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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.
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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
− 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
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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 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.
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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 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. −
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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 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.
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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 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
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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 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
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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 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
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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 x S which x , S.
Lemma1.13 Let S it will get that S if and only if
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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 −
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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 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
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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 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
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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 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
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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 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
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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 , 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
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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 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.
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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 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
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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,
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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
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−
−
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
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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 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
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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 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
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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
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− : 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
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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
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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
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
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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 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 ……
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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
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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 functions
Trigonometric 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=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
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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 2.4
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
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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=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
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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 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 −
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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 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
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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 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
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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=b
4 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 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
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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 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
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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 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
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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
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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 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
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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 “a”
Itself, 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 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
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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.
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)
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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 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”
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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 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
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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
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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 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
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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
Theorem2.18 −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
−
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
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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 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
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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
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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
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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
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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 −
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−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
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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
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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
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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.
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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
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 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
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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
−
Proof Leave the proof as exercise.
Theorem2.31 Let D is a domain of f and a it follows that
f 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 of
And let , when c lie in an open interval being the range of f
Namely, f
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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
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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 −; −
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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, − −
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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
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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
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− − − 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.
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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 =
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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
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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
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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
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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
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: Take log of both sides
: Continuity
: Indeterminate form 0.
−
: Indeterminate form
: L’H pital’s Rule
: Indeterminate form
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
: Indeterminate form
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
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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
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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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