The Basic Knowledge for Calculus (2)

8/13/2019 The Basic Knowledge for Calculus (2)

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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

Respectively

Theorem 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.



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

− and

2. 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



−

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 that

1. 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.



1. a b and 2. a

7. Prove that if a then

8. Prove that if a then b …

Each man can approach to life’s 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 to

positive

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. −



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.



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



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 Max

2. Min Min

3. 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 −Max − −

3. 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.



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 that

1. 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. Or

If 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



−

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 bound

Proof 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 completeness

Set − 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 on

a 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



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 is

Denoted 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 f

According to definition1.14 Let S= ,

Hence,

And if there is a number c which and

, It follows that



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 x S which x , S.

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



1. 1 S and

2. If n S then

Proof Let S see that are to be true.

Suppose that are to be true. And S

So, −, by well –ordering principle have to exists m −S

Which m −S and since m-1 thus m-1 − S

According to 1 S means that m 1

Hence, −, and from choosing m, we find that −S

And 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 1 S and according to will get that if k S 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 −



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 Q is 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 that

Q 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



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 society

Almost 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 i ts 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 : A B, 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: A B

Definition1.19 a function f is the one-to-one function from A into B if and only if



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 B

Test, 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 and

Show 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



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



2 −

−

And, then −

Therefore

3. − −

− −− −

And, then

− −

Therefore

4. +1

And, then

Therefore

12. THE INVERSE FUNCTIONSDefinition1.23 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



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 , respectively

4. 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 inverse

Is 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 that

1. f is a increasing function on A, if and only if



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 if

If

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. if is increasing on its domain, and then is increasing on its domain.

2. if is 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.



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 length

Such 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 then

Consider a value of x such that is meaningless.

Such as ; is a vertical asymptote

2 Horizontal asymptote; rearrange into and then

Consider 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 into

A pattern , solve equations



Will obtain m and b such that is oblique asymptote

Theorem1.18

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 one

A 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,



1 if then , 2. if then , only one

Theorem1.19 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 of

15. 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



−

−

e is irrational number which is assigned by Scottish mathematician

Leonhard Eule −, where e

Definition1.33 Let

is written by And

In calculus, is defined by which it is invented by the Scottish

Mathematician John Napier −

Definition1.34 , Let is original graph

1.

− 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 units

5. − 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



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



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 m

And 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 m

And 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-clockwise

From

− to L

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

Respectively 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 function

THE DISTANCE AND MIDPOINT FORMULAR

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



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



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

Joining , 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 plane

That is solution points of the equation.

That is such that are to be true.Intercept point on X-axis and Y-axis

Definition1.39 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 equation

1 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) p

2 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



3 A graph of =0 is said to be symmetric with respect to the origin

if, whenever

− − are points lying on its graph.

That is,

− − (x, y) p

4 A graph of p =0 is said to be symmetric with respect to the identity function

if, whenever are points lying on its graph.

That is, (x, y) p

POINTS OF INTERSECTION

Definition1.41 let =0 are all two equations

A point is an intersected point of

=0 if and only if

Satisfies both equations

That 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-intercept

2 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 is

−

Hence 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



− : substitute y by –x, rearrange term

X= : 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 the

Origin, then it is symmetric with respect to the remaining axis also.

3 Let -x, write the definition to show that the graph of =0

Is symmetric with respect to -x, and prove that the graph of =0

Is 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 circle

Remark: 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 that

A correspond to or and we find that each of a will possess



Only that is corresponding, so is a function

Remark: 1 measuring an arc anti-clockwise, if a is positive

2 measuring an arc clockwise, if a is negative

According to rule above, we can define new function as follows

1 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 formula

4 Sin − − 5 Cos − 6 Tan − −

7 Sin − 8 Cos −− 9 Tan −−

10 Sin − 11 Cos − 12 Tan



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

28 Sin 29 Cos

30 Tan

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

41 Sin 42 Cos 43 Tan

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



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 of A, B and

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

6 a area of triangle = length of the base height

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 ……



And Cos …..

Replace Sin

−− − Multiply −− − Add ; 5Cos −

Cos −, Sin −

Therefore Tan such that Tan =

Solving trigonometric equations

The solving trigonometric equations are the finding values of in domain

Satisfying the given trigonometric equations

EXAMPLE1.18 solve for in each of the following equations

1 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



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 functions

Trigonometric functions are periodic functions and have periods as follows



When b we will obtain that

Amplitude

Since, − and −

Thence, − −

Now, we have A= when y= y=

And A= when y=a y=a

For 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



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 2.4

2.5 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



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=1

When “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 f as “x” approaches 1 is 2

And it is written by

Ordinarily, if f becomes arbitrarily close to a single number “L” as “x” approaches “a” from

Either 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 of “a” than it approaches from the

Left 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 true

1 f becomes arbitrarily close to “L”

2 and “x” approaches “a”

The following is assigned by Augustin-Loius Cauchy − His − definition

Of a limit is the standard which be utilized today

Consider the first phrase; f becomes arbitrarily close to “L” means that f lies in the

Interval − in terms of absolute value as follows



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 of “a”

These bring us to the following formal − definition of a limit

Definition2.1 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 if

For 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 f

possesses

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 −



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 point

We begin by writing −=

Since, − thence −=

That is, when , we choose and we choose

It follows that whenever 0

−we have

−

Finally, letting , −0,2

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 limit

Solution let x , we are required to show that , there exists



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 exist

Then 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 exists

X where x

−, now we have

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

EXERCISE2.1



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=b

4 prove that there is not L that make

5 prove that there is not L

−

6 prove that

7 prove that −


9 let f: D D and defined by f =0, 1 when x are rational numbers and irrational

numbers

One after the other, prove that there is not L that make

The function is assigned by the German mathematician Peter Gustav DIRICHLET1859

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



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 tautology

Now, we have such that −

Whenever a- and

such that

− Whenever a , therefore

It is exempted to be exercise

Example2.4 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 establish

A 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



1 show that

=1, when f

and

2 show that − and

3 show that and

4 prove that 1 −

2

3




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 f as “x” approach a does not depend on

The value of “x” at x=a, however if it is happened that the limit is precisely f , then we say

That the limit can be evaluated by direct substitution, that is

Substitute for x

Such 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 D

If the limit of g as “x” approaches “a” exists then the limit of f also exists

And

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 that

For each there exists a such that − whenever 0 −

However, since f =g , , where x a



Therefore

Theorem2.5

Let a and b are real numbers and n , f and g be functions whose limits

Exist as x then will obtain that

1 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



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 −



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 polynomial

Repeated applications of the sum and scalar multiple properties produce

…

Finally, utilizing properties 1, 2 and 3 of theorem2.4 we obtain

… =P

Theorem2.7 if R is a polynomial function which is given by R , and “a” is



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 “a”

Itself, and if

Proof because

Hence for each restrictive there exist and such that

− Whenever −

And − whenever −

Now, letting and then we have

− − −



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 exercise

EXERCISE2.3

1 Prove that if exists and does not exist

Then does not exist

2 Prove that if then = =

3 Prove that ; m, n and is defined

4 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 exist

But exists

6 An expression if then

Show that that the expression is false

7 Prove that



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.

Theorem 2.11 if “a” is a real number l ying 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 figure

Case 1when x

From a figure, co-ordinate of point C is

While and are vertical each other

Now, by properties of a triangle and chord, we have −

The length of arc BC =x= , so − or 1- ,

O A B

C (cos x, sin x)



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 exercisesEXERCISE2.4

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 f increases or decreases without bound as x approaches “a”



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 D

1 Statement means that for each restrictive B there exists

a such that whenever −,

2 Statement − means that for each restrictive B there exists

a 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 not

Mean that the limit exists? On the contrary, it tell us how the limit fails

To exist by denoting the bounded behavior of f as “x” approach “a”

EXAMPLE2.6 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



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

−

Considering − −

1 when x , then 0 − and

2 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 0

Such 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



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 f as “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 −, D

For simplicity’s sake, we assume L is a negative and let −there exists

A 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 inequality

We have −=B−

Thus, we can conclude that 1



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



− 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

Theorem2.18 −if and only if and there exists an


Theorem2.19 if and only if and there exists an


−

Theorem2.20 − 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 that

f 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



Therefore

For each B we let there exists an such that

whenever

Similar by letting N=max it follows that whenever x implies that

f 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 − − −12

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 −

Therefore



Definition2.9

If

− −

Then the straight lines y=L and y=-L are called horizontal asymptotes of the graphs of

f 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 asymptotes

Solution 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 remaining

Properties 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

− 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.

Theorem 2.22 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.

REMARK: In the indeterminate we are able to resolve the difficulty by rewriting the

Given 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 the

right

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

Solution 1 since -1 − −, x

And , so by the squeeze theorem



2 Since

And

− −, where =0

Finally, 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 exercises1 for each there exists an N such that

− , Whenever x , when we define and t it follows

that

Whenever implies that −

The refore

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 −

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 that

1

2

3

4 ,

And similar properties hold for limits at − assign for being exercises

EXERCISE2.6

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

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 … , and

q … then

, n=m n m

5 Prove that there is not L that make

, f is the” Dirichlet function”

6 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, it’s graph is unbroken at a and there are no holes , jumps or gapsThus, 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 or

3 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

f is continuous at “a” if and only if



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 f g 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 exist

And 3 , such that

1 exists

2 exists

3

Therefore is continuous at x=a

For 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 −gb<

Therefore

Theorem2.27 Let f: and g: where , and

If f is continuous at “ a” and g is continuous at “ f then gof is continuous at “ a”

That is to say



Or

Proof g is continuous at f if and only if for each there exists a such that

− Whenever

−

f is 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 domains

1 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 a D

f 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 a D

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 If “a” 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 f rom 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 on

I

Definition2.14 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


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


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 test

By 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.

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 -2

2 As f is a polynomial function, so f is continuous on

−

3 As f f , so f is continuous from the right at 0

4 As f is a polynomial function, so f is continuous on

5 As f f , so f is continuous from the left at 2

Finally, 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

−


Theorem2.31 Let D is a domain of f and a it follows that

f is continuous at a if and only if


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 of

And let , when c lie in an open interval being the range of f

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 that

f − −

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 and

There 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 exists

A ,

− Whenever a , and

Whenever a , then it follows that

Whenever a and

Whenever

And choosing , then it follows that



That makes f f and there exists an that makes

f f where f and f are called the maximum value and

The minimum value of f on the interval respectively

Proof

EXAMPLE2.19 Let f , x find an that make f f be the

maximum

Value and the minimum value of f respectively

Solution as f is continuous on its entire domain

So, there exist an that make f f

be the maximum value and the minimum

value of f

Respective, 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.

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 either

f 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 Min are defined by

Max Max and Min Min

Use the definition to prove that if f and g are continuous at a then

Max and Min are also continuous at a

9 Let f is a function having domain D. We define the definition of a function as follows.

, when x D use the definition to prove that if f is continuous

At a then is also continuous at a

10 Prove that theorem2.33 in the case for which f is true.

11 Show that the equation − , there are solutions

APPENDIX

A: Partial fractions, In this part examines a procedure for decomposition a rational functions

Into simpler rational function, we call this procedure the method of partial fractions. This

technique

Was introduced in 1702 by John Bernoulli −a Swiss mathematician who was

instrumental

In the early development of calculus, John Bernoulli was Professor at the University of Basel

And taught many outstanding students, the most famous of whom was Leonhard Euler.

To introduce the method of partial fractions, let’s consider the factorization of the fractions in

these terms, − and −, , where are

called

Partial fractions, furthermore we have − −,

And moreover, − −



The one benefit of the partial fractions is used to evaluate the integral.

In form of rational functions recall from algebra that every polynomial with real coefficient

Can be factored into linear and irreducible quadratic factors, for instance, the polynomial,

−− − Where is a linear

Factor, − is a repeated linear facfor and is an irreducible quadratic factor.

Using the factorization, we can write the partial fraction decomposition of the rational

Expression, , where N is a polynomial of degree

Less than 5. We summarize the steps for partial fraction decomposition as follows.

Decomposition of into partial fractions

1 Divide if improper: if is an improper fraction then divide the denominator into the

numerator

To obtain , where the degree of is less than the degree of

Then apply steps 2, 3 and 4 to the proper rational expression

2 Factor denominator: completely factor denominator into factors of the form of

And , where is an irreducible.

3 Linear factors: For each factor of the form , the partial fraction decomposition

Has to include the following sum of m fractions

4 Quadratic factors: For each factor of the form , the partial fraction

Decomposition has to include the following sum of n factors

Linear factors: Algebraic technique for determining the constants in the numerators are

Demonstrate in the following examples

Example1A write the partial fraction decomposition for



We exhausted the most convenience choice for x so to find the value of A and D. we use

Any other value for x along with the calculated value of B and D. thus, using x=-2, 1, B=1

Ana D=1, we have this linear equation system as follows.

-A-C=-1 and 2A+C=-1, we solve this system, it follow that A=-2 and C=3

Finally, we have the decomposition is =

REMARK: Note that it is necessary to make as many representations for x as there are

unknowns

to be determined. For instance, In example 2A we made four representations

− − to solve for B, D and A and C respectively.

Quadratic factors: when using the method of partial fractions with linear factors, a convenient

Choice of x immediately yields a value for one of the coefficients. With quadratic factors

A system of linear equations will typically have to be solve, regardless of the choice of x

Example3A

Write the partial fraction decomposition for

Solution since

We include one partial fraction for each factor and write

Multiplying by the least common denominator, , yield the basic equation.

− +

To solve for A and B by letting x=0 and -1and obtain, A=1 and B=-1

At this points C and D are yet to be determined. We can find these remaining constants by

Choosing other values for x and solving the resulting system of linear equation

If x=-2, then, since A=1, B=-1, will obtain that D-2C=-3

If x=2, then will obtain that D+2C=1

Solving this system of two equations with two unknowns, we have

D-2C=-3… and D+2C=1… , which yields D=-1, consequently, c=1 and it follows that



− − − Finding the solution of the basic equation of the partial fraction decomposition by substituting

Values of x that made the linear factors zero. This method works well. When the partial

Fraction decomposition in values only linear factors. However, if decomposition concerns

A quadratic factor, then an alternate procedure is often more convenient. Both

Methods are outlined in the following summary.

Guidelines for solving the basic equation

Linear factors: 1 represents the roots of the distinct linear factors into the basic equation

2 For repeated linear factor use the coefficients determined in part 1 to rewrite the basicEquation, then represent other convenient values of x and solve for the remaining coefficients

Quadratic factors: 1 Expand the basic equation

2 Collect terms according to powers of x

3 Equate the coefficients of like powers to obtain a system of linear equations concerning

A, B, C,... And so on

4 Solve the system of linear equations.

The following is an example that be used the second procedure for the basic equation.Example 4A

Write the partial fraction decomposition of

Solution we include one partial fraction for each power of and write.

= +

Multiplying by the least common denominator, , yields the basic equation

=

Expanding the basic equation and collecting like terms, we have

=A

Now, we can equate the coefficient of like terms on opposite side of the equation.

A=2, B=1, B+C=2 AND B+D=2, by using the unknown values A=2 and B=1, we have.



1+C=2, then C=1 and 1+D=2, then D=1

Therefore, = +

EXERCISE A

WRITE the partial fraction decomposition of the following expression.

1 2 3 4

5 6 7

8 9 10

11 12 13

B: Indeterminate form and L ’H pital ’s Rule

The forms are described that are indeterminate, because they do not guarantee

That a limit exists, nor do they indicate what is, if one does exist. When we encountered

One of these indeterminate form, we attempted to rewrite the expression by using various

Algebraic techniques, as illustrated by the following example

1 = = =27

2 = = = =2

3 = = −=0

However, not all indeterminate forms can be evaluated by algebraic manipulation, these are

Particularly true when both algebraic and transcendental or both transcendental and

Transcendental functions are involved, for example the limit

1 , produces the indeterminate form , distributing this expression to obtain

−, merely produces another indeterminate form − of course

We are not able to estimate this limit. Similar, we have

2 =



L’H pital’ s Rule

We can use a theorem called L’H pital’s Rule for finding two limits before. This theorem

States that under certain conditions the limit of the quotient .This theorem is named

After the French mathematician Guillaume Francois Ant oine De L’H pital − , who

Published the first calculus text in 1696.To prove this theorem, we use a more general result

Called the extended Mean-Value Theorem, which states the following. If f and g are

differentiable

On an open interval and continuous on such that ,

Then there exists a point c in such that

We assign to study the proof in a general calculus text

Theorem

Let f and g be functions that are differentiable on an interval containing c, except

Possibly at c itself. If the limit of as x approaches c produces the indeterminate form

, then

Provided the limit on the right or is infiniteWe assign to see the proof in a general calculus text.

Remark: 1 the indeterminate form comes in four forms: , each of these

Forms can be applied by L’H pital’s Rule

2 Students occasionally use L’H pital’s Rule incorrectly by applying the quotient rule

To . Be sure you see that the rule concerns not the derivative of

EXAMPLE 1B

Evaluate

Solution since direct substitution result in the indeterminate form we apply

L’H pital’s Rule to obtain = = =ln 2



REMARK: in writing the string of equations in example above, we actually do not

Know that the first limit is equal to the second until we have shown that the second

Limit exists, in other words, if the second limit had not existed, It would not have been

Permissible to apply L’ H pital rule.

Another form of L’H pital’s Rule states that if the limit of as x approaches −

Produces the indeterminate form, , then =

Provided the limit on the right exists. We illustrate this form of L’H pital’s Rule in these

examples

EXAMPLE 2B

Evaluate

Solution since direct substitution results in the indeterminate form , we apply L’H pital’s

Rule

To obtain = = =o

Occasionally, it is necessary to apply L’H pital’s Rule more than once to remove an

indeterminate

Form as shown in an example below.

EXAMPLE 3B

Evaluate

Solution since direct substitution results in the indeterminate form , we apply L’H pital’s Rule

To obtain = . This limit yields the indeterminate form , so we apply

L’H pital’s Rule again to obtain

= , similar we apply L’H pital’s Rule again to obtain

= =0



In addition to the form , there are other indeterminate forms such as

0.

For instance, consider the following limits that lead to the indeterminate form 0.

=0, =0, =3, =3, =2

Since each limit is different, it is clear that the form 0. is indeterminate in the sense

That it does not determine the value of the limit. The following

Example indicate method for evaluating these forms. Basically, we attempt to convert

Each of these forms to those for which L’H pital’s Rule is applicable

EXAMPLE 4B

Evaluate

Solution since direct substitution produces the indeterminate form 0. , we rewrite the

Limit to the form as follows.

1 Form , = and

2 Form ,

=

By applying L’H pital’s Rule, we have

1 = = =0 and

2 =

In the second case yields the indeterminate form . Moreover, since the quotient seems to be

Getting more complicated, we abandon this approach and try the form, as shown in the first.

The indeterminate form arise from the limits of functions that have a variable

Base and variable exponent. When we encountered this type of function we used logarithmic

Differentiable to find the derivative. We use a similar procedure when taking limits, as indicated

In the next example.

EXAMPLE 5B



Evaluate −

Solution since direct substitution yields the indeterminate form , we proceed as follows.

We assume that the limit exists, and let y= −

Now, taking the natural logarithm of both sides, we have

= −, using the fact that logarithmic function is continuous, we write.

= −= − : Indeterminate form

= : Indeterminate form

= : Indeterminate form

= −=-1

Finally, since −, if and only if , so we conclude that

− EXAMPLE 6B

Evaluate

Solution since direct substitution yields the indeterminate form , we proceed as follows.

We assume that the limit exists, and let

So, :

Finally, since , we conclude that

L’H pital’s Rule can also be applied to one - side limits, as demonstrated in next example.

EXAMPLE 7B

Evaluate

Solution since direct substitution produces the indeterminate form , we proceed as follows.

: Indeterminate form



: Take log of both sides

: Continuity

: Indeterminate form 0.

−


: L’H pital’s Rule


0 : L’H pital’s Rule

Now, since , and it follows that

EXAMPLE 8B

Evaluate

Solution Since direct substitution produces the indeterminate form , we proceed as follows.

So, =0

Finally, since , it follows that

EXAMPLE 9B

Evaluate

Solution since substitution yields the indeterminate form −we try to rewrite the

Expression to produce a form to which we can apply L’H pital’s Rule. In this case, we

Combine the two fractions to obtain


We are able to apply L’H pital’s Rule to obtain =

This limit also yields the indeterminate form , so we apply L’H pital’s Rule again to obtain



Although the forms

are signified as indeterminate. But there are

Similar forms that you should recognize as indeterminate, such as

− − − , you are asked to verify two of these in next exercise.

In each of examples so far in this section, we have used L’H pital’s Rule to find a limit that

exists.

It is able to be also used to conclude that a limit is infinite, and this is demonstrate

In the two last example.

EXAMPLE 10B

Evaluate

Solution Since direct substitution produces the indeterminate form

Therefore, we apply L’H pital’s Rule to obtain.

Now, since we conclude that the limit of as x is also infinite.

As a final comment, we remind you that L’H pital’s Rule can be applied only to quotients Leading to the indeterminate . Please remember

EXERCISE B

1 Evaluate the following limits, using L’H pital’s Rule 1 2 3

4 5 6

7 8

9

10 11 12

13 − 14 15

16 17 18 , m, n



19 20 21

22 23 24 , where m, n

25 26 27

2 Prove that if , and , then =0

3 Prove that if , and −, then =

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