Methods of Solving Calculus Problems
-
Upload
science2010 -
Category
Documents
-
view
230 -
download
2
Transcript of Methods of Solving Calculus Problems
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 1/243
Methods of Solving Calculus Problems
1
The Educational Publisher
Columbus, 2015
METHODS OF SOLVING
CALCULUS PROBLEMS
■ Constantin Dumitrescu ■ Florentin Smarandache ■
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 2/243
C. Dumitrescu ■ F. Smarandache
2
■
C. Dumitrescu ■
F. Smarandache ■
Methods of Solving CalculusProblems ■ Theory and Exercises ■
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 3/243
Methods of Solving Calculus Problems
3
First imprint: Metode de calcul în analiza matematică
(in Romanian), INFOMED Publishing House,
Craiova, Romania, 1995
Translation to English by ANA MARIA BUZOIANU
(AdSumus Cultural and Scientific Society)
The Educational PublisherZip Publishing
1313 Chesapeake Ave.
Columbus, Ohio 43212, USA
Email: [email protected]
Editura Ferestre
13 Cantemir str.
410473 Oradea, Romania
ISBN 978-1-59973-355-5
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 4/243
C. Dumitrescu ■ F. Smarandache
4
■ Constantin Dumitrescu ■ Florentin Smarandache ■
Methods of Solving Calculus
ProblemsTheory and Exercises
The Educational Publisher
Columbus, 2015
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 5/243
Methods of Solving Calculus Problems
5
Motto:Love is the purpose, the sacred command of existence.
Driven by it, eagles search each other in spaces,
Female dolphins brace the sea looking for their grooms
Even the stars in the skies cluster in constellations.
(Vasile Voiculescu)
© Constantin Dumitrescu, Florentin Smarandache,
and the Publisher, 2015.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 6/243
C. Dumitrescu ■ F. Smarandache
6
Contents
Foreword ..................................................................................................... 9
I. Sets Theory ...................................................................................... 11
Operations with sets ........................................................................... 11
1. The Intersection ......................................................................... 12
2. The Union ................................................................................... 12
3. The Difference............................................................................ 13
4. The Complement ....................................................................... 13
5. The Symmetrical Difference .................................................... 146. The Cartesian Product ............................................................... 14
Methods for Proving the Equality of Two Sets ............................. 15
Properties of the Characteristic Function ....................................... 16
Exercises ............................................................................................... 17
II. Functions ......................................................................................... 31
Function definition ............................................................................. 31
Examples .......................................................................................... 34
The inverse of a function ................................................................... 35
Examples .......................................................................................... 37
Observation ..................................................................................... 38
The Graphic of the Inverse Function ............................................. 38
Methods to show that a function is bijective ................................. 39
1. Using the definition ................................................................... 39
2. The graphical method ................................................................ 39
3. Using the theorem ...................................................................... 404. Using the proposition 1 ............................................................ 41
5. Using the proposition 2 ............................................................ 41
Exercises ............................................................................................... 42
Monotony and boundaries for sequences and functions ............. 48
Example ........................................................................................... 50
Methods for the study of monotony and bounding ..................... 53
Exercises .......................................................................................... 54
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 7/243
Methods of Solving Calculus Problems
7
III. The limits of sequences and functions .................................. 61
The limits of functions ....................................................................... 61
Exercises .......................................................................................... 66 The limits of sequences ...................................................................... 71
Examples .......................................................................................... 72
Calculation methods for the limits of functions and sequences . 73
Joint methods for sequences and functions ............................... 73
Specific methods for sequences ................................................. 108
IV. Continuity and derivability ..................................................... 146
Continuity ........................................................................................... 146 (A) Methods for the study of continuity .................................. 146
(B) Types of discontinuity points .............................................. 149
(C) The extension through continuity ....................................... 149
(D) The continuity of composite functions ............................. 150
Derivability ......................................................................................... 154
(A) Definition. Geometric interpretation. Consequences ..... 154
(B) The derivation of composite functions .............................. 155
(C) Derivatives of order ........................................................... 158 (D) The study of derivability ...................................................... 158
(E) Applications of the derivative in economics ..................... 162
V. Fermat, Rolle, Lagrange, Cauchy Theorems ............................ 169
(A) Fermat’s theorem ....................................................................... 169
1. Enunciation ............................................................................... 169
2. Geometric and algebraic interpretation ................................ 169
(B) Rolle’s theorem ........................................................................... 172 1. Enunciation ............................................................................... 172
2. Geometric and algebraic interpretation ................................ 172
3. Consequences ........................................................................... 173
Methods for the study of an equation’s roots.............................. 177
1. Using Rolles’s theorem (its algebraic interpretation) ......... 177
2. Using Darboux’s properties ................................................... 177
3. Using Rolle’s sequence ............................................................ 177
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 8/243
C. Dumitrescu ■ F. Smarandache
8
4. The graphical method. ............................................................. 178
5. Viette’s relations ....................................................................... 178
6. Using the theorem of the average ......................................... 179 (C) Lagrange’s Theorem .................................................................. 181
1. Enunciation ............................................................................... 181
2. Geometric and algebraic interpretation ................................ 181
3. The corollary of Lagrange’s Theorem .................................. 181
(D) Cauchy’s Theorem ..................................................................... 186
1.Enunciation ................................................................................ 186
2. Geometric and algebraic interpretation ................................ 186
VI. Equalities and Inequalities ..................................................... 189
Equalities ............................................................................................ 189
Inequalities ......................................................................................... 190
Method 1. Using Lagrange’s or Cauchy’s theorem. ................ 190
Method 2: The method of the minimum ................................. 192
Method 3: (Inequalities for integrals, without the calculation of
the integrals) .................................................................................. 193
Method 4: Using the inequality from the definition of convex(concave) functions ...................................................................... 195
VII. Primitives .................................................................................. 204
Connections with other notions specific to functions ................ 204
(A) Methods to show that a function has primitives .................. 209
(B) Methods to show that a function doesn’t have primitives .. 210
Logical Scheme for Solving Problems ...................................... 221
VIII. Integration ................................................................................ 222 Connections between integration and other notions specific to
functions ............................................................................................. 227
Methods to show that a function is integrable ............................. 229
Methods to show that a function is not integrable .................... 229
Examples ........................................................................................ 229
Exercises ........................................................................................ 235
References ............................................................................................... 240
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 9/243
Methods of Solving Calculus Problems
9
Foreword
We rejoice expressing our gratitude towards all those who,
better or less known, along the years, have helped us to get to
this particular book. And they are many, many indeed whom
have helped us! Some provided suggestions, others, ideas; at
times, we have struggled together to decipher a certain elusivedetail; other times we learned from the perceptive or sometime
naïve questions of our interlocutors.
The coming into being of this book represents a very
good example of altruism, kindness and selflessly giving, of the
journey, made together, on the road to discovering life’s beauty.
He who thinks not only of his mother or father, of
himself or his family, will discover an even bigger family, will
discover life everywhere, which he will start to love more and
more in all its manifestations.
Mathematics can help us in our journey, stimulating and
enriching not just our mental capacity, our reasoning, logic and
the decision algorithms, but also contributing in many ways to
our spiritual betterment. It helps shed light in places we
considered, at first, fit only for an empirical evolution – our own
selves . We offer this succession of methods encountered in the
study of high school calculus to students and teachers, to higher
education entry examination candidates, to all those interested, in
order to allow them to reduce as many diverse problems as
possible to already known work schemes.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 10/243
C. Dumitrescu ■ F. Smarandache
10
We tried to present in a methodical manner, advantageous
for the reader, the most frequent calculation methods
encountered in the study of the calculus at this level.In this book, one can find:
methods for proving the equality of sets,
methods for proving the bijectivity of functions,
methods for the study of the monotonic sequences
and functions,
mutual methods for the calculation of the limits of
sequences and functions, specific methods for the calculation of the limits of
sequences,
methods for the study of continuity and
differentiation,
methods to determine the existence of an equation’s
root,
applications of Fermat’s, Rolle’s, Lagrange’s and
Cauchy’s theorems,
methods for proving equalities and inequalities,
methods to show that a function has primitives,
methods to show that a function does not have
primitives,
methods to show that a function is integrable,
methods to show that a function is not integrable.
We welcome your suggestions and observations for the
improvement of this presentation.
C. Dumitrescu, F. Smarandache
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 11/243
Methods of Solving Calculus Problems
11
I.
Sets Theory
A set is determined with the help of one or more properties
that we demand of its elements to fulfill.
Using this definition might trick us into considering that any
totality of objects constitutes a set. Nevertheless it is not so.
If we imagine, against all reason, that any totality of objects is
a set, then the totality of sets would form, in its own turn, a set that
we can, for example, note with . Then the family of itsparts would form a set. We would thus have ∈ .
Noting with card the number of elements belonging to ,
we will have: . However, a theorem owed to Cantor shows that we always
have
< .
Therefore, surprisingly maybe, not any totality of objectscan be considered a set.
Operations with setsDEFINITION : The set of mathematical objects that we
work with at some point is called a total set, notated with .
For example,
drawing sets on a sheet of paper in the notebook, the
total set is the sheet of paper;
drawing sets on the blackboard, the total set is the set
of all the points on the blackboard.
It follows that the total set is not unique, it depends on the
type of mathematical objects that we work with at some given point
in time.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 12/243
C. Dumitrescu ■ F. Smarandache
12
In the following diagrams we will represent the total set
using a rectangle, and the subsets of
by the inner surfaces of this
rectangle. This sort of diagram is called an Euler-Venn diagram. We take into consideration the following operations with
sets:
1. The Intersection ∩ ∈ │ ∈ ∈ }
Because at a given point we work only with elements
belonging to
, the condition
∈ is already implied, so we can
write: ∩ │ ∈ ∈ }
∩ More generally,
Let’s observe that: ∉ ∩ <> ∉ ∉
2. The Union
∪ │ ∈ ∈ }
∪
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 13/243
Methods of Solving Calculus Problems
13
As in the case of the intersection, we can consider:
We see that: ∉ ∪ <> ∉ ∉
3. The Difference │ ∈ ∉ }
We retain that: ∉ <> ∉ ∈
4. The Complement The complement of a set is the difference between the
total set and . | ∈ ∉ } | ∉ }
The complement of a set is noted with or with .̅
Let’s observe that ∉ <> ∈
More generally, we can talk about the complement of a set to
another, random set. Thus,
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 14/243
C. Dumitrescu ■ F. Smarandache
14
| ∈ ∉ }
is the complement of set
to set
.
5. The Symmetrical Difference ∆ ∪
∆
We have ∉ ∆ <> ∉ and ∉ . 6. The Cartesian Product
× , │ ∈ ∈ } The Cartesian product of two sets is a set of an ordered
pairs of elements, the first element belonging to the first set and the
second element belonging to the second set.
For example, ℝ × ℝ , | ∈ ℝ , ∈ ℝ} , and an
intuitive representation of this set is provided in Figure 1.1.
By way of analogy, the Cartesian product of three sets is a
set of triplets:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 15/243
Methods of Solving Calculus Problems
15
× × ,,| ∈ , ∈ , ∈ }. An intuitive image of
ℝ
ℝ × ℝ × ℝ is given in the figure
below.
More generally, 1
, , … , | ∈ ∈ 1, }.
Methods for Proving the Equalityof Two Sets
The difficulties and surprises met on the process of trying to
rigorously define the notion of sets are not the only ones we
encounter in the theory of sets. Another surprise is that the well-known (and the obvious, by
intuition) method to prove the equality of two sets using the
double inclusion is – at the level of a rigorous definition of sets –
just an axiom. <> ⊆ ⊆ (1.1)
By accepting this axiom, we will further exemplify two
methods to prove the equality of sets:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 16/243
C. Dumitrescu ■ F. Smarandache
16
(A) THE DOUBLE INCLUSION [expressed through the
equivalence (1.1)];
(B) UTILIZING THE CHARACTERISTIC FUNCTIONOF A SET.
We will first provide a few details of this second method that
is much faster in practice, hence more convenient to use than the
first method.
DEFINITION . We call a characteristic function of the set
the function : ⟶ 0,1} defined through:
{1 ∈ 0 ∉
As we can observe, the numbers 0 and 1 are used to divide
the elements of the total set in two categories:
(1) one category contains those elements in which the
value of is 1 (the elements that belong to );
(2) all the elements in which the value of is 0 (elements
that do not belong to
belong to the second category.
Method (B) of proving the equality of two sets is based onthe fact that any set is uniquely determined by its characteristic
function, in the sense that: there exists a bijection from set
of the subsets of to the set of the characteristic
functions defined on (see Exercise VII ).
Hence, <> .(1.2)
Properties of the CharacteristicFunction ∩ . ∪ ∩ − ∩
1
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 17/243
Methods of Solving Calculus Problems
17
∆ 2 . ∩
×, ×
∅ ≡ 0, ≡ 1 ⊆ <> for any ∈ . Let’s prove, for example . For this, let’s observe that the
total set is divided by sets and in four regions, at most:
1) for the points that do belong to neither , nor , we
have 0 and ∩ 0.
2) for the points that are in but are not in , we have 1, 0 ∩ 0 .
3) if ∉ ∈ ,
the equalities follow analogously.
4) if ∈ ∉
ExercisesI. Using methods (A) and (B) show that:
SOLUTIONS :
(A) (the double inclusion)
Let’s notice that solving a mathematical problem requires we
successively answer two groups of questions:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 18/243
C. Dumitrescu ■ F. Smarandache
18
(q) WHAT DO WE HAVE TO PROVE?
(Q) HOW DO WE PROVE?
Answering question (Q) leads us to a new (q) question. Thus,for exercise 1, the answer to the question (q) is:
(r1 ): We have to prove an equality of sets,
and the answer to the question (Q) is:
(R 1 ): We can prove this equality using two methods: using
the double inclusions or using the properties of the characteristic
function.
We choose the first method and end up again at (q). This
time the answer is:(r2 ): We have to prove two inclusions,
and the answer to the corresponding question (Q) is:
(R 2 ): We prove each inclusion, one at a time.
Then,
(r3 ): We have to prove one inclusion;
(R 3 ): We prove that an arbitrary element belonging to the
“included set” belongs to the “set that includes”.
It is obvious that this entire reasoning is mental, the solution
starts with:
a) Let ∈ , irrespective [ to continue we will read the
final operation (the difference) ], <> ∈ and ∉ ∩ [ we will
read the operation that became the last (the union)] <==>
∈ and
∉
∉
<>
∈ ∉
∈ ∉ (1.3)
From this point on we have no more operations to explain
and we can regroup the enunciation we have reached in several
ways (many implications flow from (1.3), but we are only interested
in one - the conclusion of the exercise), that is why we have to
update the conclusion: ∈ ∪ <> ∈ ∈ .
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 19/243
Methods of Solving Calculus Problems
19
We observe from (1.3) that this conclusion follows
immediately, so the inclusion is proven.
For the second inclusion: Let ∈ ∪ , [ wewill read the final operation (the reunion) ] <===>
∈ ∈ <> [ we will read the operation that became the last ]
<===>
∈ ∉ ∈ ∉
<===> ∈ and ∉ ∉
(1.4)
We have no more operations to explain, so we will describe
in detail the conclusion we want to reach: ∈ ∩ , i.e. ∈ and ∉ ∩ , an affirmation that immediately issues from (1.4).
5. (r1 ): We have to prove an equality of sets;
(R 1 ): We prove two inclusions;
(r2 ): We have to prove an inclusion;
(R 2 ): We prove that each element belonging to the included
set is in the set that includes.Let ∈ , irrespective [we will read the final
operation (the difference)], <> ∈ and ∉ [ we will
read the operation that became the last ] <>
∈ and ∉ ∈ . (1.5)
We have no more operations to explain, so we update the
conclusion: ∈ ∩ . . ∈ ∈ .
We observe that (1.5) is equivalent to:
∈ ∉ ∈ ∈ . The first affirmation is false, and the second one proves
that
∈ ∩ .
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 20/243
C. Dumitrescu ■ F. Smarandache
20
We will solve the same exercises using the properties of the
characteristic function and the equivalence (1.2).
1. We have to prove that −∩ −∪−;(r1 ): We have to prove an equality of functions;
(R 1 ): As the domain and codomain of the two functions
coincide, we have to further prove that their values coincide, i.e.
(r2 ): −∩ −∪− for any ∈ . The answer to the corresponding (Q) question is
(R 2 ): We explain both members of the previous equality.
−∩ [we will read the final operation (thedifference) and we will apply property ] = ∩∩ [we will read the operation that became the last and
we will apply property ] = . . (1.6)
The second member of the equality becomes successive:
5. In order to prove the equality −− ∩
we notice that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 21/243
Methods of Solving Calculus Problems
21
II. Prove that:
SOLUTIONS .
Method (A) (the double inclusion)
1. Let ∈ ∩ × ∩ , irrespective [ we explain the
final operation (the scalar product )] <> , , with ∈ ∩ and ∈ ∩ <>[ we explain the operations that became the last
ones ]
<> , with
∈ ,
∈ and
∈ ,
∈
[ we have no more operations to explain so we update the conclusion : ∈ × ∩ × , so belongs to an intersection (the last operation) ] <> , with ∈ , ∈ and ∈ , ∈ <===> ∈ × and ∈ × .
3. Let ∈ × , irrespective <===> ∈ ,
with ∈ and ∈ <===> ∈ , with ∈ and
∉ and
∈ [ by detailing the conclusion we deduce the following steps ]
<===> , , ∈ , ∈ and , , ∉ , ∈
===> ∈ × and ∉ × .
Reciprocally, let ∈ × × , irrespective
<===> , with , ∈ × and × <> , , ∈ and ∈ and
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 22/243
C. Dumitrescu ■ F. Smarandache
22
∉ ∉
<===>
∈ ∈ ∉ ∈ ∈ ∉
===> , ∈ × .
Method (B) (using the characteristic function)
Explaining ∆∆, in an analogue manner, we obtain
the desired equality. If we observe that the explanation
of ∆∆ from above is symmetrical, by utilizing the
commutativity of addition and multiplication, the required equality
follows easily.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 23/243
Methods of Solving Calculus Problems
23
Equalities 4)-5) were easy to prove using the characteristic
function; they are, however, more difficult to prove using the first
method. The characteristic function is usually preferred, due to the
ease of use and the rapidity of reaching the result.
III. Prove the following equivalences:
SOLUTIONS:
1. r1: We have to prove an equivalence;
R 1: We prove two implications;
r2: ∪ ⊂ > ⊂ ⊂
R 2: We prove that ⊂ ⊂ (two inclusions).
Let
∈ irrespective [we have no more operations to
explain, therefore we will update the conclusion: ∈ ; but this is
not evident from ∈ , but only with the help of the hypothesis] ∈ > ∈ ∪ > ℎℎ ℎ ℎℎℎ ∪ ⊂ > ∈ Let ∈ irrespective => ∈ ∪ ⊂ > ∈ . Reciprocally (the second implication):
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 24/243
C. Dumitrescu ■ F. Smarandache
24
r1: We have to prove that ∪ ⊂ (an inclusion)
R 1: Let
∈ ∪ irrespective (we explain the last
operation):
> ∈ > ∈ ℎ ℎℎ ∈ > ∈ ℎ ℎℎ
5. r1: We have to prove an equivalence;
R 1: We prove two implications;
r2: ∪ > ⊂ ;
R 2: We prove the conclusion (an inclusion), using the
hypothesis.
Let ∈ irrespective => ∈ ∪ > ∈
r3: We have to prove the implication: ⊂ > ∪ ;
R 3: We prove an equality of sets (two inclusions).
For the first inclusion, let ∈ ∪ irrespective [we
explain the last operation] =>
=> ∈ ∈ => ∈ ∉ ∈
[We have no more operation to explain, so we update the
conclusion]. We thus observe that from (a) it follows that ∈ ,
and from (b), with the help of the hypothesis, ⊂ , we also
obtain ∈ .
Method 2:
We have to prove two inequalities among functions.
Through the hypothesis, ∪ < and, moreover < ∪
(indeed, < ∪ <> < <> 1 > 0 – true, because the characteristic functions take two
values: 0 and 1).
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 25/243
Methods of Solving Calculus Problems
25
It follows that < ∪ < and, analogously <
∪ < .
Reciprocally, for the inverse implication, we have to provethat: ∪ < , i.e. ∩ < (x) (1.7)
in the hypothesis: < and < (1.8)
But , , can take only two values: 0 and 1. From
the eight possible cases in which (1.7) must be checked, due to (1.8),
there remain only four possibilities:
In each of these cases (1.7) checks out.
5. We have to prove that
But
For the direct implication, through the hypothesis, we
have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 26/243
C. Dumitrescu ■ F. Smarandache
26
Consequently:
0 > < ℎ 0 1 ℎ ℎ
1 0 > 1 <
Reciprocally, if < , we cannot have 1 and 0, and from (1.9) we deduce:
- equality that is true for the three remaining cases:
IV. Using the properties of the characteristic function, solve
the following equations and systems of equations with sets:
SOLUTION :
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 27/243
Methods of Solving Calculus Problems
27
We have to analyze the following cases:
a) if ∉ and ∉ , we have:∩ 1 and ∩ 1 , so for (1.10) to be
accomplished we must have
0. So, any point that doesn’t
belong to or , doesn’t belong to either.b) if ∈ and ∉ , we have:∩ 1 and ∩ 1 , so for (1.10) to be
accomplished we must have 1. Therefore, any point from is in .
c) if ∉ and ∈ and for (1.10) to be accomplished, we
must have
0, consequently no point from
is not in
.d) if ∈ and ∈ we can have 0 or 1.
So any point from ∩ can or cannot be in . Therefore \ ∪ , where ⊂ ∩ , arbitrarily.
We have to analyze four cases, as can be seen from the
following table:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 28/243
C. Dumitrescu ■ F. Smarandache
28
So ∪ .
V. Determine the following sets:
7. Let ∈ ℤ a degree polynomial and ∈ ℤ .
Knowing that , determine the set:
INDICATIONS :
1. An integer maximal part is highlighted. This is obtained
making the divisions:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 29/243
Methods of Solving Calculus Problems
29
10 is dividable by 2 1 <> ∈ 0, 1, 2}
so ∈ ℤ <> 27 is dividable by 2 1 and 4 2 11 + is a multiple of 8 <> ∈ 14,5,2,1,0,1,4,13}. so and are integer solutions of systems the shape of:
{3 1
3 1 with and as divisors of 32.
So− ∈ ℤ <> 15 is dividable by .
8. Method 1: For ≠ 1 we have
so, we must have
6 3 > 0. As
, 3 ± 2√ 3, we
deduce:
Method 2: We consider the function −++ and
, with ℝ1} the domain of . is obtained from the
variation table of .
VI. Determine the following sets, when , ∈ ℕ:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 30/243
C. Dumitrescu ■ F. Smarandache
30
where is the integer part of .
SOLUTIONS : We use the inequalities: < < 1, so:
VII. The application : ⟶ ℱ defined through is a bijection from the family of the parts belonging to to the family of the characteristic functions defined on .
SOLUTION : To demonstrate the injectivity, let , ∈
with ≠ . From ≠ we deduce that there exists ∈ so
that:
a) ∈ and ∉ or b) ∈ and ∉ In the first case 1 , case 0 , so ≠. In the second case, also, ≠ .
The surjectivity comes back to:
Let then ∈ ℱ irrespective. Then for | 1}
we have .
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 31/243
Methods of Solving Calculus Problems
31
II. Functions
Function definition A function is determined by three elements , and , with
the following significations: and are sets, named the domain
and the codomain, respectively, of the function, and is a
correspondence law from to that causes:
each element ∈ to have one,and only one corresponding element ∈ . (F)
So we can say a function is a triplet ( ,, ), the elements of
this triplet having the signification stated above.
We usually note this triplet with : → . Two functions
are equal if they are equal also as triplets, i.e. when their three
constitutive elements are respectively equal (not just the
correspondence laws).
To highlight the importance the domain and codomain have
in the definition of functions, we will give some examples below, in
which, keeping the correspondence law unchanged and modifying
just the domain or (and) the codomain, we can encounter all
possible situations, ranging from those where the triplet is a
function to those where it is a bijective function.
In order to do this, we have to first write in detail condition
that characterizes a function. We can consider this condition as being made up of two sub
conditions, namely:
( each element from the domain has a corresponding element, in the
sense of “at least one element”, in the codomain.
Using a diagram as the one drawn below, the proposition can
be stated like this:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 32/243
C. Dumitrescu ■ F. Smarandache
32
“(At least) one arrow can be drawn from each point of
the domain.”
More rigorously, this condition is expressed by:
The second sub condition regarding in the definition of a
function is:
( the element from the codomain that corresponds to is unique.
For the type of diagram in fig. 2.1, this means that the arrow
that is drawn from one point is unique.
This condition has the equivalent formulation:
“If two arrows have the same starting point then they
also have the same arrival point .”
Namely,
.
Conditions ( and are necessary and sufficient for any
correspondence law to be a function. These conditions are easy to
use in practice.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 33/243
Methods of Solving Calculus Problems
33
In Figure 2.1, the correspondence law from a) satisfies (
and doesn’t satisfy
; in b) the correspondence law doesn’t satisfy
, but satisfies . In diagram c), neither , nor aresatisfied, and in d) the correspondence law satisfies both and, so it is the (only) function.
A function is therefore a triplet ( ,, ), in which the
correspondence law satisfies and .
Let’s observe that these two condition only refer to the
domain of the function:
– from each point of the domain stems at least one arrow; – the arrow that stems from one point of the domain is
unique.
It is known that the graphic of a function is made up of the
set of pairs of points ( , ), when covers the domain of the
function.
HOW CAN WE RECOGNIZE ON A GRAPHIC IF ACORRESPONDANCE LAW IS A FUNCTION [if it satisfies
and ] ?In order to answer this question we will firstly remember (for
the case , ⊂ the answer to two other questions:
1. Given , how do we obtain – using the graphic –
[namely, the image of (or images , if more than one, and in this case
the correspondence law is, of course, not a function)] ? Answer : We trace a parallel from to 0 to the point it
touches the graphic, and from the intersection point (points) we
then trace a parallel (parallels) to 0 . The intersection points of
these parallels with 0 are the images of .
2. Reciprocally, given , to obtain the point (points) having
the property , we trace a parallel to 0 through , and
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 34/243
C. Dumitrescu ■ F. Smarandache
34
from the point (points) of intersection with the graphic, we then
trace a parallel (parallels) to
0.
Examples1. A circle with the center at the origin and with the radius
is not the graphic of a function : ℝ ⟶ ℝ, because it does not
satisfy condition ( there are points in the domain that do not have any
image , namely, all the points through which the parallel 0 doesn’t
intersect the graphic) or , because there are points in the domain that
have more than one image (all the points
∈ , have two images).
2. A circle with the center at the origin and with the radius
is not the graphic of a function : , ⟶ ℝ, because it does not
satisfy condition (this time there are no more points in the
domain that do not have an image, but there are points that have two
images ).
3. A circle with the center at the origin and with the radius
is not the graphic of a function : ℝ ⟶ 0, ∞, because it does not
satisfy condition . With the codomain 0,∞, the points have
one image at most . There are points, however that don’t have an image.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 35/243
Methods of Solving Calculus Problems
35
4. A circle with the center at the origin and with the radius
is the graphic of a function
: , ⟶ 0,1, because all the points
in the domain have one, and only one image .5. A circle with the center at the origin and with the radius
is the graphic of a function : , ⟶ 0,1.In all these examples, the correspondence law has remained
unchanged (a circle with the center at the origin and with the radius, having therefore the equation , which yields
±
/
.
By modifying just the domain and (or) the codomain, wehighlighted all possible situations, starting from the situation where
none of the required conditions that define a function were fulfilled,
to the situation where both conditions were fulfilled.
WITH THE HELP OF THE PARALLELS TO THE
COORDINATES AXES, WE RECOGNIZE THE
FULFILLMENT OF THE CONDITIONS and , THUS:
a ) A graphic satisfies the condition
if and only if any parallel to
0 traced through the points of the domain touches the graphic in at least one
point.
b) A graphic satisfies the condition if and only if any parallel to0 traced through the points of the domain touches the graphic in one point at
most.
The inverse of a function We don’t always obtain a function by inverting the
correspondence law (inverting the arrow direction) for a randomly
given function : ⟶ . Hence, in the Figure 2.3, is a function,
but − (obtained by inverting the correspondence law ) is not a
function, because it does not satisfy (there are points that have
more than an image).
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 36/243
C. Dumitrescu ■ F. Smarandache
36
With the help of this diagram, we observe that the inverse
doesn’t satisfy every time there are points in the codomain of
that are the image of at least two points from the domain of .
In other words, − doesn’t satisfy every time there aredifferent points that have the same image through .
Therefore, as by inverting the correspondence law, the
condition still be satisfied, it is necessary and sufficient for
the different points through the direct function have different
images, namely
A second situation where − is not a function is when itdoes not satisfy condition :
We observe this happens every time there are points in the
codomain through the direct function that are not the images of any
point in the domain.
So as by inverting the correspondence law, the condition still be satisfied, it is necessary and sufficient for all the points in
the codomain be consummated through the direct function , namely
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 37/243
Methods of Solving Calculus Problems
37
In conclusion,
− satisfies
if and only if
satisfies
)
− satisfies if and only if satisfies ) − is a function if and only if satisfies ). As it is known, a function that satisfies ) is called an
injective function, a function that satisfies ) is called a
surjective function, and a function that satisfies ) is
called a bijective function.
We see then that the affirmation: “A function has an inverse
if and only if it is bijective”, has the meaning that the inverse − (that always exists, as a correspondence law , even if is not
bijective) is a function if and only if is bijective.
With the help of the parallels to the axes of coordinates, we
recognize if a graphic is the graphic of an injective or surjective
function; thus:
c) a graphic is the graphic of an injective function if and only if any
parallel to
0 traced through the points of the codomain touches the graphic in
one point, at most [namely − satisfies ].d) a graphic is the graphic of a surjective function if and only if any
parallel to 0 traced through the points of the codomain touches the graphic in
at least one point [namely − satisfies ].Examples
1. A circle with the center at the origin and with the radius
is the graphic of a function : 0, ⟶ 0, ∞ that is injective but
not surjective.
2. A circle with the center at the origin and with the radius
is the graphic of a function : , ⟶ 0, that is surjective
but not injective.
3. A circle with the center at the origin and with the radius
is the graphic of a bijective function
: 0, ⟶ 0, .
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 38/243
C. Dumitrescu ■ F. Smarandache
38
So, by only modifying the domain and the codomain, using a
circle with the center at the origin and with the radius
, all
situations can be obtained, starting from the situation where noneof the two required conditions that define a function were fulfilled,
to a bijective function.
Observation − is obtained by inverting the correspondence law , namely
in other words
In the case of the exponential function, for example, the
equivalence (2.1) becomes:
because the inverse of the exponential function is noted by
− . The relation (2.2) defines the logarithm:The logarithm of a number in a given base, , is the exponent to
which the base has to be raised to obtain .
The Graphic of the Inverse FunctionIf
and
are subsets of
and the domain
of
(so the
codomain of − is represented on the axis 0, and on the axis0, the codomain (so the domain of − ), then the graphic of
and − coincides, because − only inverses the correspondence
law (it inverses the direction of the arrows).
But if we represent for − the domain, horizontally, and the
codomain vertically, so we represent on 0 and on 0 , then
any random point on the initial graphic
(that, without the
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 39/243
Methods of Solving Calculus Problems
39
aforementioned convention is the graphic for both and − ),such a point
, becomes
, .
The points , and , are symmetrical inrelation to the first bisector, so we obtain another graphic
besides if the domain of − is on .
Agreeing to represent the domains of all the functions on0 it follows that is a graphic of −.
With this convention, the graphics of and − are
symmetrical in relation to the first bisector.
Methods to show that a functionis bijective
1. Using the definition- to study the injectivity, we verify if the function
fulfills the condition ;
- to study the surjectivity, we verify if the condition is fulfilled.
2. The graphical method- to study the injectivity we use proposition c);
- to study the surjectivity we use proposition d).
Important observationIf we use the graphical method, it is essential, for functions
defined on branches, to trace as accurately as possible the graphic
around the point (points) of connection among branches.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 40/243
C. Dumitrescu ■ F. Smarandache
40
Examples
1. The function : ℝ ⟶ ℝ, {2 1 1 3 > 1 isinjective, but is not surjective. The graphic is represented in Figure
2.5 a).
2. The function : ℝ ⟶ ℝ , {3 1 22 1 > 2 is
surjective, but is not injective. The graphic is in Figure 2.5 b).
3. The function : ℝ ⟶ ℝ, {2 1 1
2 > 1 is
bijective. The graphic is represented in Figure 2.5 c).
3. Using the theorem A strictly monotonic function is injective.
To study the surjectivity, we verify if the function is
continuous and, in case it is, we calculate the limits at the
extremities of the domain.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 41/243
Methods of Solving Calculus Problems
41
ExampleLet’s show that
, : ,
⟶ ℝ
is bijective.
(i) injectivity : because is strictly positive, we
deduce that the function is strictly ascending, so, it is injective.
(ii) surjectivity : the function is continuous on all the definition
domain, so it has Darboux’s property and
from where it follows that it is surjective.
4. Using the proposition 1 The function , : ⟶ is bijective if and only if⩝ ∈ the equation has an unique solution
(see the Algebra workbook grade XII )
Example
Let ℚ × ℚ and the matrix ( ). Then the function
is bijective (Algebra, grade XII).
Let’s observe that in the proposition used at this point, the
affirmation “the equation has a solution” ensures the
surjectivity of the function and the affirmation “the solution is
unique” ensures the injectivity.
5. Using the proposition 2If , : ⟶ and . 1 then is injective and is
surjective (Algebra, grade XII).
ExampleLet ℤ × ℤ and ( − −) . Then the function
: ⟶ defined through
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 42/243
C. Dumitrescu ■ F. Smarandache
42
satisfies ∘ 1, so it is bijective ( algebra, grade XII).Let’s observe that the used proposition can be generalized, at
this point, thus:
“Let : ⟶ , : ⟶ , so that ∘ is bijective. Then
is injective and is surjective.”
ExercisesI. Trace the graphics of the following functions and specify,
in each case, if the respective function is injective, surjective or
bijective.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 43/243
Methods of Solving Calculus Problems
43
SOLUTIONS .
It is known that:
In order to more easily explain the conditions from the
inequalities over , we will proceed as follows:
1. we draw the table with the signs of the functions , and
2. using the table we can easily explain the inequalities,
because for example <> 0.
1. For 1 , 2 and 3 we
have the following table:
So:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 44/243
C. Dumitrescu ■ F. Smarandache
44
Therefore,
5. To explain we proceed as follows:
(1) we draw the variation table of the function
(2) considering in the first monotonic interval (deducedfrom the table), at the right of 1 (because we are only interested in
the values ≥ 1 ), we calculate the minimum of on the
interval ∈ 1, etc.
a) for the first monotonic interval, ∈ 1,0, the function has a single minimum point on the interval 1, ,
situated in ; its values is . Being a single minimum
point it is also the global minimum (absolute) of on the
interval 1,, so, the value of is for ∈ 1,0.b) for the second monotonic interval, ∈ 0,∞ , the
function
has a single minimum point on the interval
1,, in 0; its value is 0 0. Being a single minimumpoint, we have 0 for ∈ 0,∞, so
Let’s observe that using the same variation table we can
explain the function ≤≤ . Hence:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 45/243
Methods of Solving Calculus Problems
45
a) for ∈ 1,0 , the function , has only one
maximum, namely
1, in
1; its value is
1 1.
So 1, for ∈ 1,0.b) for ∈ 0, ∞, has two maximum points, in 1 and ; their values are 1 1 and . The values of is the global maximum, so: max1, for ∈ 0, ∞.
It follows that {1 ∈ 1,0max1, > 0
9. From the variation table of +
1. for ∈ 2, , has a single minimum on the
interval 2,, in ; its value is
+ 2. for > 0, has a single minimum, namely 2,, in 0; its value is 0 0. So
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 46/243
C. Dumitrescu ■ F. Smarandache
46
Analogously, for
10. From the variation table of
in
we deduce:
(a) if ∈ 0, √ , the function in has, for ∈
0,, a single supremum
→ > 0 0
.
(b) ∈ −√ , ∞, the function has two supreme values:
0 and 1 .
So:
II. Study the bijectivity of:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 47/243
Methods of Solving Calculus Problems
47
2. , being an uneven degree polynomial.
III. Study the irreversibility of the hyperbolic functions:
and show their inverses.
IV. 1. Show that the functions 1 ,
: , ∞ ⟶ ℝ and
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 48/243
C. Dumitrescu ■ F. Smarandache
48
are inverse one to the other.
2. Show that
−+
coincides with its inverse.
3. Determine the parameters ,, , so that ++
coincides with its inverse.
4. Show that the function : 0,1 ⟶ ℝ defined
through , if ∈
, is bijective.
5. On what subinterval is the function : , ∞ ⟶ ℝ
√ 2 1 bijective?
INDICATIONS .
4. To simplify the reasoning, sketch the graphic of function .
5. Use the superposed radical decomposition formula:
where √ .
Monotony and boundariesfor sequences and functions
A sequence is a function defined on the natural numbers’ set.
The domain of such a function is, therefore, the set of natural
numbers ℕ . The codomain and the correspondence law can
randomly vary.
Hereinafter, we will consider just natural numbers’
sequences, i.e. sequences with the codomain represented by the set
of real numbers, ℝ.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 49/243
Methods of Solving Calculus Problems
49
: ℕ ⟶ ℝ REAL NUMBERS SEQUENCE
For this kind of function, with the same domain and
codomain each time, only the correspondence law should beadded, which comes down to specifying the set of its values: 0, 1, 2, … , , …
For the ease of notation, we write, for example:
So, to determine a sequence, it is necessary and sufficient to know
the set of its values: , , … , , …. This set is abbreviated by∈ℕ.
Consequently, a sequence is a particular case of function.
The transition from a function to a sequence is made in thismanner:
( by replacing the domain with ℕ
( by replacing the codomain with ℝ
( by replacing the variable with (or , or , etc)
( by replacing with
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 50/243
C. Dumitrescu ■ F. Smarandache
50
DOMAIN CODOMAIN VARIABLE CORRESP.LAW
Function D E x Sequence ℕ ℝ n
Hereinafter we will use this transition method, from a
function to a sequence, in order to obtain the notations for
monotony, bounding and limit of a sequence as particular cases
of the same notations for functions.
We can frequently attach a function to a sequence
∈ℕ ,
obtained by replacing with in the expression of ( becomes ).
Example We can attach the function +
+ to the sequence
++.
However, there are sequences that we cannot attach afunction to by using this method. For example:
It is easier to solve problems of monotony, bounding and
convergence of sequences by using the function attached to a
sequence, as we shall discover shortly.
Using functions to study the monotony, the bounding and
the limits of a sequence offers the advantage of using the derivative
and the table of variation.
On the other hand, it is useful to observe that the monotony,
as well as the bounding and the limit of a sequence are particular
cases of the same notion defined for functions. This
particularization is obtained in the four mentioned steps (
.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 51/243
Methods of Solving Calculus Problems
51
Here is how we can obtain the particularization, first for the
monotony and bounding, then for the limit.
MONOTONIC FUNCTIONS MONOTONIC SEQUENCESa) the function : ⟶ ℝ, ⊆ℝ is monotonically increasing if:⩝ . ∈ , →
a) the sequence : ℕ ⟶ ℝ ismonotonically increasing if:⩝ , ∈ ℕ, → (1)
If condition (1) is fulfilled,
taking, particularly and 1, we deduce
+ for any (2)In conclusion 1 ⟶ 2 The reciprocal implication alsostands true. Its demonstrationcan be deduced from thefollowing example:
If 2 is fulfilled and we take 7, 11 , we have
and step by step
and .
Consequently, 1 ⟷ 2.Because conditions 1 and 2 are equivalent, and (2) is moreconvenient, we will use thiscondition to define themonotonically increasing
sequence. But we must loosefrom sight the fact that it isequivalent to that particularcondition that is obtained fromthe definition of monotonicfunctions, through the
particularizations ( ,that define the transition fromfunction to sequence.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 52/243
C. Dumitrescu ■ F. Smarandache
52
b) the function : ⟶ ℝ, ⊆
ℝ is monotonically decreasing if:
⩝ , ∈ , → ≥
b) the sequence : ℕ ⟶ ℝ ismonotonically decreasing if:
⩝ , ∈ ℕ, → ≥
It can be shown that thiscondition is equivalent to: ≥ + for any ∈ ℕ (4)
LIMITED FUNCTIONS LIMITED SEQUENCES
a) : ⟶ ℝ , ⊆ ℝ is ofinferiorly bounded if it doesn’t
have values towards ∞ , i.e.∃ ∈ ℝ,⩝ ∈ , ≥
b) : ⟶ ℝ , ⊆ ℝ is ofsuperiorly bounded if it doesn’t
have values towards ∞ i.e. ∃ ∈ ℝ,⩝ ∈ ,
c) : ⟶ ℝ , ⊆ ℝ isbounded, if it is inferiorly and
superiorly bounded, i.e.∃, ∈ ℝ,⩝ ∈ ℕ .In other words, there exists an
interval , that contains allthe values of the function. Thisinterval is not, generally,symmetrical relative to the origin,but we can consider it so, by
enlarging one of the extremities wide enough. In this case , becomes , and thelimiting condition is:∃ > 0, ,⩝ ∈ i.e.∃ > 0, ,⩝ ∈ ||
a) the sequence : ℕ ⟶ ℝ is ofinferiorly bounded if it
doesn’t have values towards∞ , i.e.∃ ∈ ℝ,⩝ ∈ ℕ, ≥
b) the sequence : ℕ ⟶ ℝ is ofsuperiorly bounded if itdoesn’t have values to wards∞ , i.e.
∃ ∈ ℝ,⩝ ∈ ℕ,
≥
c) the sequence : ℕ ⟶ ℝ isbounded if it is inferiorly andsuperiorly bounded, i.e∃, ∈ ℝ,⩝ ∈ ℕ OR∃ > 0,⩝ ∈ ℕ ||
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 53/243
Methods of Solving Calculus Problems
53
Methods for the study of monotony
and boundingMETHODS FOR
FUNCTIONSMETHODS FOR
SEQUENCES The study of monotony
1. Using the definition:
We consider and we
compare the difference to zero. This can be donethrough successive minorings and majorings or by applyingLagrange’s theorem to function on the interval , . 2. Using the variation table (inthe case of differentiablefunctions)
As it is known, the variationtable of a differentiablefunction offers preciseinformation on monotonic andbounding functions.
3. Using Lagrange’s theorem
It allows the replacement of thedifference with that is then compared tozero.
1. Using the definition: We compare the difference+ to zero, and forsequences with positive terms we can compare the quotient+/ to one. We can makesuccessive minorings andmajorings or by applyingLagrange’s theorem to theattached considered sequence.2. Using the variation tablefor the attached function
We study the monotony of thegiven sequence and, using thesequences criterion, wededuce that the monotony ofthe sequence is given by themonotony of this function, on
the interval 0,∞, ( seeMethod 10 point c )3. Using Lagrange’s theorem
for the attached function.
The study of bounding1. Using the definition 2. Using the variation table
1.. Using the definition 2. Using the variation tablefor the attached function 3. If the sequence
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 54/243
C. Dumitrescu ■ F. Smarandache
54
decomposes in a finitenumber of bounded
subsequences, it is bounded.4. Using the monotony.If a sequence is monotonic, atleast half of the boundingproblem is solved, namely:a) if the sequence ismonotonically increasing, it isinferiorly bounded by the firstterm and only the superior
bound has to be found.b) if the sequence ismonotonically decreasing, itis superiorly bounded by thefirst term, and only the inferiorbound must be found.c) IF THE SEQUENCE ISMONOTONICALLYDECREASING AND HAS
POSITIVE TERMS, IT ISBOUNDED.
ExercisesI. Study the monotony and the bounding of the functions:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 55/243
Methods of Solving Calculus Problems
55
ANSWERS:
3. We can write × and we have the
variation table:
so the function is decreasing on the interval 1,/ and increasing
on /, ∞ . It is inferiorly bounded, its minimum being −/ and it is not superiorly bounded.
5. We use the superposed radicals’ decomposition formula:
6. Let < . In order to obtain the sign of the difference we can apply Lagrange’s theorem to the given
function, on the interval [ , (the conditions of the theorem are
fulfilled): ∈ ℝ exists, so that:
so the function is increasing.
II. Study the monotony and the bounding of the sequences:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 56/243
C. Dumitrescu ■ F. Smarandache
56
also determine the smallest term of the sequence.
also determine the smallest term of the sequence.
being the intermediary value that is obtained by applying
Lagrange’s theorem to the function ln on the intervals ,1. ANSWERS:
2. The function attached to the sequence is / .
From the variation table:
we deduce, according to the sequence criterion (Heine’s criterion,
see Method 10, point c ), the type of monotony of the studied
sequence. The sequence has the same monotony as the attached
function, so it is decreasing for
≥ 3. Because the sequence is
decreasing if ≥ 3, in order to discover the biggest term, we have
to compare and . We deduce that the biggest term of the
sequence is:
3. The function attached to the sequence is .
From the variation table for ∈ 0,1:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 57/243
Methods of Solving Calculus Problems
57
using point c) from Method 10, we deduce that the given sequence
is increasing for , smaller or equal to the integer part of ln−
and is decreasing for ≥ ln− 1.
Because
we deduce that the interval [1, ln−], in which the sequence
is increasing, can be no matter how big or small.
6. The sequence of the general term is increasing, the
sequence of the general term is decreasing, and the sequence
is increasing if ∈ 0, and is decreasing for ∈ ,
, 1.
III. Using Lagrange’s theorem .1. Prove the inequality |sin| || for any real , then
show that the sequence sin , sin sin , … , sinsin…sin , is monotonic and bounded, irrespective of , and
its limit is zero.
2. Study the monotony and the bounding of the sequences:
ANSWERS:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 58/243
C. Dumitrescu ■ F. Smarandache
58
From Lagrange’s theorem applied to the function on
the interval
,0 or
0,, depending on
being either negative or
positive, we obtain | cos | 1. The sequence satisfies the
recurrence relation: + sin , so if sin > 0, we deduce that
the sequence is decreasing, and if sin < 0 the sequence is
increasing. It is evidently bounded by 1 and 1. Noting with its
limit and passing to the limit in the recurrence relation, we obtain sin , so 0.
where 1 defines the function attached to the
sequence.
b) Applying Lagrange’s theorem to the function attached to
the sequence, on the interval , 1, we deduce+
. We have to further determine the sign of the derivation:
namely, the sign of
From the table below we deduce > 0 for > 0, so + > 0.
IV. 1. Let I be a random interval and : → a function.
Show that the sequence defined through the recurrence relation+ , with being given, is:
a) increasing, if > 0 on
b) decreasing if
< 0 on
.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 59/243
Methods of Solving Calculus Problems
59
2. If the sequence ∈ℕ is increasing, and the sequence
∈ℕ is decreasing and
for any
, then:
a) the two sequences are bounded, and so convergent.
b) if → 0 then they have the same limit.
c) apply these results to the sequences given by the
recurrence formulas:
with and being given.
SOLUTIONS:
1. + > 0 in the first case.
2. the two sequences are bounded between and , so the
sequences are convergent, because they are also monotonic. Let
and represent their limits. From the hypothesis from (b) it
follows that .
(c) the sequences have positive terms and:
V. Study the bounding of the sequences:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 60/243
C. Dumitrescu ■ F. Smarandache
60
ANSWER: Each sequence is decomposed in two bounded
subsequences (convergent even, having the same limit), so they are
bounded (convergent).
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 61/243
Methods of Solving Calculus Problems
61
III.
The limits of sequences
and functions
The definition of monotony and bounding of sequences has
been deduced from the corresponding definitions of functions,
using the method shown, in which is replaced with and is
replaced with
. We will use the same method to obtain the
definition of the limit of a sequence from the definition of the limitof a function. This method of particularization of a definition of
functions with the purpose of obtaining the analogous definition for
sequences highlights the connection between sequences and
functions. Considering that sequences are particular cases of
functions, it is natural that the definitions of monotony, bounding
and the limit of a sequence are particular cases of the corresponding
definitions of functions.
The limits of functions The definition of a function’s limit is based on the notion of
vicinity of a point.
Intuitively, the set ⊂ ℝ is in the vicinity of point ∈ ℝ,
if
1 ∈ , and, moreover,
2 also contains points that are in
the vicinity of (at its left and right).
Obviously, if there is any open interval , , so that ∈, ⊂ , then the two conditions are met.
Particularly, any open interval , that contains
satisfies both conditions, so it represents a vicinity for .
For finite points, we will consider as vicinities open intervals
with the center in those points, of the type:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 62/243
C. Dumitrescu ■ F. Smarandache
62
and sets that contain such intervals.For ∞, ∞ , ∞ doesn’t make sense, so we
have to find a different form for the vicinities of ∞. In order to
accomplish this, we observe that to the right of ∞ we cannot
consider any points, so ∞ has to be replaced with ∞ .
Furthermore, because ∞ ∞, we replace ∞ with . So, we
will consider the vicinities of ∞ with the form:
, ∞ and sets that contain such intervals.
Analogously, the vicinities of ∞ have the form:− ∞,
The definition of a function’s limit in a point expresses the
condition according to which when approaches , will
approach the value of the limit.
With the help of vicinities, this condition is described by:
As we have shown, for the vicinities and from the
definition ∗∗∗, there exist three essential forms, corresponding to
the finite points and to ∞ and ∞ respectively.
CONVENTION:
We use the letter to write the vicinities of , and the letter
designates the vicinities of .
We thus have the following cases:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 63/243
Methods of Solving Calculus Problems
63
and condition ∈ and becomes >
and condition
∈ and becomes
<
All these cases are illustrated in Table 3.1.
Considering each corresponding situation to and , in the
definition
∗∗∗, we have 9 total forms for the definition of a
function. we have the following 9 situations:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 64/243
C. Dumitrescu ■ F. Smarandache
64
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 65/243
Methods of Solving Calculus Problems
65
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 66/243
C. Dumitrescu ■ F. Smarandache
66
ExercisesI. Using the definition, prove that:
ANSWERS:
In all cases, we have
.
2. We go through the following stages:
a) we particularize the definition of limit
(taking into account the form of the vicinity).b) we consider > 0, irrespective; we therefore search for .
c) we make calculation in the expression | | ,
highlighting the module | |.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 67/243
Methods of Solving Calculus Problems
67
d) we increase (decrease) the obtained expression, keeping in
mind we are ony interested in those values of
for which
| | < .
So:
e) if the expression that depends on is bounded
(continuous) in a vicinity of , we can further increase, to eliminate. We have:
because in the interval 0,2, for example, that is a vicinity of 1 , the function |+| is continuous, and therefore
bounded: ∈ ,
, if ∈ 0,2.f) we determine , setting the condition that the expression
we have reached (and that doesn’t depend on , just on ) be
smaller than
:
Any that fulfills this condition is satisfactory. We can, for
example, choose or etc.
In order to use the increase made to function , we must
also have 1, so, actually min1, (for example).
It follows that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 68/243
C. Dumitrescu ■ F. Smarandache
68
b) let > 0; we search for
because in the interval 1,3, for example, the function:
is bounded by:
f) we determine from the condition: < . We obtain
< 3, so we can, for example, take . But in order for the
increase we have made to function to be valid, we must also have 1, so, actually min1, . It follows that:
so the condition from the definition of the limit is in this case, also,
fulfilled.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 69/243
Methods of Solving Calculus Problems
69
b) let any (not necessarily positive, because ∞ , so ,∞only makes sense for any , not just for positive).
so we can take √
Then:
II. Using the definition, prove that:
ANSWERS:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 70/243
C. Dumitrescu ■ F. Smarandache
70
b) let > 0 any; we search for .
c) because isn’t finite anymore, we cannot highlight | |. We will proceed differently: we consider the inequality| | < as an inequation in the unknown (using also the
fact that → ∞ , so, we can consider |2 1| 2 1 ). We
have:
b) let > 0 any; we determine .
c) we express from the inequality > :
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 71/243
Methods of Solving Calculus Problems
71
If and are the roots of the attached second degree equation
and we assume
< , we have
> ⟺ > max100,.
So we can take max100,.
The limits of sequences The definition of a sequence’s limit is deduced from the
definition of a function’s limit, making the mentioned
particularizations:
(1) is replaced with ,
(2) is replaced with ,
(3) is replaced with .
Moreover, we have to observe that ∞ and finite
don’t make sense (for example → 3 doesn’t make sense, because, being a natural number, cannot come no matter how close to 3 ).So, from the nine forms of a function’s limit, only 3 are
particularized: the ones corresponding to ∞. We therefore have: (we transpose the definition ( )
Because in the inequality
> ,
is a natural number, we
can consider as a natural number also. To highlight this, we will
write instead of . We thus have:
∞ (we transpose the definition ( )
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 72/243
C. Dumitrescu ■ F. Smarandache
72
∞ (we transpose the definition ( )
ExamplesI. Using the definition, prove that:
ANSWERS:
1. We adapt the corresponding stages for functions’ limits, in
the case of ∞.
b) let > 0 any; we determine ∈ ℕ.
c) we consider the inequality: | | < as an inequation
with the unknown :
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 73/243
Methods of Solving Calculus Problems
73
so, − 1
b) let > 0 any; we determine ∈ ℕ as follows:
Let and represent the solutions to this second degree
equation ( < ). Then, for > , the required inequality is
satisfied. So, 1.
Calculation methods for the limits offunctions and sequences
Let’s now look at the most common (for the high school
workbook level) calculation methods of the limits of functions and
sequences.
Joint methods for sequences and functions
1.
DefinitionIn the first chapter, we have shown how to use the definition
to demonstrate the limits of both functions and sequences. We will
add to what has already been said, the following set of exercises:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 74/243
C. Dumitrescu ■ F. Smarandache
74
I. Using the definition, find out if there limits to the
following sequences and functions:
2. Giving the forced common factor The method is frequently used to eliminate the
indeterminations of type:
,
∞ ∞,
. By removing the forced
common factor we aim at obtaining as many expressions as possible
that tend towards zero. For this, we commonly employ the
following three limits:
In order to have expressions that tend towards zero we aim
therefore to:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 75/243
Methods of Solving Calculus Problems
75
obtain as many terms as possible with the form , with
< 0, if
→ ∞.
obtain as many terms as possible with the form , with > 0, if → 0.
ExamplesI. Calculate:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 76/243
C. Dumitrescu ■ F. Smarandache
76
ANSWERS:
2. Given the common factor 9 we obtain terms that have
the form , with sub unitary .
expression that tends to when tends to infinity.
II. Trace the graphics of the functions:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 77/243
Methods of Solving Calculus Problems
77
8. For any rational function, non-null, with real
coefficients, we have:
3. Amplification with the conjugateIt is used to eliminate the indeterminations that contain
radicals. If the indetermination comes from an expression with the
form:
than we amplify with:
with the purpose of eliminating the radicals from the initial
expression. The sum from 3.1 is called a conjugate of order .
If the indetermination comes from an expression with the
form:
with uneven number, the conjugate is:
An example of application for this formula is exercise 5,
below.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 78/243
C. Dumitrescu ■ F. Smarandache
78
Exercises
ANSWERS:
7. We replace, for example ⋯ , in the
given sequence and we calculate
limits of the form:
4. Using fundamental limitsHereinafter, we will name fundamental limits the following
limits:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 79/243
Methods of Solving Calculus Problems
79
OBSERVATION :
From (a) we deduce that:
From (b) we deduce that:
From (c) we deduce that:
ExercisesI. Calculate:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 80/243
C. Dumitrescu ■ F. Smarandache
80
ANSWERS:12. Noting with arcsin , we observe that
tends to zero when tends to zero, so we aim to obtain
. To
accomplish this we amplify with:
Noting with:
it follows:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 81/243
Methods of Solving Calculus Problems
81
II. Calculate:
ANSWERS:
6. In the limits where there are indeterminations with
logarithms, we aim to permute the limit with the logarithm:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 82/243
C. Dumitrescu ■ F. Smarandache
82
= (we permute again the limit with the logarithm) =
III. Calculate:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 83/243
Methods of Solving Calculus Problems
83
ANSWERS:
and noting , we obtain:
4. We consider the function:
obtained from , by replacing with . We calculate:
According to the criterion with sequences (see Method 10,
point c ), we also have: → ∞ ln 2.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 84/243
C. Dumitrescu ■ F. Smarandache
84
5. Something that is bounded multiplied withsomething that tends to zero, tends to zero
(A) If ∈ℕ is bounded and → ∞ 0, then:
→ ∞. 0
(B) If is bounded in a vicinity of and → 0,
then → . 0.
Examples
because 1 is bounded by 1 and 1 and tends to zero;
because 1 c o s is bounded by 0 and 2 and tends to zero;
where is the nth decimal fraction of the number ;
where is the approximation with exact decimal fractions of the
number .
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 85/243
Methods of Solving Calculus Problems
85
INDICATIONS:
4. ∈ℕ is bounded by 0 and 9 , being composed of
numbers;
5. ∈ℕ is bounded by 2 and 3;
6. The majoring and minoring method(A) If functions and ℎ exist, so that in a vicinity of we
have:
then:
Schematically:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 86/243
C. Dumitrescu ■ F. Smarandache
86
(B) If the sequences ∈ℕ and ∈ℕ exist so that:
, starting from the rank
and
→ ∞ → ∞ ,
then: → ∞ .Schematically:
OBSERVATION :
If ∞ , we can eliminate ℎ ( ∈ℕ ,
and if ∞, we can eliminate ( ∈ℕ .
This method is harder to apply, due to the majorings and
minorings it presupposes. These have to lead to expressions, as least
different as the initial form as possible, not to modify the limit.
We mention a few methods, among the most frequentlyused, to obtain the sequences ∈ℕ and ∈ℕ. Examples
1. We put the smallest (biggest) term of the sequence∈ℕ, instead of all the terms.
ANSWERS :
a) In the sum that expresses the sequence ∈ℕ, there are:
the smallest term
√+ and the biggest term
√+. By applying
Method 1, we have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 87/243
Methods of Solving Calculus Problems
87
2. We minorate (majorate) the terms that make up the
sequence with the same quantity.
ANSWERS :
a) This time there isn’t a smallest (biggest) term in the sum
that makes up the sequence . This is due to the fact that the
function sin isn’t monotonic. Keeping into consideration
that we have: 1 sin 1 , we can majorate, replacing sin
with 1, all over. We obtain:
Analogously, we can minorate, replacing sin with 1 ,
then, by applying the first method it follows that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 88/243
C. Dumitrescu ■ F. Smarandache
88
3. We minorate (majorate), eliminating the last terms that
make up the sequence
.
ANSWERS :
ExercisesI. Calculate the limits of the sequences:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 89/243
Methods of Solving Calculus Problems
89
ANSWERS :
We majorate using the smallest and the biggest denominator.
Thus, we obtain common denominators for the sum.
and we make majorings and minorings using the biggest and the
smallest denominator.
7. Exercises that feature the integer part The most common methods to solve them is:
(A) The minoring and majoring method using the double inequality:
that helps to encase the function (the sequence) whose limit we
have to calculate, between two functions (sequences) that do not
contain the integer part.Using this method when dealing with functions, most of the
times, we don’t obtain the limit directly, instead we use the lateral
limits.
Example : For the calculation of the limit:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 90/243
C. Dumitrescu ■ F. Smarandache
90
we proceed as follows:
(a) We use the inequalities (3.2) to obtain expressions that
do not contain the integer part:
(b) To obtain the function whose limit is required, we
amplify the previous inequalities with+
, keeping into
consideration that:
- on the left of 2 we have: + > 0, so:
- on the right of
2 we have:
+ < 0
, so:
(c) Moving on to the limit in these inequalities(the majoring
and minoring method), we obtain the limit to the left:
2
and the limit to the right: 2 , so .
(B) If the double inequality cannot be used (3.2), for → , with ∈ ℤ , we calculate the lateral limits directly,
keeping into consideration that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 91/243
Methods of Solving Calculus Problems
91
Example : For → ∞(1)/ 1 we cannot use (3.2)
because the expressions obtained for the calculation of the laterallimits don’t have the same limit. That is why we use the fact that an
the left of 1 we have 0, and on the right of 1 we have 1, so:
It follows that ∞
(C) If → ∞, we can replace , keeping into consideration
that for ∈ , 1 , we have: (obviously, → ∞
implies → ∞ ).
Example :
The variable that tends to infinity is certainly between two
consecutive numbers: 1, so:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 92/243
C. Dumitrescu ■ F. Smarandache
92
Exercises
8. Using the definition of the derivative As it is known, the derivative of a function is:
if this limit exists and is finite. It can be used to (A) eliminate the
indeterminations for limits of functions that can be written in the
form:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 93/243
Methods of Solving Calculus Problems
93
being a point in which is differentiable.
Example :
We notice that by noting , we have
and the limit becomes:
This limit has the value , because is a function
differentiable in .
We have thus reduced the calculation of the given limit to
the calculation of the value of ,, calculation that can be made
by deriving :
Observation : this method can be used like this: “let
. We have , ln . , ln . We don’t
recommend this synthetic method for the elaboration of the
solution at exams because it can misguide the examiners. That is
why the elaboration: “We observe that by noting ⋯, we
have ⋯ and the limit becomes … , because is
differentiable in …”is much more advisable.
(B) The calculation of the limits of sequences, using attached
functions. The function attached to the sequence is not obtained,this time, by replacing with , in the expression of , because for
→ ∞ we can’t calculate the derivative, but by replacing with
(which leads to the calculation of the derivative in zero).
Example : For → ∞ √ 2 1 we consider the function
−
obtained by replacing
with
in the expression of
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 94/243
C. Dumitrescu ■ F. Smarandache
94
. We calculate → 0 . For this we observe that by noting
, we have 1 and the limitbecomes: → 0 −− ,0 , because is a differentiable
function in zero. We have: , 2 ln 2, so ,0 ln 2. Then ,
according to the criterion with sequences, we obtain → ∞
ln 2.
Exercises
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 95/243
Methods of Solving Calculus Problems
95
II. For > 0 calculate:
ANSWER:
4. We divide the numerator and the denominator with .
III. Show that we cannot use the definition of the derivative
for the calculation of the limits:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 96/243
C. Dumitrescu ■ F. Smarandache
96
ANSWERS :
2. We consider the function −− , obtained by
replacing the sequence on with . We calculate
→ 0 .
To use the definition of the derivative, we divide the numerator and
the denominator with , so:
We observe that by noting 2 , we have 0 1
and the limit from the numerator becomes: → 0 −
,
because
is differentiable in zero.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 97/243
Methods of Solving Calculus Problems
97
, 2 ln 2 , so, according to the criterion with
sequences
,
0 ln 2. Analogously, by noting
ℎ 3 ,
the limit from the numerator becomes: → 0 − ℎ,0 ,
because ℎ is differentiable in zero.ℎ 3 ln 3 , so, according to the criterion with
sequences ℎ,0 ln 3, and .
5. The limit contains an indetermination with the form 1,
so we will first apply method 4.
We now consider the function 2
obtained by replacing with in the expression above. We
calculate → 0 . For this, we observe that by noting
we have 0 2 and the limit can be written:
→ 0
− ,0 because is differentiable in zero.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 98/243
C. Dumitrescu ■ F. Smarandache
98
Now , ln ln , so ,0 ln ln
According to the criterion with sequences:
and the initial limit is √ . .
9. Using L’Hospital’s theorem The theorem. [l’Hospital (1661-1704)] If the functions and
fulfill the conditions:
1. are continuous on , and differentiable on , } 2. 0
3. , is not annulled in a vicinity of
4. there exists →
,, , finite or infinite,
then there exists →
.
This theorem can be used for:
(A) The calculation of limits such as → when these
contain an indetermination of the form or
.
(B) The calculation of limits such as → .
when these contain an indetermination of the form ∞. 0.
This indetermination can be brought to form (A) like this:
(indetermination with the form )
(indetermination with the form )
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 99/243
Methods of Solving Calculus Problems
99
Observation : sometimes it is essential if we write the
indetermination or
.
Example :
If we write:
(indetermination ) and we derive, we obtain:
(indetermination with the form with a higher numerator degree).
If we highlight the indetermination:
and we derive, we obtain
0.
(C) The calculation of limits such as → ,
with indetermination ∞ ∞.
This indetermination can be brought to form (B) giving
or as forced common factor.
We have, for example:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 100/243
C. Dumitrescu ■ F. Smarandache
100
and because
we distinguish two cases:
(namely,
doesn’t tend to
1 ), we have:
the indetermination is of the form (B).
(D) The calculation of limits such as
→ , with
indeterminations 1, 0, ∞.
These indeterminations can be brought to the form (C),
using the formula:
that for , for example, becomes , so:
The last limit contains an indetermination of the form (C).
For sequences, this method is applied only through one of
the two functions attached to the sequence (obtained by replacing
with
or by replacing
with
in the expression of
). According
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 101/243
Methods of Solving Calculus Problems
101
to the criterion with sequences it follows that the limit of the
sequence is equal to the limit of the attached function.
Attention : by replacing with we calculate the limit in ∞,
and by replacing with, we calculate the limit in 0.
ExercisesI. Calculate:
(a polynomial increases faster to infinity than a logarithm)
(a polynomial increases slower to infinity than an exponential)
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 102/243
C. Dumitrescu ■ F. Smarandache
102
II. Show that l’Hospital’s rule cannot be applied for:
ANSWER :
1. For and sin we have:
- doesn’t exist, so the condition 4 from the theorem isn’t
met. Still, the limit can be calculated observing that:
10. Using the criterion with sequences (the HeineCriterion)
The criterion enunciation:
with the properties:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 103/243
Methods of Solving Calculus Problems
103
Observation : If ∞ , condition doesn’t make sense
anymore.
This criterion cannot be used to eliminate indeterminations
because the sequence ∈ℕ from the enunciation is random, so,
by modifying the sequences (an infinite number of sequences), we
cannot eliminate the indetermination. Still, the criterion can be used
at:
(A) The calculation of limits that don’t contain indeterminations.
Example : Using the criterion with sequences, show that:
Solution : (a) we have to show that:⩝ ∈ℕ , with the properties:
(b) let ∈ℕ be any sequence with the
properties ,,
(c) in → ∞ we use the properties of the
operations of limits of sequences to highlight → ∞ (it is
possible because there are no indeterminations), then we use the
fact that → ∞ :
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 104/243
C. Dumitrescu ■ F. Smarandache
104
(B) To show that a function doesn’t have a limit in a point.
To accomplish this we can proceed as follows:1 . we find two sequences: ∈ℕ and ∈ℕ with the
properties ,, so that:
or:2 . we find a single sequence with the properties , ,
so that → ∞ doesn’t exist.
Example :
→ ∞ sin doesn’t exist
1. let 2 and 2 . We have:
So the limit doesn’t exist. 2 . let . We have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 105/243
Methods of Solving Calculus Problems
105
(C) The criterion with sequences can be used in the calculation of the
limits of sequences as follows:
(a) let be the function attached to the sequence, by thereplacement of, for example, with in the expression of ( or
of with, but then we calculate
→ 0 )(b) we calculate
→ ∞ ( → 0 , respectively).
(c) according to the criterion with sequences lim means: for any sequence
∈ℕ with the properties:
we have → ∞
We observe that the sequence fulfills the conditions and and, moreover, for this sequence we have:
ExercisesI. Using the criterion with sequences, calculate:
II. Show that the following don’t exist:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 106/243
C. Dumitrescu ■ F. Smarandache
106
7. If has a limit to its left (right) in , then:
8. If → ∞ , then → ∞ , but not the
other way around.
9. → ∞ doesn’t exist if : → ℝ is a periodic, non-
constant function. Consequences: the following do not exist:
Answer : 9. This exercise is the generalization of exercises 1.-6.
For the solution, we explain the hypothesis:
We now use method 1 . The following sequences fulfill
conditions and :
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 107/243
Methods of Solving Calculus Problems
107
and:
III. Determine the limit points of the functions: , , ∘ , ∘ , for:
Answer : let ∈ ℝ randonm. We check if has a limit
in . For this we observe it is essential if ∈ℕ is rational or
irrational. We firstly consider the situation where the sequence
∈ℕ is made up only of rational points (or only of irrationalpoints).
Let ∈ℕ be a sequence of rational points with the
properties ,,. We have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 108/243
C. Dumitrescu ■ F. Smarandache
108
So the function doesn’t have a limit in any point for
which
≠ 2, i.e.
and
±2. For
√ 2 we consider a random
sequence ∈ℕ with the properties ,,. 1decomposes in
two subsequences: , ∈ℕ made up of solely rational terms and, ∈ℕ made up of irrational terms. Because:
we deduce that:
The same goes for √ 2.
Specific methods for sequences
11. Any bounded and monotonic sequence
is convergent Applying this method comes back to the study of monotony
and bounding, and for the determination of the limit we pass over
to the limit in the recurrence relation of the given sequence. If such
a relation is not initially provided, it can be obtained at the study of
the monotony.
Example : !
(a) For the study of monotony and bounding we cannotapply the method of the attached function ( ! doesn’t make
sense). We employ the method of the report (we begin by studying
the monotony because 1 if the sequence is monotonic, at least
half of the bounding problem is solved, as it has been already
proven; 2 if the sequence isn’t monotonic, we can’t apply the
method so we won’t study the bounding.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 109/243
Methods of Solving Calculus Problems
109
We therefore have:
So the sequence is decreasing.
(b) The bounding. The sequence is bounded between
0 and
, being decreasing and with positive terms. Therefore → ∞
exists.
(c) For the calculation of we observe that studying the
monotony we have obtained the recurrence relation:
Moving on to the limit in this equality, we obtain:
EXERCISES :
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 110/243
C. Dumitrescu ■ F. Smarandache
110
where ∈ℕ fulfills conditions:
ANSWERS :
3. We observe the recurrence relation:
a) The monotony:
In order to compare this difference to zero, we make a
choice, for example + ≥ , that we then transform through
equivalences:
So + ≥ ∈ 1,2.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 111/243
Methods of Solving Calculus Problems
111
We have thus reached the conclusion that the sequence is
increasing if and only if it is bounded between
and
. We
obviously have > 0 > 1 so we still have to prove that < 2. This inequality can be proven by induction:
Verification :
So we have proven the monotony and the bounding.
Therefore → ∞ exists. To determine we move on to the
limit in the recurrence relation (this time this type of relation is
given initially):
As 1 is impossible, because > 0, we deduce 2.
6. We decompose the fraction+
in a difference of
consecutive terms:
To determine the four parameters:
a) we use the identification method and
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 112/243
C. Dumitrescu ■ F. Smarandache
112
b) we state the condition that the terms from the right
member in
3.4 be consecutive.
c) the denominators are consecutive, so we state the
condition that the numerators be in the same relation as
consecutives; the denominator of the first fraction is obtained from
the denominator of the second one, by replacing with 1, so
we must have the same relation between the numerators:
We have thus obtained the system:
with the solution:
It follows that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 113/243
Methods of Solving Calculus Problems
113
consequently:
7. We have:
so:
8. We apply the conclusion from 7.
II. Study the convergence of the sequences defined through:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 114/243
C. Dumitrescu ■ F. Smarandache
114
Indications :
4. sequence with positive terms (induction) and decreasing.
Let and be the limits of the subsequences of even and
uneven value. Show that .
12. Using the Cesaro-Stolz and Rizzoli LemmasLemma: (Cesaro-Stolz) Let ∈ℕ be a random sequence and
∈ℕ a strictly increasing sequence, having the limit
infinite. If → ∞ −− exists – finite or infinite,
then: → ∞ . This lemma can be used to calculate the limits of sequences
that can be expressed as a fraction:
and of the limits of the form:
presupposing that this limit exists and ∈ℕ is strictly
monotonic.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 115/243
Methods of Solving Calculus Problems
115
CONSEQUENCES:
Indeed,
Example :
because for , we have:
(B) The limits of the arithmetic, geometric and harmonic
averages of the first terms of a sequence having the limit , have
the value also:
Observation : a variant for the Cesaro-Stolz lemma for the case
when , → 0 has been proven by I.Rizzoli [ Mathematic Gazette ,
no 10-11-12, 1992, p. 281-284]:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 116/243
C. Dumitrescu ■ F. Smarandache
116
Lemma: (I.Rizzoli) If ∈ℕ, ∈ℕ are two sequences
of natural numbers that fulfill the conditions:
(ii) the sequence ∈ℕ is strictly monotonic
(increasing or decreasing)
(iii) there exists:
Then: → ∞ . EXERCISES :
8. ∙∙ ….∙+∙∙…∙++⋯+−+∙−+∙… ∙
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 117/243
Methods of Solving Calculus Problems
117
( Indications : amplification with the conjugate, the
minoring/majoring method, the Cesaro-Stolz lemma).
5. Let + ln1 , > 0. Using the Cesaro-Stolz
lemma show that
→ ∞. 2.
(Indications : )
III. Calculate the limits of the following sequences,
presupposing they exist:
knowing that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 118/243
C. Dumitrescu ■ F. Smarandache
118
where > 0 and is Euler’s constant.
Answers :
1. We apply the Cesaro-Stolz lemma for:
We have:
and through the hypothesis it follows that:
6. We consider:
We have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 119/243
Methods of Solving Calculus Problems
119
and keeping into account that:
(we can use the attached function), we obtain:
13. Utilization of Lagrange’s theorem Theorem: (Lagrange, (1736-1813)). If : , → ℝ is
continuous on , and differentiable on,,
then:
∃⊂∈ , ..−
− ,.
This theorem can be used for:
(A) The calculation of the limits of sequences in which we
can highlight an expression with the form:
for the attached function.
Example :
Solution :
a) we highlight an expression of the form 3.6, considering:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 120/243
C. Dumitrescu ■ F. Smarandache
120
b) we apply Lagrange’s theorem on function on the
interval
,1:
c) we replace in the expression−
− with ,. We
obtain:
d) we use the inequalities: < < 1 to obtainconvenient majorings (minorings). In our case, we have:
It follows that:
(B) The calculation of the limits of sequences that can be put
under one of the formulas:
being a function that is subject to Lagrange’s theorem on the
intervals: ,1, ∈ ℕ.
Observation : the sequences of the form are always
monotonic and bounded (so convergent) if is a function
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 121/243
Methods of Solving Calculus Problems
121
differentiable on ℝ and so that and , have different
monotonies.
Demonstration : to make a choice, let’s assume is increasingand , is decreasing:
1. the monotony:
2. we apply Lagrange’s theorem to function on the interval ,1:
so:
( ,- decreasing), wherefrom we deduce the sequence is decreasing.
3. the bounding: being decreasing, the sequence is superiorily
bounded by . We have only to find the inferior bound. For this,
we write 3.8 starting with 1 and we thus obtain the
possibility of minoring (majoring) the sequence .
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 122/243
C. Dumitrescu ■ F. Smarandache
122
We obtain a convenient replacement for the sum:
We write the relation 3.7 starting 1 and we add the
obtained equalities:
it follows that:
consequently, the sequence is inferiorly bounded. It is thus
convergent. Its limit is a number between 1 and .
ExercisesI. Using Lagrange’s theorem, calculate the limits of the
sequences:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 123/243
Methods of Solving Calculus Problems
123
II. Study the convergence of the sequences:
(the limit of this sequence is called Euler’s constant ∈0,1 )
(the partial sums of the harmonic sequence ∑ = )
(the partial sums of the generalized harmonic sequence)
Indication : 4. In order to have one of the forms 1 or 2, we
determine , keeping into account that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 124/243
C. Dumitrescu ■ F. Smarandache
124
We apply for this function the steps from the demonstration
at the beginning of this paragraph.
III. Show that:
Indication : 1. we must show that:
for:
The harmonic sequence
The sequence 1
⋯ is called a harmonic
sequence because it has the property: any three consecutive terms
are in a harmonic progression. Indeed, if we note , we have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 125/243
Methods of Solving Calculus Problems
125
For a long time, it was believed that this sequence has a
defined sum . In antiquity they sought to obtain the approximate
value of , by calculating the sum of as many terms of the sequence
as possible. Today it is a fact that ∞ (the harmonic sequence is
divergent). We mention a few of the methods that help prove this,
methods that use exercises from high school manuals, or exercises
of high school level:
1. Lagrange’s theorem (exercise I.2)
2. The inequality:
that is proven by induction in the X grade. Indeed, if had a finite
value:
then, by noting:
we would have:
and because → ∞, we deduce → ∞ 0.
But:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 126/243
C. Dumitrescu ■ F. Smarandache
126
so doesn’t tend to zero. Therefore, isn’t finite either.
3. Using the limit studied in the XII grade:
4. Using the inequality:
5. Using the definition of the integral (Riemann sums) (see
method 16, exercise II).
14. Sequences given by recurrence relations
(A) The linear Recurrence
1. The first degree linear recurrenceIs of the form:
The expression of the general term follows from the
observation that:
so:
Example :
We have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 127/243
Methods of Solving Calculus Problems
127
2. Second degree linear recurrenceIs of the form:
In order to find the expression of the general term, in this
case, we will use the expression of the general term from the first
order recurrence. We saw that for this recurrence we have: . For the second order recurrence, we search for the general
term with the same form:
∙ ,
being an undetermined constant.
By replacing in the recurrence relation, we obtain:
from where, dividing by − , we obtain the characteristic
equation:
We consider the following cases:
a) ∆> 0 (the characteristic equation has real and distinctroots and )
In this case, having no reason to neglect one of the roots, we
modify the expression , considering it of the form:
and we determine the constants and so that the first two
terms of the sequence have the initially given values.
Example :
(Fibonacci’s sequence). The characteristic equation is:
with the roots:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 128/243
C. Dumitrescu ■ F. Smarandache
128
Considering of the form:
from the system:
we obtain:
so:
Observation : +√ is known from antiquity as the golden
ratio. It is the limit of the sequence:
( radical) (method 11 applies). This number is often found in
nature (the arrangement of branches on trees, the proportion of the
human body etc. (For details, see, for example Matila Ghika:
Esthetics and art theory , Encyclopedic and Scientific Press, Bucharest,
1981).
b)
∆ 0 (the characteristic equation has equal roots
). In this case we consider of the form:
(namely ∙ , with - first degree polynomial) and we
determine and , stating the same condition, that the first two
terms have the same initially given values.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 129/243
Methods of Solving Calculus Problems
129
Example :
Taking ∙ , we obtain the characteristic equation:
Considering:
from the system:
we obtain: 4, 6, so:
c) ) ∆< 0 (the characteristic equation has complex roots)
Let these be:
We have:
but, in order to use only natural numbers, we show that we can
replace and respectively with:
Then we can write:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 130/243
C. Dumitrescu ■ F. Smarandache
130
3. Linear recurrence of degree h (bigger than 2)Is of the form:
with the first ℎ terms given.
Proceeding as with the second order recurrence, we combine ∙ and we state the condition that the recurrence relation
be met. We obtain the characteristic equation:
We distinguish the following cases:
a) if the characteristic equation has all the roots real and
distinct: , , … , , we will consider of the form:
and we determine the constants , ∈ 1, ℎ , stating the condition
that the first terms of the sequence have the initially given values.
b) if a root, for example, is multiple of order , ( ⋯ ), we replace the sum:
from the expression of with:
− being a polynomial of degree 2 , whose coefficient is
revealed by stating the condition that the first terms have the
initially given values.
c) if a root, for example, , is complex, then its conjugate isalso a root of the characteristic equation (let for example, ).
In this case, we replace in the expression of the sum:
with:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 131/243
Methods of Solving Calculus Problems
131
and if the root is a multiple of order ( ⋯ and
we replace in the expression of the terms that contain the
complex root with:
EXERCISES :
I. Determine the expression of the general term and calculate
the limit for:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 132/243
C. Dumitrescu ■ F. Smarandache
132
II. 1. Determine so that the sequence given by the
recurrence relation:
has a limit and calculate that limit.
2. Write the recurrence relations and the general terms of the
sequences for which: 1 , 2 , and the roots of the
characteristic equation are:
3. Let : ∙ ∙ , with , , , ∈ ℝ and , ≠ 0, || ≠ | |. Determine so that:
a) the sequence
is convergent;
b) → ∞ ∞
c) → ∞ ∞
4. Let ∙ ∙ ∙ , ≥ 1, with , , , ∈ ℝand , ≠ 0 . Determine so that the sequence is
convergent.
(B) Nonlinear recurrence
1. Recurrence of the form + ∙ ,
with given The expression of the general term is obtained by observing
that if is a root for the equation α ∙ l β (obtained by
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 133/243
Methods of Solving Calculus Problems
133
replacing an1and an with in the recurrence relation), then the
sequence
a ∈ℕ is a geometric progression.
Example:
From the equation + we deduce .
Then the sequence a ∈ℕ is a geometric progression.
Let 1. We have:
The ratio of the geometric progression is, consequently: . We obtain:
2. Recurrence of the form + ∙ ( being a random function)
The general term is found based on the observation that, if∈ℕ is a sequence that verifies the same recurrence relation,
then any sequence ∈ℕ verifies the given relation of the form:
In practice we chose of the form:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 134/243
C. Dumitrescu ■ F. Smarandache
134
the coefficients of being undetermined. We determine these
coefficients by stating the condition that
∈ℕ verifies the
recurrence relation.Particular case: a recurrence with the form:
with a polynomial of degree .
Example :
We search for the sequence with the form:
We state the condition that ∈ℕ verifies the given
recurrence relation:
We obtain:
By identification, it follows that:
so:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 135/243
Methods of Solving Calculus Problems
135
3. Recurrence of the form + (with : , →ℝ
continuous function)
Theorem: 1. If the function : , → ℝ iscontinuous and increasing on ,, then:
a) if > , the sequence ∈ℕ is increasing,
b) if < , the sequence ∈ℕ is decreasing.
The limit of the sequence is a fixed point of , i.e. a
characteristic equation .
2. If
is decreasing on
, , then the subsequences of
even and uneven value, respectively, of ∈ℕ aremonotonic and of different monotonies. If these two
subsequences have limits, then they are equal.
Example :
We have:
from the variation table we deduce that is decreasing on 0, √ 2
and increasing on √ 2,∞. By way of induction, we prove that
> √ 2 , and, because 5,1 < , then the sequence is
decreasing.
The bounding can be deduced from the property ofcontinuous functions defined on closed intervals and bounded of
being bounded.
The characteristic equation has the roots , ±√ 2, so √ 2.
Particular case: the homographic recurrence:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 136/243
C. Dumitrescu ■ F. Smarandache
136
We have:
called a homographic function and:
so the monotony of
depends on the sign of the expression
∙ ∙ .
If ≠ are the roots of the characteristic equation , by noting:
it can be shown that the general term of the sequence is given by
the relation:
Example :
Prove that:
From here it follows that the sequence is convergent (it
decomposes in two convergent sequences). It is deduced that the
limit of the sequence is 1.
ExercisesI. Determine the general term and study the convergence of
the sequences:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 137/243
Methods of Solving Calculus Problems
137
Determine coefficients and so that the sequence has a
finite limit. Calculate this limit.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 138/243
C. Dumitrescu ■ F. Smarandache
138
15. Any Cauchy sequence of natural numbersis convergent
One of the recapitulative exercises from the XI grade
analysis manual requires to be shown that if a sequence ∈ℕ of
real numbers is convergent, then:
A sequence that satisfies condition 3.8 is called a Cauchy
sequence or a fundamental sequence. It is proven that a natural
numbers’ sequence is convergent if and only if it is a Cauchy
sequence. This propositions enables the demonstration of theconvergence of sequences, showing that they are Cauchy sequences.
In this type of exercises the following condition is used:
which is equivalent to 3.9 but easier to use.
Example :
It has to be proven that:
(b) let > 0. We check to see if exists so that property
takes place.
(c) we have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 139/243
Methods of Solving Calculus Problems
139
(d) we state the condition: < and we obtain > :
, so
we can take: 1.
ExercisesI. Show that the following sequences are fundamental:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 140/243
C. Dumitrescu ■ F. Smarandache
140
Show that |+ | < + for any , ∈ ℕ . Deduce
from this that the sequence is a Cauchy sequence.
INDICATIONS :
and we decompose in simple fractions.
If
is uneven, we deduce:
and if it is even:
So:
From the condition:
+ < , we obtain: 1 1.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 141/243
Methods of Solving Calculus Problems
141
9. We proceed as we did in the previous exercise, where we
had:
16. Using the definition of the integralIt is known that the Riemann sum attached to the function : , → ℝ, corresponding to a division:
and to the intermediate points:
is:
If the points are equidistant, then:
and we have:
A function is integrable if:
exists and is finite. The value of the limit is called the integral offunction on the interval ,:
To use the definition of the integral in the calculation of the
limits of sequences, we observe that for divisions ∆ formed with
equidistant points, we have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 142/243
C. Dumitrescu ■ F. Smarandache
142
So we can proceed as follows:a) we show that the general term can be written in the
form:
with as a continuous function on , , and being the points
of an equidistant division.
b)
, being continuous, it is also integrable, and:
Example :
a) we write in the form:
highlighting the common factor. We have:
b) in:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 143/243
Methods of Solving Calculus Problems
143
we make the equidistant points:−
appear:
and we deduce the function . We have:
c) we arrange the equidistant points on a straight line:
and we deduce the division:
and the interval ,.d) we observe that is the Riemann sum corresponding to
function (continuous, thus integrable) and to the deduced division∆, on interval ,. As the points of ∆ are equidistant, we have:
Exercises:I. Using the definition of the integral, calculate the limits of
the following sequences:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 144/243
C. Dumitrescu ■ F. Smarandache
144
Indications :
6. By making logarithms, we obtain:
II. a) Write the Riemann sum
corresponding to the function:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 145/243
Methods of Solving Calculus Problems
145
the division:
and the intermediate points:
b) Calculate:
c) Calculate:
and deduce that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 146/243
C. Dumitrescu ■ F. Smarandache
146
IV. Continuity and derivability
ContinuityDEFINITION : the function : → ℝ is continuous in ∈ if:
1. it has a limit in ∈,
2. the limit is equal to
,
i.e.
THE CONSEQUENCE: Any continuous function in a
point has a limit in that point.
(A) Methods for the study of continuity
1. Using the definitionExample:
is continuous in 1.
Indeed, let’s show that:
Let there be any > 0. We must determine . We have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 147/243
Methods of Solving Calculus Problems
147
because on the interval 0,2, for example, 2 3 is bounded
between
3 and
7. We determine
from the condition:
We can take, for example 2. Then:
therefore
is continuous in
1.
2. Using the criterion with the lateral limits is continuous in:
where
are the left and right limit, respectively in
.
EXAMPLE: Let’s study the continuity of the function:
for ∈ 0,2. SOLUTION: Because we are required to study the
continuity, with a certain point being specified, we must make the
study on the whole definition domain. The points of the domain
appear to belong to two categories:
1. Connection points between branches, in which continuity
can be studied using the lateral limits.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 148/243
C. Dumitrescu ■ F. Smarandache
148
2. The other points, in which the function is continuous,
being expressed by continuous functions (elementary) (but this
must be specified each time).In our case:
(a) in any point ≠ 1, the function is continuous being
expressed through continuous functions.
(b) we study the continuity in ≠ 1. We have:
The value of the function in the point is 1 , so is
continuous in:
To explain the module we use the table with the sign of the
function sin :
1. if ∈ 0, ∪ ,2, the equation becomes:
2. if ∈ ,
, we obtain sin 1, so .
For this four values of the function is continuous also in
point 1, so it is continuous on ℝ.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 149/243
Methods of Solving Calculus Problems
149
3. Using the criterion with sequences This criterion is obtained from the criterion with sequences
for the limits of functions, by replacing with andeliminating the condition ≠ (which is essential in the case of
limits, because we can study the limit of a function in a point where doesn’t exist).
EXAMPLE:
Proceeding as we did with Method 10, with the two
adaptations mentioned just above, it follows that is continuous in
the points for which:
(B) Types of discontinuity points The point ∈ in which is not continuous is said to be
a first order discontinuity point if the lateral limits in exist and
are finite.
Any other discontinuity point is said to be of a second order.
(C) The extension through continuity
To extend the function : → ℝ means to add new pointsto the domain , where we define the correspondence law willingly.
If is the set of added points and ℎ is the correspondence
law on , the extension will be:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 150/243
C. Dumitrescu ■ F. Smarandache
150
For example:
can be extended to the entire set ℝ by putting:
There are many extensions to a function. Nevertheless, if
is continuous on
and has a finite limit in a point
∈ , there is
only one extension:
that is continuous, named the extension through continuity of in
point .
So, for our example, because → 0
1, the function:
is the only extension of sin/ that is continuous on ℝ.
(D) The continuity of composite functionsIf is continuous in and is continuous in , then
∘ is continuous in
.
So the composition of two continuous functions is a
continuous function. Reciprocally it isn’t true: it is possible that
and are not continuous and ∘ is.
For example:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 151/243
Methods of Solving Calculus Problems
151
isn’t continuous in any point, but the function ∘
is the identically equal function to 1, so it is continuous in any point
from ℝ.
ExercisesI. Study, on the maximum definition domain, the continuity
of the functions defined by:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 152/243
C. Dumitrescu ■ F. Smarandache
152
II. Specify the type of discontinuity points for the functions:
Determine the extensions using the continuity, wherepossible.
III. Study the continuity of the functions , , ∘ , ∘ ,
for:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 153/243
Methods of Solving Calculus Problems
153
IV. 1. Let , : ℝ → ℝ be a continuous function so that
for any
∈ ℚ. Show that
. (Manual)
2. Let : → ℝ , ⊆→ ℝ be a function with Darboux’s
property. If has lateral limits in any point , then is continuous
on . (Manual)
3. : ℝ → ℝ so that:
for any , ∈ ℝ. Prove that > 0 exists so that for any with
|| we have
|| < . Deduce the existence of a fixed point
for . ( is a fixed point for if . (Manual)
4. Let : , → ℝ, continuous. Then:
(Manual)
A function with the property 4.1 is called an uniform
continuously function on ,. The uniform continuity is therefore defined on an interval,
while continuity can be defined in a point.
Taking in 4.1 it is deduced that any uniform
continuous function on an interval is continuous on that interval.
Exercise 4 from above affirms the reciprocal of this
proposition, which is true if the interval is closed and bounded.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 154/243
C. Dumitrescu ■ F. Smarandache
154
Derivability
(A) Definition. Geometric interpretation.Consequences
Definition The function : → ℝ is differentiable in ∈ if the
following limit exists and is finite:
The value of this limit is noted by ,( and it is called the
derivative of in .
Geometric interpretation The derivative of a function in a point
is the slope of the
tangent to the graphic of in the abscissa point .
When tends to , the chord tends towards the
tangent in
to the graphic, so the slope of the chord
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 155/243
Methods of Solving Calculus Problems
155
tends to the slope of the tangent. Therefore, ,( is the slope ofthe tangent in the abscissa point to the graphic.
The geometric interpretation of the derivative thus follows
from the next sequence of implications:
Consequences1. Any function differentiable in a point is continuous in that
point.
2. The equation of the tangent to the graphic of a
differentiable function.
It is known that equation of a straight line that passes
through the point , is:
3. The equation of the normal (the perpendicular on the
tangent) is deduced from the condition of perpendicularity of two
straight lines ∙ 1 and it is:
(B) The derivation of composite functions
From formula 4.2 it is deduced that:
So, in order to derive a composite function we will proceed
as follows:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 156/243
C. Dumitrescu ■ F. Smarandache
156
(a) we derive the last function that is composed (in our case
and we replace in this function the variable
with the
expression issuing from the composition of the other function (inour case ∘ ).
(b) by neglecting the derived function in the previous step (in
our case ), we derive the function that became last (for us and
we also replace the variable with the expression that resulted from
the composition of the functions that have no yet been derived (in
this step we only have .
(c) we continue this procedure until all the functions arederived.
EXAMPLE:
1. By applying formula 4.2 we have:
We can obtain the same results much faster, using the
generalization of formula 4.2 presented above.
Therefore, the composite functions are:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 157/243
Methods of Solving Calculus Problems
157
CONSEQUENCES :
1. The derivative of the inverse function.
From −() , by applying formula 4.2, it follows
that:
or by noting :
EXAMPLE: For ln , : 0, ∞ → ℝ we have:
and as ln > we obtain:
2. High order derivatives of composite functions.
From () ,. , , deriving again in
relation to it follows that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 158/243
C. Dumitrescu ■ F. Smarandache
158
We have therefore obtained the second order derivative of
the composite function and the procedure can go on.
EXAMPLE: Let’s calculate ,, if − , being a function two times differentiable on ℝ.
We have:
and
(C) Derivatives of order For the calculation of the derivative we can use one
of the following methods:
1. We calculate a few derivatives ( ,, ,,, ,,,, … in order to
deduce the expression of , which we then demonstrate by way
of induction.
2. We use Leibniz formula for the derivation of the productof two functions:
The formula can be applied to a random quotient, because:
(D) The study of derivabilityIn order to study the derivability of a function:
(a) we distinguish two categories of points that the domain
has: points in which we know that the function is differentiable
(being expressed by differentiable functions) and points in which we
pursue the study of derivability (generally connection points
between branches).
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 159/243
Methods of Solving Calculus Problems
159
(b) for the study of derivability in this second category of
points, we use one of the following methods:
(1) we calculate the lateral derivatives using thedefinition
(2) we calculate the lateral derivatives using the
Corollary of Lagrange’s theorem: If is
differentiable in a vicinity of and if it
continues in and if there exists:
then, has a derivative on the left (on the right,
respectively) in and:
The continuity condition in that is required in thehypothesis of the corollary is essential for its application, but it
doesn’t constitute a restriction in the study of derivability,
because if isn’t continuous in , it isn’t differentiable either.
EXAMPLE: Let’s determine the parameters , ∈ ℝ so
that the function:
is differentiable. What is the geometric interpretation of the result?
Answer : Method 1. (using the lateral derivatives)
a) in any point ≠ , the function is differentiable, being
expressed through differentiable functions.
b) we study the derivability in
.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 160/243
C. Dumitrescu ■ F. Smarandache
160
Because of the proposition: non continuous ==> non
differentiable, we study the continuity first.
so: continuous in <> 1. (4.3)
If the continuity condition in
isn’t met, the function is
also not differentiable in this point, so we study the derivability
assuming the continuity condition is fulfilled.
For the calculation of this limit we can:
(1) apply the definition of the derivative,
(2) permute the limit with the logarithm,
(3) use l’Hospital’s rule. (1) We observe that by noting , we have 1 and the limit becomes:
(because is differentiable in )
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 161/243
Methods of Solving Calculus Problems
161
Therefore, is differentiable in:
and from the continuity condition we obtain 1 . We thus
have:
The geometric interpretation of the result: the straight
line
1
is tangent to the curve
ln .
If the branch of a function is expressed through a
straight line , the function is
differentiable in the point of connection
between the branches, if and only if the
respective straight line is tangent in the point of
abscissa to the graphic of the other branch.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 162/243
C. Dumitrescu ■ F. Smarandache
162
Method 2: (Using the corollary of Lagrange’s theorem)
(a) In any point
≠ , the function is differentiable, being
expressed by differentiable function and we have:
(b) We study the derivability in . We first state the
continuity condition:
In order to use the corollary of Lagrange’s theorem wecalculate the lateral limits of the derivative:
From the corollary we deduce , . Analogously:
If is differentiable in if and only if . From thecontinuity condition we deduce 1.
(E) Applications of the derivativein economics
(See the Calculus manual, IXth grade )
1. Let be the benefit obtained for an expenditure of lei. For any additional expenditure of ℎ lei, the supplementary
benefit on any spent leu is:
If ℎ is sufficiently small, this relation gives an indication of
the variation of the benefit corresponding to the sum of lei.
If this limit exists:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 163/243
Methods of Solving Calculus Problems
163
it is called the bound benefit corresponding to the sum of lei.
2. Let be the total cost for the production of units of
a particular product. Then, the cost per each supplementary unit of
product is:
and the limit of this relation, when ℎ tends to zero, if it exists, is
called a bound cost of production for units of the consideredproduct.
EXAMPLE: The benefit obtained for an expenditure of
lei is:
For any additional expenditure of ℎ lei, calculate the
supplementary benefit per spent leu and the bound benefit
corresponding to the sum of 1000 lei. Answer: The supplementary benefit per spent leu is:
For 1000 this is equal to 1997ℎ and
ExercisesI. The following are required:
1. The equation of the tangent to the graphic ln √ 1 in the abscissa point 1.
2. The equation of the tangent to the curve √ , that is parallel to 0.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 164/243
C. Dumitrescu ■ F. Smarandache
164
3. The equation of the tangent to the curve ,
parallel to the first bisector.
4. The equation of the tangent to the curve ++, that is
parallel to the chord that unites the abscissae points 1 and 3.
5. Determine and so that and − are
tangent in 1. Write down their common tangent.
6. Show that the straight line 7 2 is tangent to the
curve 4. (Manual)
II. 1. Calculate the derivative of the function:
2. Let
: ∞, 0 → ℝ,
3. Determine a
subinterval ⊆ ℝ, so that : ∞, 0 → is bijective. Let be itsinverse. Calculate ,1 and ,,1. (Manual)
3. If:
with > 0 for any , ∈ , the function is constant on .
4. If has a limit in point , then the function:
is differentiable in .5. If is bounded in a vicinity of , then is differentiable in . Particular case:
Indications:
1. Let be a primitive of the function:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 165/243
Methods of Solving Calculus Problems
165
We have:
3. For ≠ the inequality from the enunciation is
equivalent to:
from where, for ⟶ we obtain 0, so is constant on .
III. Calculate the derivatives of order for:
10. from the expression of the derivative
of order , deduce Newton's binomial formula.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 166/243
C. Dumitrescu ■ F. Smarandache
166
IV. 1. Show that:
2. Applying Leibniz’s formula for:
show that:
Find similar formulas, using the functions:
3. Let 0,1 and the functions , : ⟶ ℝ , ∈ , ∈ . Study the
derivability of the functions and and calculate ∈ ,
∈ . (Manual)
4. If
is a sequence of differentiable function, having a
limit in any point , then:
V. Study the derivability of the functions:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 167/243
Methods of Solving Calculus Problems
167
is indefinitely differentiable on ℝ and:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 168/243
C. Dumitrescu ■ F. Smarandache
168
∈ ℝ.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 169/243
Methods of Solving Calculus Problems
169
V.
Fermat, Rolle, Lagrange,
Cauchy Theorems
(A) Fermat’s theorem
1. EnunciationIf
: , → ℝ is continuous on
, and differentiable
on ,, then in any point of extremum from , (fromthe interior of the interval), the derivative is annulled.
(Fermat, 1601-1665)
Observation : ∈ , point of extremum > , 0. If is a point of extremum at the ends of the interval, it is
possible that the derivative isn’t annulled in .
Example:
has two points of extremum: in 1 and in 2, and:
2. Geometric and algebraic interpretationa) The geometric interpretation results from the
geometrical interpretation of the derivative: if the conditions of
Fermat’s theorem are fulfilled in any point from the interior of theinterval, the tangent to the function’s graphic is parallel to the axis0.
b) The algebraic interpretation : if the conditions of
Fermat’s theorem are fulfilled on , , any point of extremum
from , is a root to the equation , 0.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 170/243
C. Dumitrescu ■ F. Smarandache
170
Exercises1. If
, , … are positive numbers, so that:
for any ∈ ℝ, then we have:
2. Let , , … , , , … be positive real numbers, so
that:
for any ∈ ℝ. Then :
(generalization of the previous exercise)
3. If is continuous on , and differentiable on ,,
and
0, then:
(i) if , is increasing, it follows that 0 on ,;(ii) ) if ,, is decreasing, it follows that ≥ 0 on ,.4. If ≥ for any > 0, then .
5. <
if 0 < < < .
SOLUTIONS :
1. The exercise is a particular case of exercise 2.
2. We highlight a function that has a point of extremum
(global) on ℝ, observing that the inequality from the hypothesis canbe written:
Because this inequality is true for any real , it follows that 0 is
a global point of minimum for . According to Fermat’s theorem,
in this point the derivative is annulled, so
,0 0. We have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 171/243
Methods of Solving Calculus Problems
171
3. (i) Let’s assume there exists ∈ , so that > 0.
We can even assume that is a point of maximum, because such a
point exists, according to Rolle’s theorem, so we have
,
0.
Between and at least one point exists in which , < 0 , because, if, let’s say , ≥ 0 for any ∈ , , it
follows that is increasing on , and so < 0
implies 0. Then, from , 0 , we deduce the
contradiction:
4. ≥ <> . ln ≥ . ln <> ≥ (for any > 0 ). So is the abscissa of the maximum of the function:
We have:
It is sufficient to prove that is increasing, i.e.
, > 0.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 172/243
C. Dumitrescu ■ F. Smarandache
172
(B) Rolle’s theorem
1. EnunciationIf : , → ℝ is continuous on ,, differentiable
on , and , then there exists ∈, so that , 0. (Rolle, 1652-1719)
Fermat’s theorem states that in a point of extremum from
the interior of an interval the derivative is annulled, but it doesn’t
mention when such a point exists.
Rolle’s theorem provides a sufficient condition for theexistence of at least one such point, adding to the hypothesis from
Fermat’s theorem the condition: .
2. Geometric and algebraic interpretationa) The geometric interpretation : if the conditions of
Rolle’s theorem are met, there exists at least one point in the
interval
,in which the tangent to the graphic is parallel to
0.
b) The algebraic interpretation : if the conditions of
Rolle’s theorem are met, the equation , has at least one
root in the interval ,.
The algebraic interpretation of the theorem highlights a
method used to prove that the equation 0 has at least one
root in the interval ,. For this, it is sufficient to consider a
primitive
of
, for which the conditions of Rolle’s theorem are
fulfilled on ,. It results that the equation , 0 has at least
one root in the interval ,, i.e. has a root in the interval,.
A second method to show that the equation 0 has
at least one root in the interval , is to show just that is
continuous and . < 0.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 173/243
Methods of Solving Calculus Problems
173
3. Consequences1. Between two roots of a differentiable function on an
interval, there exists at least one root of the derivative.
2. Between two roots of a function’s derivative on an
interval, there exists, at most a root of the function.
Consequence 2. allows us to determine the number of roots of
a function on an interv al with the help of its derivative’s roots
(Rolle’s sequence):
Let , , … , be the derivative’s roots.
Then has as many real, simple roots as there are variationsof sign in the sequence:
(if we have 0, then is a multiple root)
ExercisesI. Study the applicability of Rolle’s Theorem for the
functions:
What is the geometric interpretation of the result?
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 174/243
C. Dumitrescu ■ F. Smarandache
174
Indication : 1. We obtain ∈ 4,5 . In any point from the
interval
4,5, the graphic coincides with the tangent to the graphic.
II. Given:
Show that the equation:
has at least one solution in the interval 0,2. (Manual)
2. If:
then the equation:
has at least one root in the interval 0,1.
3. (i) If:
then the equation:
has at least one root in the interval 0,1.
(ii) in what conditions the same equation has at least one
root in the interval 1,0?
(iii) The same question for the equations:
on the interval 1,1.
4. Let : ℝ → ℝ be differentiable and < < ⋯ < ,
roots of . Show that ,has at least 1 roots.
(Manual)
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 175/243
Methods of Solving Calculus Problems
175
Consequences:
a) A polynomial function of degree
has, at most,
real,
distinct zeros.b) If all the roots of a polynomial are real and distinct, its
derivative has the same property.
c) If all the roots of a polynomial are real, then its derivative
has the same property.
5. Given:
Show that the equation 0 has distinct roots inthe interval 1,1.
6. If is times differentiable on and it has 1 distinct
roots on , then has at least one root on .
7. Let , : , → ℝ, continuous on ,, differentiable
on , with ≠ 0 and , ≠ 0 on [a,b], and
.Show that
∈ , exists, so that:
8. Let : , → ℝ, continuous on ,, differentiable on,. Then between two roots of there exists at least a root of. ,.9. If the differentiable functions and have the property:
on an interval, then between two of its roots there is a root of and vice versa.
Consequence: If ,. cos . sin ≠ 0 , in any
length interval bigger than there is at least one root of .
10. If , : , → ℝ, continuous on ,, differentiable
on , , ≠ 0, , ≠ 0 and 0, then ∈, so that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 176/243
C. Dumitrescu ■ F. Smarandache
176
SOLUTIONS :
7. The required equality can also been written:
This equality appears in points that are roots of the
derivative of ℎ , so it is sufficient to demonstrate that the
function
ℎ fulfills the conditions of Rolle’s theorem on
,.
8. The equation . , 0 comes from theequalizing with zero the derivative of the function , so it is sufficient to show that satisfies the conditions of
Rolle’s theorem.
9. Let , be roots of . By hypothesis, we have ≠0 and ≠ 0. If, for example, there can be found a root of
between
, , this means that the function
ℎ
satisfies
the conditions of Rolle’s theorem and so ℎ, is annulled in at least
one point between and , which contradicts the hypothesis.
has at least one root in , <> ℎ, 0 has at least one root
in , , where ℎ .
We will present next, two methods for the study of an
equation’s roots. The first of these uses Rolle’s Theorem and the
other is a consequence of the fact that any continuous function has
Darboux’s property.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 177/243
Methods of Solving Calculus Problems
177
Methods for the study of an equation’s
roots1. Using Rolles’s theorem (its algebraicinterpretation)2. Using Darboux’s properties (if has Darboux’s
property (particularly if it is continuous) on , and:
. 0, then
has a root in
, ).
Examples :1) If : , → , is a continuous function, then , ∈, exists so that and . (Manual)
2) Let : 0, 2 → ℝ be a continuous function so that 0 2. Show that ∈ 0, exists so that . (Manual)
Consequences : If a traveler leaves in the morning from spot
and arrives, in the evening in spot , and the next day he leavesand reaches again spot , show that there is a point between and where the traveler has been, at the same hour, in both days.
Indication : If is the space covered by the traveler, we
have: 0 24 and so ∈ 0,12 exists, so that: 12.
By formulating the problem like this, the result might seem
surprising, but it is equal to saying that two travelers that leave, one
from spot heading to spot , and the other from to , meet on
the way.
3. Using Rolle’s sequence Example : Let’s study the nature of the equation’s roots:
being a real parameter.
Answer : The derivative’s roots are:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 178/243
C. Dumitrescu ■ F. Smarandache
178
and we have:
The results are illustrated in the following table:
4. The graphical method. The equation , 0 is
reduced to the form
. The number of roots is equal to the
number of intersection points between the graphics: and .
Example :
The number of roots of the given equation is equal to the
number of intersection points between the graphics:
5. Viette’s relations Example : Show that the equation:
with , , ∈ ℝ has, at most, two real roots.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 179/243
Methods of Solving Calculus Problems
179
Answer: we have:
so the equation has at least two complex roots.
6. Using the theorem of the average Theorem: If : , → ℝ is a Riemann integral (so it is
bounded), ∈ , exists, so that:
Example : Let : 0,1 → ℝ be continuous, so that:
Show that the equation has a root ∈ 0,1.
Answer: <> 0 , so by noting:ℎ we have to show that the equation ℎ 0 has
a root in the interval
0,1. We have:
But: ℎ 0, according to the theorem of the average,
with ∈ , . The function , being continuous, has Darboux’s
property, so, for
there exists
∈ 0,1so that
ℎ
.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 180/243
C. Dumitrescu ■ F. Smarandache
180
Exercises1. The equations:
can’t all have real roots if , , ∈ ℝ.
Indication :
2. If : 0,1 → ℝ is continuous and with the property:
then, the equation sin 0 has one root in 0,1.3. If : 0,1 → ℝ is continuous and with the property:
then, the equation 0 has one root in0,1.
4. If : 0,1 → ℝ is continuous and > 1 exists for which:
the equation: 1 1 has one root in 0,1.
Indication : We apply the theorem of the average to the
function:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 181/243
Methods of Solving Calculus Problems
181
(C) Lagrange’s Theorem
1. Enunciation:Let : , → ℝ be a continuous function
on , and differentiable on , . Then ∈, exists, so that:
(Lagrange, 1736 – 1813)
2. Geometric and algebraic interpretation
a) Geometric interpretation:If the conditions of the theorem are met, there exists a point ∈ , in which the tangent to the graphic is parallel to the
chord that unites the graphic’s extremities.
b) Algebraic interpretationIf the conditions of the theorem are met, the equation:
has at least one root in the interval ,.
3. The corollary of Lagrange’s TheoremIf is continuous on an interval and differentiable on\} and in there exists the derivative’s limit:
then is differentiable in and:
This corollary facilitates the study of a function’s derivability
in a point in an easier manner that by using the lateral derivatives.
Examples:
1. Study the derivability of the function:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 182/243
C. Dumitrescu ■ F. Smarandache
182
Solution : In any point ≠ 1 , the function is differentiable,
being expressed by differentiable functions. We study the
derivability in 1 using the corollary of Lagrange’s theorem.
- the continuity in 1:
1 ln 1 0 , so is continuous on ℝ (and
differentiable on ℝ1} ).- derivability. For the study of derivability in 1, we have:
(we know that , exists just on ℝ 1 ).Exercises
I. Study the applicability of Lagrange’s theorem and
determine the intermediary points for:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 183/243
Methods of Solving Calculus Problems
183
Indication :
2. The continuity condition in is:
and the derivability condition (the corollary to Lagrange’s theorem)is:
Point is the solution of the equation:
II. 1. Let . Apply the theorem of the
finite increases on the interval , , finding the intermediary
point . Deduce from this a way to build the tangent to the
parabola, in one of its given points.
2. We have , for any ∈ , if and only if:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 184/243
C. Dumitrescu ■ F. Smarandache
184
3. (The generalization of Rolle’s theorem). If is continuous
on
,and differentiable on
,, then
∈ , exists so that:
4. Supposing is twice differentiable in a vicinity of point, show that for ℎ, small enough, there exists , ∈ so that:
(Manual)5. Let : 0, ∞ → ℝ be a differentiable function so that:
and let > 0 be fixed. Applying the Lagrange theorem on each
interval:
show there exists a sequence ∈ℕ, having the limit infinite andso that:
6. The theorem of finite increases can also be written:
Apply this formula to the functions:
and study the values corresponding to the real point .
SOLUTIONS :
1. + , so, the abscissa point being given, in order to
build the tangent in point ( , ) of the graphic, we consider two
points, symmetrical to
:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 185/243
Methods of Solving Calculus Problems
185
The tangent passes through (
, ) and is parallel to the
chord determined by the points:
2. The necessity: For 1 is verified, and for ≠ ,
we have:
inequalities which are true due to the hypothesis.
Reciprocally, let ∈ , randomly. Making tend to , for:
we have:
3. We distinguish three cases:
In this case we must prove that ∈ , exists, so that: , > 0. If we had, let’s say, , 0 on , , it would follow
that is decreasing, so > .
We are situated within the conditions of Rolle’s theorem.
(3) The case > is analogous to 1.4. a) We apply Lagrange’s theorem to on ℎ, ℎ.b) We apply Lagrange’s theorem twice to the function on the interval ℎ, .6. a) 1/2 , c) is obtained ℎ ℎ , so cannot be
determined.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 186/243
C. Dumitrescu ■ F. Smarandache
186
(D) Cauchy’s Theorem
1.EnunciationLet If , : , → ℝ be continuous on, and
differentiable on , so that:
Then ∈ , exists, so that:
(Cauchy, 1789-1857)
2. Geometric and algebraic interpretationa) The geometric interpretation : in the conditions of the
theorem, there exists a point in which the relation of the tangents’
slopes to the two graphics is equal to the relation of the chords’
slopes that unite the extremities of the graphics.
b) The algebraic interpretation : in the conditions of thetheorem, the equation:
has at least one root in ,.
ExercisesI. Study the applicability of Cauchy’s theorem and determine
the intermediary point
for:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 187/243
Methods of Solving Calculus Problems
187
II. If has derivatives of any order on , , by applying
Cauchy’s formula to the functions:
a) , ℎ , show that ∈ , exists,
so that:
show that ∈ , exists, so that:
show that ∈ , exists, so that:
show that ∈ , exists, so that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 188/243
C. Dumitrescu ■ F. Smarandache
188
2. Applying the conclusion from point show that there
exists no functions
: ℝ → ℝ twice differentiable, so that:
≥ 0 and ,, < 0 (there are no non-negative and strictly concave functions on the
entire real axis).
SOLUTION :
2. From ,, < 0 for any ∈ ℝ , we deduce that , isstrictly increasing and so we cannot have , 0 for any ∈ ℝ.
Therefore, for any ∈ ℝ exists, so that , ≠ 0.
But then, from the relation:
it follows that:
We have two situations:
(1) if
,
> 0, making
tend to
∞ in
1, we have:
consequently:
2. If , < 0, we obtain the same contradiction if we
make tend to ∞.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 189/243
Methods of Solving Calculus Problems
189
VI.
Equalities and Inequalities
EqualitiesIt is known that if the derivative of a function is equal to
zero on an interval, the respective function is constant on that
interval. This observation allows us to prove certain equalities of
the form:
.or, in particular, and .
Indeed:
and in order to prove this equality it is sufficient to prove that:
( 0.
Observation : If the derivative is equal to zero on a reunion of
disjunctive intervals, the constant may vary from one interval to the
next.
To determine the constant , we can use two methods:
1. We calculate the expression in a
conveniently chosen point from the considered interval.
2. It the previous method cannot be applied, we can
calculate:
also being a conveniently chosen point (one end of the interval).
Example: Show that we have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 190/243
C. Dumitrescu ■ F. Smarandache
190
Solution : For:
we have ℎ, 0 for any ∈ ℝ1, domain that is a reunion of
disjunctive intervals. We calculate the value of the constant on each
interval.
(1) For > 1, we choose the point 0 where we can
easily make the calculations:
(2) For < 1 , we can’t find a point in which to easilycalculate the value of ℎ, but we observe that we can calculate:
InequalitiesMethod 1. Using Lagrange’s or Cauchy’stheorem.
In some inequalities, we can highlight an expression of the
form:
and the points , being conveniently chosen. In this case, we canuse Lagrange’s theorem to prove the respective inequality, by
replacing the expression:
from the inequality, with ,. The new form of the inequality can
be proven taking into account the fact that < < and the
monotony of
,.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 191/243
Methods of Solving Calculus Problems
191
A similar method can be used if in the inequality an
expression of the following form is highlighted:
using Cauchy’s theorem.
EXAMPLES :
1. Using Lagrange’s theorem, show that:
if 0 < < .Solution : We go through these steps:
(a) We highlight an expression of the form:
observing that:
and dividing by . The inequality becomes:
(b) we apply Lagrange’s theorem to the function ln
on the interval , . ∈ , exists so that:
(c) the inequality becomes:
This new form of the inequality is proven considering the
fact that < < and that , is decreasing.
2. Using Lagrange’s theorem, show that > 1 for any ≠ 0.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 192/243
C. Dumitrescu ■ F. Smarandache
192
Solution :
To use Lagrange’s theorem, we search for an interval
,
and a function so that we can highlight the relation −− in our
inequality. For this we observe that, if ≠ 0 , we have two
situations:
(1) > 0. In this case we can consider the interval ,.a) We highlight, in the given inequality, an expression of the
form:−
− , observing that, if > 0, we have:
(b) We apply Lagrange’s theorem to the function:
on the interval 0,. ∈ 0, exists so that:
(c) The inequality becomes:
This form of the inequality is proven considering the fact
that 0 < < and the derivative , is increasing (so < ).(2) If < 0 the demonstration is done analogously.
Method 2: The method of the minimum This method is based on the observation that if is a global
point of minimum (the smallest minimum) for a function ℎ on a
domain and if ℎ ≥ 0 (the smallest value of ℎ is non-
negative), then ℎ ≥ 0 for any ∈ (all the values of ℎ are
non-negative). We can demonstrate therefore inequalities of the
form ≥ , i.e. ≥ 0, in other words, of the
form:
ℎ ≥ 0.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 193/243
Methods of Solving Calculus Problems
193
Example : Using the method of the minimum, show that >
1 for any
≠ 0.
Solution : In order to use this method, we proceed as follows:(a) We write the inequality under the form: ℎ ≥ 0:
(b) Let ℎ 1 . Using the variation table, we
calculate the global minimum (the smallest minimum) of ℎ:
From the table it can be observed that ℎ has a single
minimum, in 0, so this is also the global minimum. Its value isℎ0 0. So, for ≠ 0, we have ℎ > 0.
Method 3: (Inequalities for integrals, withoutthe calculation of the integrals)
To prove an inequality of the form:
without calculating the integral, we use the observation that if
, or
respectively, are values of the global minimum and maximum of on the interval ,, we have:
and so, by integrating, we obtain:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 194/243
C. Dumitrescu ■ F. Smarandache
194
so, in order to prove the required inequalities, the following
inequalities (numerical) still have to be proven:
Example : Show that:
Solution: (a) The inequality can be written under the form:
(b) Using the variation table, we calculate the
global minimum and maximum of the function −
on the interval 0,1:
We thus have 2√ and 1
(c) We integrate the inequalities
on the interval , and we obtain:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 195/243
Methods of Solving Calculus Problems
195
Method 4: Using the inequality from the
definition of convex (concave) functionsLet be an interval on a real axis.
Definition : The function : → ℝ is convex on if between any
two points , ∈ , the graphic of is situated
underneath the chord that unites the abscissae points and .
With the notations from the adjacent graphic, this condition
is expressed through the inequality: .
In order to explain and , we observe that:
1. is in the interval , if and only if it can be written
in the form:
Indeed, if ∈ , , it is sufficient that we take / ) to meet the required equality.
Reciprocally, if has the expression from ..1, in order
to show that ∈ , , the inequality system must be checked:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 196/243
C. Dumitrescu ■ F. Smarandache
196
The first inequality is equivalent to 1 , and the second
inequality is equivalent to
0.
Observation : The condition ∈ 0,1 from ..1 istherefore essential.
The inequality from the convexity becomes:
2. To express , we observe that:
so:
from which it follows:
By replacing
and
, we obtain:
: → ℝ is convex <>
: → ℝ is said to be concave if, for any , ∈ , between
the abscissae points
and , the graphic of the function
is
above the chord that unites these points, namely:
It is known that we can verify the convexity and concavity of
a function twice differentiable on an interval with the use of the
second derivative:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 197/243
Methods of Solving Calculus Problems
197
Using the two ways of expressing convexity, we can
formulate and prove certain inequalities.
Example : Show that for any and from ℝ, we have:
Solution : The function
is convex on
ℝ,,
>0, so:
from where, for we obtain the inequality from the
enunciation.
ExercisesI. Show that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 198/243
C. Dumitrescu ■ F. Smarandache
198
5. For the intervals , ⊂ ℝ let’s consider the continuous
and injective functions:
and the constant in the interior of . From the
condition:
() it follows that
−
for any that fulfills the condition ∈ . Usingthis observation, construct equalities of the form ℎ for:
Answer : (a) We have:
We determine function from the condition () that becomes:
For , for example, we have:
and from the condition:
we deduce:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 199/243
Methods of Solving Calculus Problems
199
and so we have:
II. Using Lagrange’s theorem, prove the inequalities:
6. Using Cauchy’s theorem applied to the functions:
show that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 200/243
C. Dumitrescu ■ F. Smarandache
200
III. Using the method of the minimum, demonstrate the
inequalities:
IV. Without calculating the integrals, show that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 201/243
Methods of Solving Calculus Problems
201
V. Without calculating the integral, show which one of the
following integrals has the highest value:
Indication : Using the method of the minimum, we prove
inequalities of the form < or > on the
considered intervals. We then integrate the obtained inequality.
VI. 1. If : → ℝ is convex, then for any , , … , ∈ ,
we have the inequality (Jensen’s inequality):
(Manual)
2. Using Jensen’s inequality applied to the convex function , prove the inequalities of the averages:
3. Show that for any , , … , ∈ , ], we have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 202/243
C. Dumitrescu ■ F. Smarandache
202
What are the maximum domains that contain zero and in
which these inequalities take place? Give examples of other
domains, that don’t contain the origin and that host inequalities likethose illustrated above. In what domains do inverse inequalities take
place?
4. Using the concavity of the logarithmic function, prove that
for any , , … , > 0 we have:
5. Show that the function , > 1 , > 0 isconvex. Using this property show that for any non-negative
numbers , , … , , the following inequality takes place:
6. Applying Jensen’s inequality to the function ,
show that for any , , … , ∈ ℝ, we have:
What analogous inequality can be deduced from the convex
function , for > 0? And from the concave function , for 0?
7. Applying Jensen’s inequality for the convex function
, for > 0, show that, if:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 203/243
Methods of Solving Calculus Problems
203
is a polynomial with all real and positive roots, we have:
What can be said if the polynomial has all the roots real and
negative?
SOLUTIONS :
By comparing the right member of the required inequality
with the right member of Jensen’s inequality, we deduce that the
points , , … , are deduced from the conditions:
We have:
With these points, Jensen’s inequality, for , becomes:
Making the calculations in the left member we obtain the inequality
of the averages.
3. On the interval 0, , the functions sin and cos are
concave. The biggest interval that contains the origin and in whichsin is concave, is the interval 0,. On ,2, for example, the
same function is convex, so the inequality from the enunciation is
changing.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 204/243
C. Dumitrescu ■ F. Smarandache
204
VII. Primitives
Connections with other notionsspecific to functions
Definition : The function : → ℝ has primitives on the interval if there exists a function : → ℝ, differentiable
and
,
on
.
It is known that two primitives differ through a constant:
The set of all primitives is called an undefined integral of the
function and is noted by:
In order to enunciate the method employed to show that
a function has primitives and the method to show that afunction doesn’t have primitives, we recall some of the
connections that exist between the main notions of calculus studied
in high school, relative to functions:
1.
THE LIMIT
2.
THE CONTINUITY
3.
THE DERIVABILITY
4. DARBOUX’S PROPERTY
5.
THE PRIMTIVE6. THE INTEGRAL
The connections between these notions are given by the
following properties: ) Any continuous function in a point has a limit in this point:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 205/243
Methods of Solving Calculus Problems
205
) Any differentiable function in a point
is continuous in this point:
) Any continuous function on , has Darboux’s
property on ,: ) Any continuous function on , has primitives on,:
) Any continuous function on , is integrable on,: ) Any function that has primitives on , has Darboux’s
property on ,:
) If a function has Darboux’s property on
,, then in
any point in which the limit (the lateral limit) exists, it is equal to .
Consequences: 1. A function with Darboux’s property (so a
function that admits primitives) can’t have an infinite limit (lateral
limit) in a point .
The functions for which at least one lateral limit is
infinite or different from doesn’t have primitives.
2. If has Darboux’s property on , and is adiscontinuity point, AT LEAST ONE LATERAL LIMIT
DOESN’T EXIST.
Due to the implications stated above, we can form the Table :
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 206/243
C. Dumitrescu ■ F. Smarandache
206
OBSERVATION: If is a discontinuity point for , one of
the following situations is possible:
(1) the lateral limits exist and they are finite
(2) a lateral limit is finite, the other is infinite
(3) both lateral limits are infinite
(4) one lateral limit is infinite, the other doesn’t exist
(5) one lateral limit is finite, the other doesn’t exist
(6) both lateral limits don’t exist
FROM ALL THE FUNCTIONS DISCONTINUED IN
AT LEAST ONE POINT, THE ONLY ONES THAT CAN
HAVE PRIMITIVES ARE THOSE FOR WHICH AT LEASTONE LATERAL LIMIT DOESN’T EXIST (AND THE OTHER,
IF IT IS EXISTS, IS FINITE).
Also, let’s observe that, although in case 1 the function
doesn’t have primitives on , (doesn’t have Darboux’s
property), still, if, moreover, the lateral limits are equal, then has
primitives on , \} through the function – the extension
through continuity of
′ restriction to the domain
, \}.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 207/243
Methods of Solving Calculus Problems
207
Example :
has Darboux’s property for any ∈ 1,1 , but doesn’t have
primitives if ≠ 0.
Indeed, assuming ad absurdum that has primitives, let be
one of its primitives. Then must be of the form:
In order to calculate ∫ sin , we observe that sin
comes
from the derivation of the function cos . Indeed,
so:
The function
cos
has primitives on
0,∞
because it is continuous on this interval, but doesn’t have primitives
on 0,∞ (we aim for the interval to be closed at zero in order to
study the continuity and derivability od ).
We observe that, because → 0 > 0 0, we can consider
its extension through continuity:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 208/243
C. Dumitrescu ■ F. Smarandache
208
This function, being continuous on 0,∞, has primitives on
this interval. Let be one of these primitives. Then:
We state the condition that be continuous in 0. We
have:
so we must have 2. 0 and so, for the primitive, we
obtain the following formula:
We state the condition that be differentiable in zero:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 209/243
Methods of Solving Calculus Problems
209
because is differentiable in zero. We have ,0 0 0 ( is a primitive of
), so
,
0 0. Consequently,
is not
differentiable in zero if ≠ 0.Observation : For the study of derivability we have mentioned
two methods: using the definition and using the corollary of
Lagrange’s theorem. Let’s observe that in the example from abo ve,
we can’t use the corollary of Lagrange’s theorem because → 0 > 0
, doesn’t exist.
The table at page 193 allows us to formulate methods toshow that a function has primitives and methods to show that a
function doesn’t have primitives.
We will enunciate and exemplify these methods, firstly with
exercises from the calculus manual, XII grade, in order to highlight
the necessity of familiarizing oneself with the different notions
frequently used in high school manuals.
(A) Methods to show that a functionhas primitives
The first three methods are deduced from the table, using
implications of the form → : ) We show that the function is differentiable.
) We show that the function is continuous.
) We construct the primitive. ) We show that the function is a sum of functions that
admit primitives.
Keep in mind that method
is preferred over
because derivability is easier to prove that continuity, and the
method is used only if the function doesn’t have a limit in a
discontinuity point (this is the only situation where it can still have
primitives).
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 210/243
C. Dumitrescu ■ F. Smarandache
210
(B) Methods to show that a function
doesn’t have primitives The first three methods are deduced from the table, using
the equivalence: → ⟷ ⟶ .
We therefore have:
This way, we deduce the following methods to show that a
function doesn’t have primitives: ) We show that in a discontinuity point, both lateral
limits exist and are finite or at least one lateral limit is infinite.. ) We show that the function doesn’t have Darboux’s
property..
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 211/243
Methods of Solving Calculus Problems
211
) We assume ad absurdum that has primitives and we
obtain a contradiction.
) If is the sum of two function, one of themadmitting primitives, the other not admitting primitives, it follows
that doesn’t have primitives.
Examples :
has primitives because is continuous.
Solution : We observe that 2.
comes from the
derivation of the function , so, a primitive of must have
the form:
We state the condition that be continuous in zero:
so and we have:
We state the condition that be differentiable in zero:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 212/243
C. Dumitrescu ■ F. Smarandache
212
(something bounded multiplied with something that tends to zero,
tends to zero), so is differentiable in zero. But we must also have,0 0, equality which is also met, so has primitives.
We have:
where and:
and the functions and ℎ are primitives.
) a) doesn’t have primitives on ℝ
because in points is discontinuous and both lateral limits are
finite.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 213/243
Methods of Solving Calculus Problems
213
doesn’t have primitives because
→ 0 > 0 ∞.
The functions in which this method is used in the
manual are of the form:
with
and
ℎ, continuous functions,
≠ ℎ. We will prove on this
general case that doesn’t have Darboux’s property. Thedemonstration can be adapted to any particular case.
We must prove there exists an interval whose image through isn’t an interval. We use the two conditions from the hypothesis.
Solving the exercise : We must prove there exists an interval
whose image through isn’t an interval, using the two conditions
from the hypothesis.
From the condition ≠ ℎ we deduce there exists at leastone point in which ≠ ℎ. Then, for small enough,
the intervals , and ℎ , ℎ
are disjunctive (see the Figure below).
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 214/243
C. Dumitrescu ■ F. Smarandache
214
From the conditions and ℎ continuous in we deduce:
Considering, as we said small enough for the intervals , and ℎ , ℎ to be
disjunctive, and min, }, it follows that the image throughthe function of the interval , isn’t an interval,
because:
This method presupposes the same steps as in
method .
(a) we look for primitives for ′ branches and we obtain a
first expression of ;
(b) we state the condition that be continuous in the
connection point (points) between the branches;(c) we state the condition that be differentiable in these
points:
- if is differentiable and , ,
we have covered method .
- if isn’t differentiable in or , ≠ , we have covered method .
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 215/243
Methods of Solving Calculus Problems
215
doesn’t have primitives on ℝ.
Indeed, we have ℎ with:
and is continuous so it has primitives and ℎ has lateral limits,
finite in zero, so it doesn’t have primitives. If, let’s say by absurd,
had primitives, it would follow that ℎ has
primitives.
ExercisesI. Using the method , show that the following
functions have primitives in ℝ:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 216/243
C. Dumitrescu ■ F. Smarandache
216
II. Using methods , study if the following
functions have primitives:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 217/243
Methods of Solving Calculus Problems
217
if : ℝ → ℝ and if it satisfies: 0 <> 0.
10. The product from a function ℎ that admits primitives
and a function differentiable with the continuous derivative is a
function that admits primitives.
Indications : If is the primitive of ℎ, then . , is continuous,
so it admits primitives. Let be one of its primitives. We have. , ℎ. .
being a differentiable function with a continuous derivative on ℝ.
III. Using method , show that the following functions
don’t have primitives:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 218/243
C. Dumitrescu ■ F. Smarandache
218
IV. Using method , show that the following functions
don’t have primitives:
Answer :
1. The functions 3 and ℎ 2 1 are
continuous on ℝ and ≠ ℎ . We will show that doesn’t have
Darboux’s property. Let be a point in which we have ℎ ), for example 2. We have 2 6, ℎ2 9.ℎ 2 <>
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 219/243
Methods of Solving Calculus Problems
219
2 <>
Let there be so that the intervals:
are disjunctive. For example 1 and let there be min, . Then the image of the interval 2 , 2 through isn’t an interval because for:
we have 3 ∈ 5,7 and for:
we have 2 1 ∈ 8,10. V. Using methods , study if the following
functions admit primitives:
Indication :
so ℎ, with:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 220/243
C. Dumitrescu ■ F. Smarandache
220
Indication :
We obtain 0.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 221/243
Methods of Solving Calculus Problems
221
Logical Scheme for Solving ProblemsSolving problems that require studying if a function has
primitives can be done according to the following logical scheme:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 222/243
C. Dumitrescu ■ F. Smarandache
222
VIII. Integration
An integral is defined with the help of a limit. It is much
easier to understand the theoretical aspects regarding the integral if
one has understood why it is necessary to renounce the two broad
categories of limits considered up to this point:
1. the limit when tends to ,2. the limit when tends to infinity,
and a new category of limits is introduced:
3. the limit when the division’s norm tends to zero.
The specifications that follow are aimed exactly at clarifying
this aspect.
The theory of the integral has appeared out of a practical
necessity, in order to calculate the area that lays between the graphic
of a function and the axis 0.Given a bounded function and (for now) non-negative : , → ℝ we can approximate the desired area with the help of
certain rectangles having the base on 0.
For this, we consider the points:
The set of these points form a division of the interval
,
and they are noted with ∆:
The vertical strips made using the points have an area as
hard to calculate as the initial area, because of the superior outline.
We obtain rectangles if we replace the superior outlines with
horizontal segments.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 223/243
Methods of Solving Calculus Problems
223
In order that the obtained approximation be reasonable, it is
only natural that these horizontal segments meet the graphic of thefunction. We observe that such a horizontal is uniquely determined
by a point situated between − and . In this way, we have
obtained the rectangles with which we aimed to approximate the
desired area.
The areas of the rectangles are:
The sum of these areas is called the Riemann sum attached
to the function , to the division ∆ and to the intermediary points . This sum:
approximates the desired area.
The necessity of introducing the third category of limits is
due to the necessity that the approximation be as accurate as
possible.
And now, a question: Is the approximation better when:
(a) the rectangles increase in number, or when
(b) the rectangles are narrower and narrower?
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 224/243
C. Dumitrescu ■ F. Smarandache
224
We observe that the rectangles “increase in number” if and
only if
tends to infinity, but making
tend to infinity we do not
obtain a “better approximation”. Indeed, if we leave the firstrectangle unchanged, for example, and raising the number of points
from its location to its right as much as we want (making tend to
infinity), the approximation remains coarse. The approximation becomes
much “better” if the rectangles are “ thinner and thinner ” .
In order to concretize this intuitive intuition, we observe that
the rectangles are thinner and thinner if the “thickest of them”
becomes thinner and thinner.
The biggest thickness of the rectangles determined by a
division ∆ is called the norm of the division ∆.
Therefore, “better approximations” <=> “thinner and
thinner rectangles” <=> “the biggest thickness tends to zero” <=>
“the norm of ∆ tends to zero” <=> ||∆|| → 0. The area of the stretch situated between the graphic of the
function and the axis 0 is, therefore:
This limit is noted with ∫ and it is called the
integral of function
on the interval
,.
For functions that are not necessarily non-negative on ,,the integral is represented by the difference between the area
situated above the axis 0 and the area situated below the axis 0.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 225/243
Methods of Solving Calculus Problems
225
Consequently, in order to obtain the area of the stretch
situated between the graphic and the axis 0, in such a case, wehave to take into consideration the module of :
Coming back to the affirmations and from above,
we observe that: “the rectangles are thinner and thinner” => “the
rectangles increase in number”, namely > .In other words, |∆| → 0 → ∞.
Reciprocally, this statement isn’t always true. That is why we
cannot renounce |∆| → in favor of → ∞ . Nevertheless,
there is a case where the reciprocal implication is true: when the
points of the division are equidistant. → ∞ > |∆| → 0 for equidistant points.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 226/243
C. Dumitrescu ■ F. Smarandache
226
In this case, the rectangles have the same dimension of the
base, namely
− −
for any
1,2, … , , and the
Riemann sum becomes:
For divisions such as these, the following equivalence takes
place:
|∆| → 0 <> → ∞ that allows, as it was revealed by Method 16 for the calculation of
limits, for the utilization of the definition of the integral for the
calculation of sequences’ limits.
Before we enumerate and exemplify the methods for the
study of integration, let’s observe that, if the function is
continuous, among the Riemann sums ∆, that can be
obtained by modifying just the heights of the rectangles (modifying
just the intermediary points ), there exists a biggest and smallest
Riemann sum, namely the Riemann sum having the highest value is
obtained by choosing the points ∈ −, for which:
and the Riemann sum having the smallest value is obtained by
choosing the points for which:
By noting with ∆ and ∆ , respectively, these sums (called
the superior Darboux sum, the inferior Darboux sum, respectively),
we have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 227/243
Methods of Solving Calculus Problems
227
The points
that realizes the supreme, respectively, the
infimum of on the intervals −, exists because it is known
that a continuous function, on a closed and bounded interval is
bounded and it touches its bounds. It is known (see Theorem 37
from the Calculus manual, grade XI ) that a function is integrable if and
only if:
Connections between integration andother notions specific to functions
In order to formulate such connections, we fill the list of the
propositions
enunciated in the previous chapter, with the
following: any function integrable on , is bounded, any function monotonous on , is integrable, any function continuous on , is integrable, any function that has on , a finite number of
discontinuity points of first order is integrable.
The demonstration of
follows from the fact that if we
modify the values of a function integrable on ,, in a finitenumber of points, we obtain also an integrable function and from
the property of additivity in relation to the interval of the integral:
Using the proposition stated above, we can make the
following table:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 228/243
C. Dumitrescu ■ F. Smarandache
228
and the table containing implications between all the notions that
we have dealt with is the following (
and
being one of the
lateral limits):
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 229/243
Methods of Solving Calculus Problems
229
Methods to show that a function is
integrable (Riemann) on , using the definition; we show that it is continuous; we show that it has a finite number of discontinuity
points of first order; we show it is monotonic; we show that the function is a sum of integrable
functions.
Methods to show that a function is not
integrable (Riemann) on , using the definition (we show that ∆, doesn’t
have a finite limit for ||∆|| → 0 ); we show that the function is not bounded; we show that the function is the sum of an integrable
function and a non-integrable function.
Examples (a) Using the definition of the integral, show that any
differentiable function with the derivative bounded on
, is
integrable on ,.Solution : Let : , → ℝ be differentiable. We have to
show that ∆, has a finite limit when ||∆|| → 0 . The given
function being differentiable, it is continuous, so it has primitives.
Let be one of its primitives. We can apply Lagrange’s theorem to
function , on any interval −, . There exists therefore ∈− , so that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 230/243
C. Dumitrescu ■ F. Smarandache
230
It follows that:
We calculate the two sums separately:
We show that the second sum tends to zero when ||∆|| → 0.
For this, we take into consideration that , is bounded, in other
words, > 0 exists, so that | , < | for any from , and
we apply the theorem of Lagrange to the function , but on the
intervals of extremities and . There exists between and so that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 231/243
Methods of Solving Calculus Problems
231
It follows therefore that:
For the second sum, we therefore have:
The condition from this exercise’s hypothesis is very strong.
As it is known, it can be shown that a function that fulfills only the
continuity condition is also integrable. Still, the method used in this
exercise can be easily adapted to most exercises that require
showing that a function is integrable.
(b) Show that
= is integrable on any interval
,. ( Manual ).
Solution : We will adapt the previous demonstration. The
given function is differentiable and has the derivative , cos
bounded. Being differentiable, it is continuous and it has primitives.
Let be one of its primitives, cos and let:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 232/243
C. Dumitrescu ■ F. Smarandache
232
be a division of , . We apply Langrange’s theorem to the
function
on each interval
−, . There exists therefore
∈− , so that:
The function has a bounded derivative: | cos | < 1 for
any from ℝ , so for any from , . We have to show that∆, has a finite limit when ||∆|| → 0. We will show that this
limit is (as it is known) cos cos . We have:
Applying the theorem of Lagrange to the function sin, on the interval determined by the points and , we deduce
there exists between and so that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 233/243
Methods of Solving Calculus Problems
233
The function : 0,3 → ℝ defined by:
is integrable. Indeed, it is continuous in any point ≠ 2 from the
definition domain, being expressed through continuous functions.
We study the continuity in
2:
The function is continuous on the interval 0,3, so it is integrable.
, : 0, 2√ 3 → ℝ can be discontinuous
just in the points where
is discontinuous. These are of the form
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 234/243
C. Dumitrescu ■ F. Smarandache
234
, ∈ ℤ. From these, only points 0,1,2,3, are in the function’s
domain. We study the limit in these points. We have:
so the function has a finite number of discontinuity points of first
order and is therefore integrable.
For the function we can say it is integrable
either by using the previous criterium, either using the fact that it is
monotonic on any interval ,. The function : 1,1 → ℝ through || is integrable because it is the sum of three integrable
functions: is integrable because it has a finite
number of discontinuity points (a single point, 0 ) and the
discontinuity is of the first order; the function || is
integrable because it is continuous, and the function isintegrable because it is monotonic (or because it has a single f
discontinuity point, of the first order).
isn’t integrable because, by choosing in the Riemann sum:
the intermediary points , rational, we have: 0, so in this
case:
and if are irrational, we have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 235/243
Methods of Solving Calculus Problems
235
isn’t integrable (Riemann) on 0,1 because it is not bounded.
is the sum of the functions:
If would be integrable, from it would follow
that so would be integrable on 0,1, as the sum of
two integrable functions.
ExercisesI. Using the definition, show that the following functions are
integrable:
II. Study the integration of the functions:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 236/243
C. Dumitrescu ■ F. Smarandache
236
8. The characteristic function of the set 0,1 ⊂ ℝ.
III. For which values of the parameters and are the
functions below integrable? For which values of the parameters do
they have primitives?
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 237/243
Methods of Solving Calculus Problems
237
INDICATIONS :
1.
0
, so for any value of the parameter, the
function has only one discontinuity, of first order on 0,1.2. For any value of ∈ 0,1 the function has a limited
number of discontinuity points of first order.
4. The function comes from the derivation of
sin .
IV. 1. Show that any continuous function on an interval, is integrable.
2. Adapt the demonstration from the previous point to show
that the following functions are integrable on the interval 0,1:
ANSWERS :
1. We will show that:
We have:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 238/243
C. Dumitrescu ■ F. Smarandache
238
where:
We now use two properties of continuous functions on a
closed and bounded interval:
(1) a continuous function on a closed, bounded interval is
bounded and it touches its bounds; so, there exists , ∈ , sothat , .
Therefore:
In order to evaluate the difference , we use
the property:
(2) a continuous function on a closed and bounded interval is
uniformly continuous (see Chapter ), namely:
In order to replace with and with , we have to
consider a division ∆ having a smaller norm than (which is
possible because we are making the limit for ||∆|| → 0 ). Therefore,
assuming ||∆|| < we have:
i.e. < and so:
Because is arbitrary (small), we deduce that:
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 239/243
Methods of Solving Calculus Problems
239
so the function is integrable.
2. We show that
∆ ∆ tends to zero when
||∆||
tends to zero. We have:
where:
because on the interval 0,1 the function is increasing.
So:
The function is continuous on 0,1 so it isuniformly continuous, namely:
Considering the division ∆ so that ||∆|| < (possibly
because ||∆|| → 0 ), we have − < , so:
Because is arbitrarily small, we deduce that:
i.e.
is integrable on
0,1.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 240/243
C. Dumitrescu ■ F. Smarandache
240
References
T. ANDREESCU & collective: Mathematical problems from competitions and
exams sustained in 1983, printed at the Polygraphic Enterprise “Banat”,
Timisoara, 1984.
L. ARAMA, T. MOROZAN: Problems of differential and integral calculus ,
Ed. Tehnica, Bucharest, 1978.
C. AVADANEI, N. AVADANEI, C. BORS, C. CHIREA: From
elementary to superior mathematics , Ed. Academiei R.S.R, Bucharest, 1977.
D. M. BATINETU: Mathematical problems for the secondary high school , Ed.
Albatros, Bucharest, 1979.
D. M. BATINETU-GIURGIU, V. GHIORGHITA, I. V. MAFTEI, I.
TOMESCU, F. VORNICESCU: Problems given at mathematical Olympiads
for high schools , Ed. Stiintifica, Bucharest, 1992.
D. BUSNEAG, I. MAHTEI: Homework for circles and mathematical contests for students , Ed. Scrisul Romanesc, Craiova, 1983.
A. CORDUNEANU, G. RADU, L. POP, V. GRAMADA:
Mathematical problems workbook, Ed. Junimea, Iasi, 1972.
GH. CALUGARITA, V. MANGU: Mathematical problems for the
elementary and secondary level in high school , Ed. Albatros, Bucharest 1977.
M. COCUZ: Mathematical problems workbook, Ed. Academiei R.S.R,Bucharest, 1984.
N. DONCIU, D. FLONDOR: Algebra and calculus, workbook with
problems , E.D.P., Bucharest, 1978.
M. DANGA: Homework and mathematical problems , Ed. Tehnica,
Bucharest, 1991.
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 241/243
Methods of Solving Calculus Problems
241
B. R. GELBAUM, J. M. H. OLMSTED: Counterexamples in the analysis ,
Ed. Stiintifica, Bucharest, 1973.
I. GIURGIU, F. TURTOIU: Mathematical problems workbook, E.D.P,
Bucharest, 1980.
I. ILIESCU, B. IONESCU, D. RADU: Mathematical problems for the
admission in the higher education , E.D.P., Bucharest, 1976.
C. IONESCU-TIU, L. PIRSAN: Differential and integral calculus for the
admission in college , Ed. Albatros, Bucharest, 1975.
AL. LEONTE, C.P.NICULESCU: Workbook of algebraic and calculus
problems , Ed. Scrisul Romanesc, Craiova, 1984.
I. A. MARON: Problems in calculus of one variable , Mir Publishers,
Moscow, 1973.
A. S. MURESAN, V. MURESAN: Algebraic and calculus problem, admission
contests 1981-1990, Ed. Tehnica, Bucharest, 1991.
C. NASTASESCU, C. NITA, M. BRANDIBURU, D.JOITA: Algebraicexercises and problems , E. D. P., Bucharest, 1981.
C. P. NICOLESCU: Calculus , Ed. Albatros, Bucharest, 1987.
C. P. NICOLESCU: Mathematical synthesis , Ed. Albatros, Bucharest,
1990.
G. RIZESCU, E. RIZESCU: Homework for high school mathematical circles ,
Ed. Tehnica, Bucharest, 1977.
V. SCHNEIDER, GH. A. SCHNEIDER: Calculus problems workbook for
grades XI-XII , Ed. Hyperion-Apollo, Craiova, 1993.
GH. SIRETCHI: Calculus problems , Typography of the Bucharest
University, 1975.
N. TEODORESCU, A. CONSTANTINESCU, M. MIHAI, L.
PIRSAN, E. PERJERIU, A. POPESCU-ZORICA, P. RADOVICI-
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 242/243
C. Dumitrescu ■ F. Smarandache
MARCULESCU, M. TENA : Problems from the mathematical gazette,
methodological and selective edition , Ed. Tehnica, Bucharest, 1984.
R. TRANDAFIR, AL. LEONTE: Mathematical problems and exercises
workbook, Ed. Junimea, Iasi, 1975.
*** Exercises workbook for the admission in the higher education, Mathematics,
Physiscs and Chemistry , Ed. St. Encicl., Bucharest, editions 1984 and 1989.
*** High school manuals, different editions.
*** The Mathematical Gazette collection .
242
8/20/2019 Methods of Solving Calculus Problems
http://slidepdf.com/reader/full/methods-of-solving-calculus-problems 243/243
In this book, we discuss a succession of methods encountered in
the study of high school calculus to students and teachers, to higher
education entry examination candidates, to all those interested, in
order to allow them to reduce as many diverse problems as possible
to already known work schemes.
We tried to present in a methodical manner, advantageous for the
reader, the most frequent calculation methods encountered in the
d f h l l hi l l