotonic on an interval if it is either non-increasing

on the entire interval or non- decreasing on the entire interval.

That is, 1 if then , 2. if then , only oneA function f is strictly monotonic on an interval if it is either increasing

on the entire interval or decreasing on the entire interval.

That is,


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1 if then , 2. if then , only oneTheorem1.19 if f is strictly monotonic on its entire domain, then it is one to one and has to

Possesses the inverse

Proof it is exempted the proof as an exercise. Hint by the contra positive of15. EVEN, ODD AND A TYPE OF FUNCTIONS.

Definition1.32 let f is real valued function of real variables, will obtain that

1 The function is even if 2 The function is odd if 3 A polynomial function amount to functions lying in pattern

,Where is a degree of this, and are coefficients having is the leading.4 A rational function amount to functions lying in this pattern. , such as 5 An exponential function amount to functions lying in this pattern , ,

Property: 1

2.

,

3 () 4 If 0 , then is decreasing function on its entire domain.5 If 1 , then e is increasing function on its entire domain.6 A logarithmic function amount to functions being the inverse of the exponential

Property: 1

2 then l is decreasing function on its entire domain.3 then l is increasing function on its entire domain.4 Let A, B


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e is irrational number which is assigned by Scottish mathematicianLeonhard Eule , where e Definition1.33 Let

is written by And ( ) In calculus, is defined by which it is invented by the ScottishMathematician John Napier

Definition1.34 , Let is original graph1.

Is horizontal shift h units to the right.

2. Is vertical shift h units to the left.3. Is vertical shift k units downward.4. Is vertical shift k units5. Is reflection with respect to the X- axis.6. Is reflection with respect to the Y- axis.

EXERCISE 1.6

1. Let

, find

and show that

1.1 is a function from 1.2 is one-one function from 1.3 is onto function from 1.4 If (() ) And then has a period is equal to four.

2. Let find


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3. Let , find 4. Let

find

.

5. According to point above, find 6. Let is one-one function from A to B, show that

6.1 6.2 7. Let f and g are real valued function of real variables

7.1 If f is increasing and there exists an inverse function then is increasing.7.2 If f is decreasing and there exists an inverse function then is decreasing.7.3 If f and g are increasing function and there exist an gof and fog thengof and fog are increasing function.

7.4 If f and g are decreasing function and there exist an gof and fog then

gof and fog are decreasing function.

8. If , prove that 9. If , show that 10. Prove that if f has an inverse, then the inverse is unique.

11. If f has an inverse, then 12. Prove that f has an inverse if and only if it is one-one function.13. Prove that if f and g are one-one function, then , 14. If f is strictly monotonic on its entire domain, then it is one-one

And hence, possesses an inverse.

15. Show that the function

is odd.

16. Show that is even.17. Show that the product of two odd or two even are even.

18. Show that the products of odd and even are even.

Desperate man always do desperately, we have to make and assign hence and hope


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With these group for comeback efficiently

16. BASIC ANALISYS GEOMETRY

STRAIGHT LINES AND DISTANCE IN A PLANE

Definition1.35 the slope of a non-vertical line passing through the points Are ; Equation of lines

Theorem1.20 let is a point lying on a line of slope mAnd are any other points on the line, then

, that is

Proof by utilizing property similar triangle

Theorem1.21 the graph of the equation is a line having a slope of mAnd a y- intercept at

Proof by theorem above

PARALLEL AND PERPENDICULAR LINES

Definition1.36 a angle of inclination of L amount to which measure anti-clockwiseFrom

to L

Theorem1.22 if L is non-vertical line and there exist m, are a slope and angle of inclinationRespectively then

Proof by definitions of slope and tangent

Theorem1.23 two distinct non-vertical lines are parallel if and only if

Their slopes are equal

Proof by definitions of function and one-one function

Theorem1.24 two distinct non-vertical lines are perpendicular if and only if

Their slope are rated by the following equation, Proof by utilizing and one-one functionTHE DISTANCE AND MIDPOINT FORMULAR

Theorem1.25 the distance d between the points in a plane


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Is given by Proof by these two points, the right-triangle can be formed and utilize the Pythagorean

Theorem

Theorem1.26 the midpoint of the line segment joining the points Is

Proof by property of similar triangle

CIRCLES

Definition1.37 let is a point in the plane and the set of all points Such that r is the distance between

is called a CIRCLE.

The point are called a centre and a radius of the circle respectively.Theorem1.27 the point lie on the circleof radius r and a centre

If and only if Proof by utilizing theorem1.25

EXERCISE 1.7

1 prove that the diagonal of a rhombus intersect at right-angle.

2 prove that the figure formed by connecting consecutive midpoints of the sides of

Any quadrilateral is a parallelogram.

3 prove that if the points lie on the same line as Then,

, assume 4 prove that if the slope of two non-vertical lines is negative reciprocals

Of each other, then the lines are perpendicular.

5 Prove that the distance between the points

and the line

Ax+ By+ C=0 is || 6 The distance between Ax+ By+ =0 and Ax+ By+ C=0 is ||7 Use the midpoint formula to find the three points that divide the line segment

Joining into four equal parts


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8 Prove that is one of the points of trisection of the line segmentJoining

, also, find the midpoint of the line segment joining

, To find the second point of trisection.9 Prove that the line segment joining the midpoints of the opposite sides of a

Quadrilateral bisects each other.

10 Prove that an angle is inscribed in a semi-circle is a right-angle.

11 Prove that the perpendicular bisector of a chord of a circle

Passes through the centre of the circle

12 let are diameters of a circle, prove that are parallel.Good youngsters should recognize when to start what has to be done,Good elders should know when to stop what is being done.

7. GRAPH, INTERCEPT, SYMMETRY AND INTERSECTION

Graph of an equation having two variables x and y

Definition1.38 let , its graph is the sets of all points in the planeThat is solution points of the equation.

That is such that are to be true.Intercept point on X-axis and Y-axisDefinition1.39 let is an equation

1 The point is X-intercept of its graph, when is to be true.2 The point is Y-intercept of its graph, when is to be true.

Definition1.40 let is an equation1 A graph of

=0 is said to be symmetric with respect to the Y-axis

if, whenever are points lying on its graph.That is, (x, y)p2 A graph of =0 is said to be symmetric with respect to the X-axis

if, whenever are points lying on its graph.That is, (x, y)p


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3 A graph of =0 is said to be symmetric with respect to the originif, whenever

are points lying on its graph.

That is, (x, y)p4 A graph of p =0 is said to be symmetric with respect to the identity functionif, whenever are points lying on its graph.That is, (x, y)p

POINTS OF INTERSECTION

Definition1.41 let =0 are all two equationsA point

is an intersected point of

=0 if and only if

Satisfies both equationsThat is Moreover, the intersected point of two its graph can be found by

Solving the equations simultaneously

EXAMPLE1.15 let 1 find X-intercept and Y-intercept2 show that it possesses symmetries with respect to the X-axis

The origin and the identity

Solution 1 let x=o will obtain has a solution set isHence Y-intercepts are Let y=0 will obtain has a solution set is Hence X-intercepts are Therefore it possesses X-intercepts and Y-intercept

2 The following expressions are to be true

Therefore it possesses symmetries with respect to the X-axis, the Y-axis

The origin and the identity consecutive

EXAMPLE1.16 finds all intersected points of graphs of Solution To find these two points, we proceed as follows : given equation


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: substitute y byx, rearrange termX=

: solve the equation

The corresponding values of y are obtained by representing x= and Into either of the original equations, we choose the equation y=-x, then

The values of y are - consecutive

Therefore the two intersected points are EXERCISE 1.8

1 Prove that if the graph of

=0 is symmetric with respect to the X-axis, Y-axis, then

It is symmetric with respect to the origin, given an example to show that the converse

Is not to be true

2 Prove that if the graph of =0 is symmetric with respect to the one axis and theOrigin, then it is symmetric with respect to the remaining axis also.

3 Let -x, write the definition to show that the graph of =0Is symmetric with respect to -x, and prove that the graph of =0Is symmetric with respect to

-x if and only if it is symmetric

With respect to the X-axis, the Y-axis and the origin.

Law can be evaded but khamma can not be evaded.

18. REVIEW TRIGONOMETRIC FUNCTIONS

Definition1.42 the relation r= | is called an unit circleRemark: 1

2 its graph has centre is the point 3 its X-intercepts are points

4 its Y-intercepts are points Let is a real number and measure a length of arc from a point According to an arc to be long || units, there exists a point such thatA correspond to or ( ) and we find that each of a will possess


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Only that is corresponding, so is a functionRemark: 1 measuring an arc anti-clockwise, if a

is positive

2 measuring an arc clockwise, if a is negativeAccording to rule above, we can define new function as follows1 Cosine |||

And it is written by Cos 2 Sine |||

And it is written by Sin According to two trigonometric functions will obtain four trigonometric

Functions as follows

3 Secant || And it is written by Sec , where Cos

4 Cosecant || And it is written by Csc

5 Tangent

||

And it is written by tan , where Cos 6 Cotangent || And it is written by Cot , where Sin

TRIGONOMETRIC IDENTITIES

Pythagorean identities

1 2 3 Reduction formula4 Sin 5 Cos 6 Tan 7 Sin 8 Cos 9 Tan

10 Sin 11 Cos 12 Tan


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13 Sin 14 Cos 15 Tan 16 Sin

17 Cos

18 Tan

19 Sin 20 Cos 21 Tan 22 Sin 23 Cos 24 Tan 25 Sin 26 Cos 27 Tan

The sum or the difference of two angles


The sum or the difference of the sum or the difference of two angles

31 Sin 32 Sin 33 Cos 34 Cos

The sum of the difference of an angle

35 Sin 36 Sin 37 Cos

38 Cos

Double angle formula

39 Sin2 40 Cos2 41 Tan2

Half- angle formula


Laws of Cosines and Sine

44 45 47 All points can be proved by the reader from the following hints.

1 in the unit circle, 2 the unit circle possesses symmetric with respect to the X-axis, Y-axis


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The origin, f and g 3 in a single circle or a same circle, chord AB equal to chord CD if and only if

Arc AB equal to arc CD

4 the Pythagorean Theorem and the distance d= 5 when has a, b, c are lengths of opposite sides ofA, B and

Respectively, we can assign arbitrarily A, B, C are standard position angles

6 a area of triangle = length of the baseheight

Evaluation of trigonometric functions by

1 decimal approximation with calculator or a table of trigonometric values

2 exact evaluations utilizing trigonometric identities and formulas from geometry

We demonstrate the second method first

EXAMPLE1.17 evaluate the following trigonometric functions

1 Cos 2 Sin 3 Sec 4 Tan, when Cos and Sin

5 if Sin and Sin , 0 Find 1 Sin

Solution 1 Cos 2 Sec 3 Sin 4 Tan2

5 since, Sin So, Consider Sin


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And Cos ..Replace Sin

Multiply Add ; 5Cos

Cos , Sin Therefore Tan such that Tan =Solving trigonometric equationsThe solving trigonometric equations are the finding values of in domainSatisfying the given trigonometric equations

EXAMPLE1.18 solve for in each of the following equations1 Sin 2 3 Tan

4 Cos

Solution 1 Sin And then So, , similarly Now, we have intersected points are That is Therefore a solution set is

2 rearranged term factorized by cross method Thus such that , therefore a solution set is


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| 3

: replace

,Cos2

: multiply both sides by Cos2 , so 2 Therefore a solution set is |

4 Cos : pull the common factors of two termS

: Theorem

Now, the answer is Therefore a solution is

Graph of trigonometric functions

In the XY-coordinate system we usually use the variable X in place of as follows Where x can have any real number

Domain and range of trigonometric functions

Since of are, hence ,

Periodic functionsTrigonometric functions are periodic functions and have periods as follows


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When b

we will obtain that

|| || || || || || || || || || || || || || || || || Amplitude

Since, and Thence, Now, we have A=

|| when y= y=And A=

|| || when y=a y=aFor remaining trigonometric functions have not amplitude, not applicable

Asymptotes

The equation x=n are vertical asymptote lines of y=And y= as Sin

The equation x= are vertical asymptote lines of y=

And y= as For

have not asymptote lines, not applicable

Graph of trigonometric functions

We can sketch and illustrate graphs of each of trigonometric functions by its domain, range,

amplitude,

Asymptote and period which these are left as exercises

EXERCISE1.9


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1 evaluate each of the following expressions

1.1

1.2 1.3 1.4 1.5

2 solve each of the following equations

2.1

2.3 2.42.5 Sin2x+Cos2x+Sinx+Cosx+1=0 2.6

3 show that 4show that

5 show that 6 show that if when p is the length of an interval that is

The recurred graph then

6.1 6.2 7 let 7.1 show that f is the period function.

7.2 show that there exists the primitive period is 1.

8 show that the trigonometric functions Sine, Cosine, Tangent, Secant, Cosecant

And Cotangent are the period functions and there exist the primitive period

Are 4 show that the trigonometric functions Sin , Cos , Tan , Sec Csc possess the primitive are || || || || || ||One after the other and the graph of trigonometric functions above are obtained by


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Shift its original graphs according to X-axis is ; All remaining times of each person always have value for itself and society

As each person can utilize the remaining times make the good work

And assign to refer to society

LOSSON TWO

LIMITS AND CONTINUTIES

1 An introduction to limits

Let f , consider for all values other than x=1When x approaches 1 from left or right will make f

Although x can not be equal to 1, but we can move arbitrarily close to 1 and as result

f moves arbitrarily close to 2, we say that the limit of fas x approaches 1 is 2And it is written by Ordinarily, if f becomes arbitrarily close to a single number L as x approaches a fromEither sides, then we say that the limit of f as x approaches a is L and it is written by The next three cases for which a limit does not exist

1 f approaches a different number from the right side ofa than it approaches from theLeft side, such as ||

2 f increases or decreases without bound as x approaches a such as 3 f oscillates between two fixed values as x approaches a such as

According to if and only if the following two phrases are true1 f

becomes arbitrarily close to L

2 and x approaches a

The following is assigned by Augustin-Loius Cauchy His definitionOf a limit is the standard which be utilized today

Consider the first phrase; f becomes arbitrarily close to L means that f lies in theInterval in terms of absolute value as follows


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f

| |

Similarly, the second phrase; x approaches a means that there exists a positive number Such that x lies in either in the interval That is x ( )

| | | | | |

The first inequality, | |, Expresses the fact that x , while the second,| | Says that x is within a distance ofaThese bring us to the following formal definition of a limitDefinition2.1 let f be a function defined on an open interval containing a And let L be a real number, the statement Means that for all x , for each there exists

A

such that

| | whenever 0

| |

The present, the mathematician assigned obviously the definition of a limit

By a limit point

Definiotion2.2 let D and a , we say that a is a limit point of D if and only ifFor all that appoint it will obtain ( )

EXAMPLE2.1 let A= We find that all points in A, B are limit points of A, B one after the other.

And all points in C are limit points of C except 6 as there exists

=0.5 such that

Definition2.3 let f: D , when D and a is a limit point of D we will say that fpossesses

A limit to be a number L as x approaches a if and only if for all x , for each , there exists a such that | | whenever | |


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And it is denoted by EXERCISE2.1 prove that

by

definition of a limit

Solution let D=, we have to show that there exists a such that Whenever 0 | | , 1 is a limit pointWe begin by writing = || Since, | | || thence = || That is, when , we choose and we choose It follows that whenever 0 | | we have Finally, letting , (

)It follows that whenever | |

EXample2.2 utilize the definition of a limit t0 prove that Solution let x

we have to show that

there exists a

Such that whenever | | Considering, since| | |||| And | | , So, | | and || , we can say that , we choose It follows that whenever | | we have And if , we will choose , it follows that whenever | | We have , finally letting will obtain that Whenever | |

Example2.3 prove that by the definition of a limitSolution let x , we are required to show that , there exists


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a such that whenever | | We have to begin by writing |||||| Since || | | | | | | and || ; Hence will obtain that whenever | | We can say that when , we have to choose And if , we have to choose such that Finally, letting it follows that whenever | | ,We have

Theorem2.1 let f: when and a is a limit point of D, and if existThen the value of a limit has unique

Proof according to the given information, we have

If Suppose that

imply that

| | 0, we choose

| |

And, because , there exists a that make| | | |Whenever | | Similar there exists a that make| | | | Whenever | | Thus, letting and then a is a limit point such that there existsX where x

( ) , now we have

| | | | | | | | | | | |=| |, therefore | | | | which be a contradiction, so

EXERCISE2.1


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1 prove the following statements by the 1

2

3 4 |||| 5 6

2 let L and M are real numbers, prove that if then 3 prove that | | if and only if a=b4 prove that there is not L that make 5 prove that there is not L

6 prove that 7 prove that 8 prove that there is not L that make 9 let f: D D and defined by f=0, 1 when x are rational numbers and irrationalnumbers

One after the other, prove that there is not L

that make

The function is assigned by the German mathematician Peter Gustav DIRICHLET If goodness, happiness, success, love, accuracy, and, and, can be knew, I shell tell them

That down to you

2 LEFT-HAND LIMITS AND RIGHT-HAND LIMITS

Definition2.4 let f: When D and a is a limit point of D We say that there exists a limit to be a real number L as x approaches left-hand

Of a if and only if , for each there exists a such that| | Whenever a- and we write Definition2.5 let f:D , When D and a is a limit point of D

We say that there exists a limit to be a real number L as x approaches right-hand

limit


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If and only if , for each , there exists a such that

| | Whenever a

and we write

Theorem2.2 let f: D and a is a limit point of either D or D Where L then will obtain that Proof let if and only if 0 such that | |

Whenever 0| | And since is tautologyNow, we have such that | | Whenever a-

and

such that

| | Whenever a , therefore It is exempted to be exerciseExample2.4 let a Pricewise function f as follows

Show that and

Solution let f: So,

We have to show that there exists a such that | | Whenever 2 , since our choice for depend on , we try to establishA connection between the in-equality| | and By simplifying the first in-equality, we will get| | | | | |

Finally, we choose , this choice works, because Implies that | | The remaining is exempted as exercise

EXERCISE2.2


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1 show that

=1, when f

and

2 show that and 3 show that || and || 4 prove that 1

2 || 3

5 prove that there is not L

that make

6 prove that there is not L that make || 7 prove that there is not L that make

3 PROPPERTIES OF LIMITS

A strategy for finding limits, limits of algebraic functions and limits of trigonometric functions

We have pointed out already that a limit of fas x approach a does not depend onThe value of x at x=a, however if it is happened that the limit is precisely f, then we sayThat the limit can be evaluated by direct substitution, that is

Substitute for xSuch well-behaved functions are said to be continuous at a, an important application of

Direct substitution is shown in the following theorem

Theorem2.3 let a and f g for all x in D, where a is a limit point of DIf the limit of gas x approaches a exists then the limit of f also existsAnd

Proof since f

=g

,

x

a which a is a limit point of D

So, f=g, ( ) for each restrictive And if assume that by the definition of the limit, it follows thatFor each there exists a such that | | whenever 0 | | However, since f=g, , where xa


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Now, we have | | whenever | | Therefore we can conclude that

also

That is EXAMPLE2.5 let f and g , show that f and g haveThe same values for all x other than x=2

Solution by factorizing the numerator of f, we have

f () Hence, if x ,we can cancel like factor to obtainf () =g x We see that where x f=g so

A strategy for finding limit

1 Learn to recognize which limits can be evaluated by direct substitution

2 if the limit of f as x can not be evaluated by direct substitution, try to discoverA function g that agrees with f for all x other than x=a Of g

can be evaluated by direct substitution )

3 apply theorem before to conclude that Limits of algebraic functions Theorem2.4 if a and b are real numbers and n then the following properties

Are true 1 2 3 4 Proof we prove the property 3 and exempt the proof of the others as an exercise

Assume that , so | | Hence we choose

| |there exists a

such that

| | | | Whenever | | Let x and then we choose x=a+0.1Considering | | | | | | whenever | | Now, it arise of a contradiction


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Therefore Theorem2.5

Let a and b are real numbers and n , f and g be functions whose limitsExist as x then will obtain that1 if then || || | |2 if then 3 scalar multiple: 4 if then there exists a such that

|| Whenever

| |

5 sum or difference: 6 products: () 7 if then there exists a such that|| || Whenever | | 8 if then 9 quotients:

10 Powers: () 11 nth-roots:

Proof 1since there exists a such that| | Whenever | | And as || || | | Whenever | | Finally, we have || ||

2 since

if and only if for each restrictive

there exists a

Such that | | Whenever 0 | | And then | | | | | | || || Therefore, we can write

3 assume that , let then and since || ; b


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We know that there exists a such that

| |

||Whenever

| |

Considering | | || ||| | |||| || || | | | | ; || Finally, we say that for each there exists a such that| |

Whenever

| | it follows that

4 assume that will obtain that there exists a

Such that | | Whenever | | And since || || | | and then|| || Whenever | |

5 let and For all

then, since

we know that there exists a

Such that | | Whenever | | and| | Whenever | | We complete by choosing such that | | and| | Whenever | | Finally, we operate by the triangle inequality as follows

( ) | | | | This implies that ( ) 6 let and M

We have to express that there exists a such that| | Whenever | |


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Considering | | | |=

| | ||| | ||| |

So, there exists a such that || || And | | || | | ||| |And since

|| || there exist | | || , Whenever | | and| | || , Whenever | | Finally, let we choose such that

| | || || + || Whenever | | Therefore () 7 since then, since || there exists a

Such that | | || Whenever | | And since || || | | || Now, we have

|| ||

whenever

| |

For the remaining are left the proof as exercise

Theorem2.6 If P is polynomials having degree n, a is a real number then

Proof let P is a polynomialRepeated applications of the sum and scalar multiple properties produce

Finally, utilizing properties 1, 2 and 3 of theorem2.4 we obtain =PTheorem2.7 if R is a polynomial function which is given by R , and a is


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A real number that make q , then Proof the proof is left as exercise

Theorem2.8 If f and g are functions that make Then ()

Proof considering gof=g()=, where u=f and x f For each restrictive , we have to discover a such that() Whenever | |

Since

g

, we know that there exists a

such that

| | Whenever | | Now, will obtain that | | whenever | | , replace x by u

Moreover, since when , we know that there exists a Such that | | whenever | | Finally, we have () | | | | | | | | Then will get that() | |

Therefore () Theorem2.9

If h f g, where a be a limit point of D, except possibly at aItself, and if

Proof because

Hence for each restrictive there exist and such that| | Whenever | | And | | whenever | | Now, letting and then we have| | | | | |


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Considering | | And

| |

But, since h f g, it follows that And, then | | | | Finally according to the - definition of a limit, we have Or considering | | | | | | | | | | | | | | | | | | | | | |=| | | | Whenever

| |

Theorem2.10 define If f , where a is a limit point of D that make | | And then

Proof hint: assume that then discover a contradiction, leave the proof as exerciseEXERCISE2.3

1 Prove that if exists and does not existThen

does not exist

2 Prove that if then () = = 3 Prove that ; m, n and is defined4 Let f be a piecewise function, where

F= , find the values of a and b that make

Exist

5 Discover two functions f and g that make do not existBut ( ) exists

6 An expression if|| || then Show that that the expression is false

7 Prove that


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8 Prove that 9 Prove that

10 Prove that if then 11 Prove that if || 4 LIMITS OF TRIGONOMETRIC FUNCTIONS

The limits of basic trigonometric functions can be evaluated also by direct substitution.

Theorem2.11 if a is a real number lying in domain of the given trigonometric

Functions, then the following properties are true

1

2

3 4 5 6 7 8 9 10

Two special limits of trigonometric functions

11 12 13 If and there exists a Such that f ( ) then () Proof 1 Since if and only if for each there exists a

Such that | | whenever || So, we have to find the bearing between | | and ||For restricting Considering the unit circle: from the following figureCase 1when x

From a figure, co-ordinate of point C is While and are vertical each otherNow, by properties of a triangle and chord, we have The length of arc BC =x=||, so | | || or 1- ,

O AB

C (cos x, sin x)


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Case 2 when x Considering 1

=1

Therefore | | || According to the two cases above conclude that | | || Finally, for each | | || || That is 2

Since , x , x And So, we have = = =0That is 6 Now, by letting h=x-a and observing that x , we can write

= =

11 by the squeeze theoremConsidering the unit circle: from the following figureCase 1 when x From the figure,

and

are vertical at a point A

Such that So, co-ordinate of a point B is . Now, we can see thatArea of a triangle OACarea of circular sector OACarea of a triangle OABThat is

B

C

O A


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While x we find that Hence by the squeeze theorem,

And while x , it will obtain that And then Hence by the squeeze theorem, Finally, from 1 and 2 we can conclude that For remaining properties are exempted as exercises

EXERCISE2.4

1 Determine the following limits 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15

2 Use the squeeze theorem to find 2.1 c=0, 2.2 c=a, b| | | |

3 Prove that

4 prove that

5 Prove that there is not that make 6 Prove that there is not that make

5 INFINITE LIMITS

A limit in which fincreases or decreases without bound as x approaches a


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Is called that be an infinite limits

Definition2.6 let f is real valued function of a real variable, where

f: D when D and a is a limit point of D1 Statement means that for each restrictive B there existsa such that whenever | | ,

2 Statement means that for each restrictive B there existsa such that whenever | | ,

Remark: according to definition above, we can define the infinite limit from the left

By replacing 0

| | by a-

a, and we can define the infinite limit

From the right by replacing 0 | | by a a+ as follows.1 means that for each restrictive there exists a Such that whenever a ,

2 means that for each restrictive there exists a Such that whenever a- ,

3 means that for each restrictive there exists a Such that

whenever a

,

4 means that for each restrictive there exists a Such that whenever a- , Be sure to notice that the equal sign in the statement lim f does notMean that the limit exists? On the contrary, it tell us how the limit fails

To exist by denoting the bounded behavior of fas x approach aEXAMPLE2.6 Prove that 1 , 2 , 3

Solution 1 For each B

we have to show that there exists a

such that

| | Consider

Finally, we see that for each B we have a Therefore


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3 we have to show that for each there exists a such that | |

Considering | | 1 when x , then 0 and2 when x then 0 Finally, we can conclude that fro each B there exists a That make

whenever 0 | | Therefore Theorem2.12 if and only if and there exists a Such that whenever a

Theorem2.13 if and only if and there exists a Such that whenever a

Theorem2.14

if and only if

and there exists a

Such that whenever a Theorem2.15 if and only if and there exists a Such that whenever a

Proof we prove only the theorem2.12 and leave the proof of the remaining theorem as

exercises Since if and only if for each restrictive B there exists a 0Such that f

whenever a

So, when B=1 there exists a such that f whenever a And for each restrictive , we choose B= there exists a such that

f whenever a Now, by letting , we have f whenever a


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That is whenever a therefore

Since

for each restrictive B

, we choose

there exists a

Such that whenever a but as there exists a such that

f Finally, we let and then we have f and Whenever a now, it follows that whenever a Therefore,

,

Remark: according to the four theorems above, we obtain that

1 and there exists a Such that whenever | | and

2 and there exists a Such that whenever | |

Definition2.7

If fapproaches infinity or negative infinity as x approaches a from the leftOr right, then we call the line x=a that is a vertical asymptote of graph of f

That is the line x=a is a vertical asymptote of the graph of f if and only if or EXAMPLE 2.7 Show that 1 and there exists a

Such that f whenever , Solution 1 For each B

we have to discover

A such that whenever -1 , Considering

Now, while and , it is implied that , Therefore


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2 For each we have to discover a such that

Whenever -1

,

Considering Now, while and -1 , it is implied that Therefore and we have x=-1 is a vertical asymptote.

Theorem2.16 If a and L are real numbers and f, g are functions that make

and then the following properties are true.

1The sum or the different: ( ) 2 The product: 3 The quotient: Similarly, properties hold for one-sided limits and for functions for which

The limits of fas x approaches a is Proof We prove the first property and leave the proofs of the remaining properties as exercise.

( ) if and only if for each B there exists a Such that ( ) whenever 0 | | , DFor simplicitys sake, we assume L is a negative and let there existsA such that f whenever 0 | | And also, let there exists a such that | | whenever 0| | By letting , we conclude that

And

| |

whenever 0

| |

Now, since | | , and adding this to the first inequalityWe have =B Thus, we can conclude that ( )


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The proof for can prove similarlyEXERCISE2.5

1 Prove the following given limits by definition and theorem

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 2 Prove that , when m is an even number

3 Prove that , when m is an odd number4 Prove that , when m is an odd number5 Prove that 6 Prove that

6 LIMITS AT INFINITY

Let , we see that the value of f approaches 2 as x increases without bound , similarly fapproaches 2 as x x , we denote these limits at infinityBy Definition2.8 Let f is a real valued function of a real variable, where

f: D

, when D

and for each k

will have x

that make x

1 Statement means that for each there exists an Such that| | , whenever x 2 Statement means that for each there exists an

Such that| | , whenever x 3 Statement means that for each there exists an


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Such that f , whenever x 4 Statement

means that for each

there exists an

Such that f , whenever x 5 Statement means that for each there exists an Such that f , whenever x

6 Statement means that for each there exists an Such that f , whenever x

EXAMPLE2.8 Show that 1 2 3

4

5

Solution 1 Since So, for each there exists an N such that Whenever x That is

2 since

So, for each there exists an N such that Whenever x That is

3 Since So, for each B there exists an N= such that

,

whenever x

That is 4 Since

So, for each B there exists an N= such that


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Whenever x That is

5 Since So, for each B there exists N such that , Whenever x That is

Theorem2.17 if and only if and there exists an Such that f whenever x


Theorem2.19 if and only if and there exists an Such that f

whenever x


Proof We will prove the theorem2.17 and exempt the proof of the remaining

Theorems as exercises Since for each there exists an N such thatf

whenever x

, then since

will has an

such that

f whenever x and for each B , then since so will has Such that f , whenever x By letting N=max it follows that whenever x , we have And f


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Therefore

For each B

we let

there exists an

such that

whenever

Similar by letting N=max it follows that whenever x implies thatf and f That is

EXAMPLE2.9 According to example above, show that

1 2 3 Solution 1

We say that when have to possess an N=1+ that make Whenever Therefore

2

We say that when have to possess an N= that make Whenever

Therefore 3

We say that when have to possess N= that make Whenever


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Therefore Definition2.9

If Then the straight lines y=L and y=-L are called horizontal asymptotes of the graphs off and g respectively

Remark: According to the definition above, it follows that the graph of a function of x can have

At most two horizontal asymptotes one to the right and one to the left

EXAMPLE2.10 Verify that f || has two horizontal asymptotesSolution f || f And then

We conclude that the lines y=1 and y=-1 are horizontal asymptotes of the graph of y=||

Theorem2.21 Let then the following properties are true.1 2

3 4 5 Similar, properties hold for limits at -

Proof We will verify only the second property and exempt the proof of the remainingProperties as exercises

1 since So, for each then, since there exist an and such that| | , whenever x and| | , whenever x And then N=max implies that


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| | And | | , whenever x Finally, we apply the triangle inequality to conclude that

( ) | | | | This implies that When evaluating limits at infinity, the following theorem is helpful.

Theorem2.22 if s is a positive rational number and a is any real number then , Furthermore if is defined when Proof 1 When let s=, where p then

Since Now, for each there exists N = such that

Whenever

Therefore The proof of in the case a=0, a

and the second part of the theorem are similar

EXEMPLE2.11 Evaluates the following limits.

1 2 Solution 1

2 To resolve this difficulty, we have to divide both the denominator and the numerator

By x, after this division, the limit may be evaluated as follows

= Hence, the line y=3 is a horizontal asymptote to the right and by taking the limit as x We can see that y=3 is also a horizontal asymptote to the left.


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REMARK: In the indeterminate we are able to resolve the difficulty by rewriting theGiven expression in an equivalent form, in general, we suggest dividing by the highest

Power of x in the denominator, this is illustrated in the following example.

EXAMPLE2.12 Evaluate the following limit.

1 2 In this case, we conclude that the limit does not exist because the numerator increase

Without bound while the modified denominator approaches 2

REMARK: We see obviously that if f is a rational function then a horizontal asymptote to theright

And to the left are same however, function that are not rational function may approach

Different horizontal asymptote to the left and to the right, as shown in the following

example

EXAMPLE2.13 determine the following limit 1 2 Solution 1 for x , we have x=. Thus dividing both the numerator and the denominator

By x produces = = 2 for x we have x=. Thus dividing both the numerator and the denominator

By x produces =

=

We see that the line y=2and y=-2 are horizontal asymptote to the right and to the

Left one after the other of the graph of f EXAMPLE2.14 Determine the following limits by the squeeze theorem

1 2


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Solution 1 since -1 , x And

, so

by the squeeze theorem

2 Since And , where =0Finally, by the squeeze theorem, we have

Theorem2.23 Let f is a real valued function of a real variable, where L, t 1 2 3 4 5 6

Proof We will prove the first and sixth points and leave the proof of the remaining as exercises

1 for each there exists an N such that| | , Whenever x , when we define and t it followsthat

Whenever implies that Therefore for each there exists a such that Whenever 0 When we define N and . It follows that whenever

N

implies that

| |

Therefore 6 for each there exists such that Whenever , when we define and It follows that whenever implies that


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There fore

For each

there exists a

such that

whenever , when we define and It follows that whenever implies that Therefore

REMARK: If define will obtain that1

( )

2 3 ()() 4 ,

And similar properties hold for limits at assign for being exercisesEXERCISE2.6

1 Find the indicated limit.

1 2 3 4 5 6 7 8 9 10 11 12

13

14

15 16 17 ( ) 18 ( )19 ( ) 20 ( )

2 Verify that each of the following functions has to horizontal asymptotes


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1 f || 2 f 3 f || 4 f ||3 Prove the following limits by the concerned definition

1 2 3 4 5 6 7 8 9

4 Prove that if p , andq then

, n=m nm5 Prove that there is not L that make , f is the Dirichlet function6 Prove that there is not L that make 7 Prove that 8 Prove that Hint: by utilizing the squeeze theorem

7 CONTINUITY

To say that a function is continuous at x=a means that there has no interruption in the graph f

At a that is, its graph is unbroken at a and there are no holes, jumps or gaps

Thus, it appears that the continuity of a function at x=a can be destroyed by

Anyone of the following conditions

1 The function is not defined at x=a

2 The limit of f does not exist at x=a or3 The limit of f

exists at x=a but it is not equal to f

This brings us to the following definition

Definition2.10 Let f:D when D and a , we say that f is continuous at point aIf and only if the following three conditions are true together

1 f is defined 2 exists 3 REMARK: if a and a is a limit point of D then


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f is continuous at a if and only if That is for each

there exists a

such that

| | Whenever | | Theorem2.24 Let f:D When and a where a is a limit point of Df is continuous at a if and only if for each there exists a such that| | Whenever | |

Proof Since a is a limit point of D and f is continuous at a, according to definition soWe can investigate that for each there exists a such that

| | Whenever

| | : third point

And first point | | whenever | | is trueTherefore | | Whenever | | As a is a limit point of D and for each there exists a such that| | Whenever | | , we can investigate that1 fis defined as | | whenever | | is true2 exists is true according to designed expression3

is true as

| | implies that

| |

Example2.15 Let = , when x and when x=3Consider whether f is continuous at x=3, 4 Let , when and , when x3Consider whether is continuous at x=3

Solution Since f and , , it follows that

Does not exist, That is f is not continuous at x=3

At x=4 obtain f =1, Therefore f is continuous at x=4 At x=3, obtain f and thenWe find that , also that is


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Thus f is continuous at x=3

8 PROPERTIES OF CONTINUITY

The next, will say to continuity of algebraic functions, composite function

Inverse function and continuous of some functions

Theorem2.25 Let f and g are functions from D to when D and a If f and g are continuous at x=a then

1 fg is continuous at x=a 2 bf is continuous at x=a, when a 3 f.g is continuous at x=a 4

is continuous at x=a, g Proof 1 Since f and g are continuous at x=a. so, we have

1 f and g are defined 2 and existAnd 3 , such that1 exists2 exists3 Therefore is continuous at x=aFor the remaining properties are exempted the proof as exercises

Theorem2.26 If and g is continuous at b then () Proof We have to show that () | |

As g is continuous at b, so for each there exists a Such that| | whenever | | and For there exists a such that| | , whenever | | Finally, letting u=f

, it follows that whenever

| | implies that

() Therefore ()

Theorem2.27 Let f: and g: where , and If f is continuous at a and g is continuous at fthen gof is continuous at a


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That is to say Or

() ( )

Proof g is continuous at f if and only if for each there exists a such that () Whenever | | fis continuous at a if and only if for there exists a such that| | , whenever | | Finally, by letting u=f,it follows that whenever | | implies that| | Implies that () ( ) | | Therefore

() ( )

Or Theorem2.28 The following functions are continuous at Entire points in their domains1 Power functions: f 2 Radical functions: p 3 Polynomial functions: q 4 Rational functions: r , 5 Trigonometric functions: Proof we will prove 3rd property and others properties are left the proof as exercises.3 as q ,

q , And

= , Therefore q is continuous function on its entire domain.

The following, we will say to the continuity on an interval.

Definition2.11 Let f:D , when D and aDf is continuous from the right if and only if the following three conditions are true together.

1 f is defined. 2 exists and 3 Definition2.12 Let f:D when D and aD


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F is continuous from the left if and only if the following three conditions are true together.

1 f

is defined. 2

exists and 3

REMARK: In the case that D is an interval with a be terminal point

1 Ifa is left terminal point of an interval then we agree on the continuity at a

Signify the continuity from the right at a

2 If a is right terminal point of an interval then we agree on the continuity at a

Signify the continuity from the left at a

Definition2.13 Let f:D , when D and open interval I f is continuous on I if and only if f is continuous at all x

And if there is which f is not continuous at then we say that f is not continuous onIDefinition2.14 Let f:D , when D , we will say that

1 f is continuous on if and only if the following two conditions are true together.1.1 f is continuous from the right at a and

1.2 f is continuous on 2 f is continuous on

if and only if the following two conditions are true together.

2.1 f is continuous from the left at b and

2.2 f is continuous on 3 f is continuous on if and only if the following three conditions are true together.3.1 f is continuous from the right at a

3.2 f is continuous from the left at b and

3.3 f is continuous on E XAMPLE2.16 Test that h

is whether continuous on its entire domain.

Solution For convenience we can create functions f and g such that gof h and testBy theorem2.27, let f , g , we can state that From the obtained conditions above and properties of polynomial and radical functions,

We have 1 f and g are continuous on interval and respectively.


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2 , it follows that f 3 g is continuous on f

be an interval

By theorem2.27, we can conclude that h is continuous on its entire domain.

EXAMPLE2.17 Test that f is continuous whether on its entire domain, where

f Solution we have to examine continuity as follows

1The continuity from the right at -2 2The continuity on 3The continuity from the right at 0 4The continuity on and

5The continuity from the left

1 As f f, so f is continuous from the right at -22 As f is a polynomial function, so f is continuous on 3 As f f, so f is continuous from the right at 04 As f is a polynomial function, so f is continuous on 5 As f f, so f is continuous from the left at 2Finally, we can conclude that f is continuous on

Theorem2.29 If f is continuous at a and f

, then there exists a

such that

f | | Proof Leave the proof as exercise.Theorem2.30 If f is continuous at a and f then there exists a such that | | Proof Leave the proof as exercise.

Theorem2.31 Let D is a domain of f and a it follows thatf is continuous at a if and only if

Proof Leave the proof as exercise.

Theorem2.32 If f is continuous on its domain then is continuous on its domain.

Proof Assume that a is any number lying in open interval in the domain ofAnd let , when c lie in an open interval being the range of f


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Namely, f And for each

, we suppose that f

f

and f

f

So, let f and f And if let y lying in the interval ( )Then by the one-to-one nature of f and , it follows thatf ( ) Hence, by letting , it follows that whenever y

-

Whenever

| | Whenever | | Finally, we can conclude that This implies that is continuous at a

Theorem2.33 Let f and g are continuous on an open interval containing a. If f g andThere exists an open interval containing a such that g , in the interval,Then the graph of the function is given by F

Has a vertical asymptote at a

Proof When we have We have to show that . We consider the case for which f And there exists b such that g whenever a Now, for each then, since there existsA

,

| | Whenever a , and|| Whenever a , then it follows that Whenever a and Whenever


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And choosing , then it follows that

Whenever a

Therefore we have And the line x=a is a vertical asymptote of the graph of F

Theorem2.34 The second important geometric property of the graph of a continuous function that we

Will establish is that if a continuous function has a domain consisting of an interval, and

If its graph contains points that are both above and below a line y=c, then, in fact, the

Graph intersects the line y=cSuppose that the function f: is continuous. Let c be number strictly betweenf and f; that is, f f or f f( )Then there is a point in the open interval at which f

Proof to see in text book for advance calculus

Example2.18 Let f , when x Test that there is whether x

that make f

Solution since f is continuous on and f and f By letting c=6 will obtain that 2 according to the intermediate value theorem,There exists that makes f From , so Therefore there is that make f be 3, 2

REMARK: 1 note that the intermediate value theorem tells us that at least one c exists, but

It does not give us a method for finding c. Such theorems are called existence theorem.

2 if f f have different sign then at least one k exists k such thatf k , such f= : , f=-3, f=5, there exist an Such that f and f

Theorem2.35 If f is the function having continuity on an interval then f has the maximum value


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And minimum value on the interval that is to say, there exists an That makes f

f

and there exists an

that makes

f f where f and f are called the maximum value andThe minimum value of f on the interval respectivelyProof EXAMPLE2.19 Let f , x find an that make f fbe themaximum

Value and the minimum value of f respectively

Solution as f is continuous on its entire domain

So, there exist an that make f fbe the maximum value and the minimumvalue of fRespective, where and such that f and f

EXERCISE2.7

1 Show that the following functions are continuous on its domain.

1 f

2 g

3 h

4 i 5 j () 6 k ()() 2 show that the following functions are continuous on its entire domain.1 f ; g 3 h 4 i

; 4

3 Find the values of a and b that make f is continuous at 1 when

F= 4 Find the values of a and b that make f is continuous at -2and 1 when

f 5 Show that the following functions are continuous on the assigned interval.


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1 f ; 2 g ; 2 h

;

4i

;

by the definition of continuity

6 Prove that if f is continuous and has no zero on , then eitherf or f 7 Show that the Dirichlet function is discontinuous at every real numbers.

8 Definition Let f and g are functions having domain D

Function Max and function

The Basic Knowledge for Calculus

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