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7. Subsequences. Cauchy criterion 61
Let us show that {yk} converges to c. We must prove that
∀ε > 0 ∃N ∈ N ∀k > N |yk− c| < ε.
Choose an arbitrary ε > 0. There exists a natural N such that
1
N
< ε. By
constructing the sequence {yk}, the following estimate holds for all k > N:
|yk− c| <
1
k
<
1
N
< ε.
Therefore, by the definition of the limit, the sequence {yk} converges to
the point c.
Corollary 7A/15:54 (08:57)
Theorem (corollary of the Bolzano–Weierstrass theorem).
Any sequence contains a subsequence that has a finite or an infinite limit.
Proof.
If the given sequence {xn} is bounded, then, by the Bolzano–Weierstrass
theorem, we can extract a convergent subsequence from it.
It remains to consider the case when the given sequence is not bounded.
We show that in this case a subsequence having the limit ∞ can be obtained
from the given sequence.
Let us write a condition meaning that the sequence {xn} is not bounded,
applying the negation operation to the definition of a bounded sequence:
∃M > 0 ∀ n ∈ N |xn| ≤ M ,
∀M > 0 ∃ n ∈ N |xn| > M .
We describe the process of constructing the required subsequence of {yk}.
Choose M = 1. Then there is an index n1such that |x
n
1
| > 1. Let
y1= x
n
1
.
Choose M = 2 and exclude from consideration all the elements of the
original sequence {xn} with indices less than or equal to n1. The rest of the
sequence is still unbounded, so there is an index n2> n1such that |x
n
2
| > 2.
Let y2= x
n
2
.
Continuing this process, we obtain a subsequence {yk} of the sequence
{xn}, for whose elements the estimate |yk| > k holds. It is easy to prove that
lim
k→∞yk
= ∞.
An example of an unbounded
sequence with no limit 7A/24:51 (04:55)
From the fact that the sequence is not bounded, it does not follow that it
has an infinite limit. For example, the following sequence is unbounded:
62 M. E. Abramyan. Lectures on differential calculus
n
n
(−1)
n
o=n
1, 2,
1
3
, 4,
1
5
, . . .o.
However, this sequence has no limit, since it has an infinite number of elements contained both in an arbitrary neighborhood of the point 0 and in an
arbitrary neighborhood of the point +∞. At the same time, we can extract
from it a subsequence converging to 0, as well as a subsequence approaching +∞.
Fundamental sequences. Cauchy criterion
for sequence convergence
Fundamental sequences: definition 7A/29:46 (04:25)
Definition.
The sequence {xn} is called the fundamental sequence, or the Cauchy
sequence, if
∀ε > 0 ∃N ∈ N ∀m > N, n > N |xm− xn| < ε. (1)
The Cauchy criterion
for sequence convergence 7A/34:11 (05:20), 7B/00:00 (26:41)
Theorem (Cauchy criterion for sequence convergence).
The sequence {xn} is convergent if and only if it is a fundamental sequence:
lim
n→∞
xn= A ∈ R⇔ ({xn} is the fundamental sequence).
Proof.
1. Let us prove the necessity.
Given: lim
n→∞xn
= A ∈ R.
Prove: {xn} is a fundamental sequence, that is, it satisfies (1).
We use the definition of the limit of a sequence in the language ε–N :
∀ε > 0 ∃N ∈ N ∀n > N |xn− A| <
ε
2
.
The same is true for an arbitrary number m > N:
∀m > N |xm− A| <
ε
2
.
Assuming m > N , n > N , we get:
|xm−xn| = |xm−A + A − xn| ≤ |xm−A| + |A − xn| <
ε
2
+
ε
2
= ε.
Thus, for {xn}, condition (1) is satisfied, therefore, {xn} is a fundamental
sequence. The necessity is proven.
7. Subsequences. Cauchy criterion 63
2. Now we prove sufficiency.
Given: {xn} is a fundamental sequence, that is, it satisfies (1).
Prove: the sequence {xn} is convergent.
First, we prove that the sequence {xn} is bounded.
If we put ε = 1 in condition (1), then we get:
∃N ∈ N ∀ m > N, n > N |xm− xn| < 1. (2)
Let us choose some m0> N. Then
|xn| = |xn− x
m
0
+ x
m
0
| ≤ |xn− x
m
0
| + |x
m
0
|.
If n > N, then the first term in the right-hand side of the resulting inequality is estimated from above by 1, by virtue of (2). The second term does
not depend on n. Thus, for all n > N we get the estimate:
|xn| < 1 + |x
m
0
| = M.
We have found that there exists a value M such that for all n > N the
estimate |xn| < M holds. This means that the set of elements of the sequence
{xn} with numbers greater than N is bounded.
But the set of remaining elements {x1, x2, . . . , xN} is finite and, therefore, also bounded. Therefore, the entire sequence {xn} is bounded (you
can compare this proof with the proof of the theorem on the boundedness of
a converging sequence).
We have proved that the sequence {xn} is bounded. Therefore, the
Bolzano–Weierstrass theorem holds for it. By virtue of this theorem, there
exists a convergent subsequence {yk} of the sequence {xn}:
lim
k→∞
yk= A, where yk= x
n
k
, 1 ≤ n1< n2< ··· < nk< . . .
Now we prove that the limit A of the subsequence {yk} is also the limit of
the given sequence {xn}. For this, we must show that
∀ε > 0 ∃N ∈ N ∀n > N |xn− A| < ε. (3)
Let us select some ε > 0. For this ε, due to the fundamentality of the
sequence {xn}, condition (1) is satisfied:
∃N1∈ N ∀m > N1, n > N1|xm− xn| <
ε
2
. (4)
In addition, we know that the subsequence {yk} has a limit A, so for the
same ε, by definition of the limit, we get:
∃N2∈ N ∀k > N2|yk− A| <
ε
2
.
The last inequality can be rewritten as follows:
64 M. E. Abramyan. Lectures on differential calculus
|x
n
k
− A| <
ε
2
. (5)
Put N = max{N1, N2} and show that condition (3) is fulfilled for all
n > N , that is, for all n > N the estimate |xn− A| < ε holds. We choose
k0= N + 1 and transform the expression |xn− A| as follows:
|xn− A| = |xn− x
n
k
0
+ x
n
k
0
− A| ≤ |xn− x
n
k
0
| + |x
n
k
0
− A|. (6)
Since k0= N + 1 > N ≥ N2, the last term in the resulting sum, by virtue
of (5), can be estimated as follows:
|x
n
k
0
− A| <
ε
2
. (7)
To estimate the first term |xn−x
n
k
0
|, we note that, by virtue of the lemma
on the indices of subsequence elements, the index n
k
0
can be estimated as
follows: n
k
0
≥ k0> N ≥ N1. Then for n > N ≥ N1we have n
k
0
> N1,
n > N1, therefore, by virtue of (4),
|xn− x
n
k
0
| <
ε
2
. (8)
Given estimates (7) and (8) in inequality (6), we finally obtain:
|xn− A| <
ε
2
+
ε
2
= ε.
Thus, condition (3) is satisfied and, therefore, lim
n→∞xn
= A. The suffi-
ciency is proved.
Examples of applying the Cauchy criterion3.
To prove that the sequence {xn} does not have a finite limit, it suffices to
show that {xn} is not a fundamental sequence, that is, that the negation of
the condition (1) is fulfilled:
∃ε > 0 ∀N ∈ N ∃m > N, n > N |xm− xn| ≥ ε. (9)
1. Let us show that the sequence {xn} = {(−1)n} is not a fundamental
sequence.
Let ε = 1. For any positive integer N, we choose m = 2N > N and
n = 2N + 1 > N. Then
|xm− xn| = |(−1)2N− (−1)
2N+1
| = |1 −(−1)| = 2 > 1 = ε.
Thus, condition (9) is satisfied, the sequence is not fundamental and, therefore, it does not have a finite limit. Recall that we have already established
this fact (see Chapter 2).
2. Let us show that the sequence {xn} =1 +
1
2
+ ··· +
1
n
is not a fun-
damental sequence.
3
In video lectures, there are no such examples.
7. Subsequences. Cauchy criterion 65
Let ε =
1
2
. For any positive integer N, we choose m = 4N > N and
n = 2N > N. Then
|xm− xn| =
1 +
1
2
+ ··· +
1
4N
−1 +
1
2
+ ··· +
1
2N
=
=
1
2N + 1
+
1
2N + 2
+ ··· +
1
4N
.
The last sum contains 2N terms, and each term can be estimated from
below by
1
4N
:
1
2N + k
≥
1
4N
, k = 1, 2, . . . , 2N .
Therefore, we finally get:
|xm− xn| =
1
2N + 1
+
1
2N + 2
+ ··· +
1
4N
≥ 2N ·
1
4N
=
1
2
= ε.
Thus, condition (9) is satisfied, the sequence is not fundamental and, there-
fore, it has no finite limit.
Since this sequence increases and has no finite limit, it is unbounded due
to the convergence criterion for monotone sequences. So, the sequence is
unbounded and increasing, therefore its limit is +∞.
8. The limit of a function
Definition and uniqueness of the limit of a function
Definitions of the limit of a function
and their equivalence 7B/26:41 (19:39)
By UA, VA, we will denote, as before, the neighborhood of the point A.
Recall the definition of a symmetric neighborhood of the point A of radius ε:
U
ε
A
def
= {x ∈ R : |x − A| < ε}.
Definition.
A punctured neighborhood◦Uaand a punctured symmetric neighborhood◦U
ε
a
of the point a are defined as follows:
◦
U
a
def
= Ua\ {a},
◦
U
ε
a
def
= {x ∈ R : 0 < |x − A| < ε}.
Definition 1 of the function limit (in the language of neigh-
borhoods).
Let f : E → R be a function defined on E ⊂ R, let a be the limit point of
the set E. We say that the function f has a limit, as x → a, equal to A ∈ R
and write it in the form lim
x→a
f(x) = A if for any neighborhood UAof the
point A there exists a punctured neighborhood◦Vaof the point a such that
for any x belonging to the intersection E ∩◦Va, f (x) belongs to UA:
∀UA∃◦Va∀x ∈ E ∩◦Vaf(x) ∈ UA.
Definition 2 of the function limit (in the language of sym-
metric neighborhoods).
The function f has a limit, as x → a, equal to A ∈ R if for any symmetric
neighborhood U
ε
A
of the point A there exists a punctured symmetric neigh-
borhood◦V
δ
a
of the point a such that for any x belonging to the intersection
E ∩◦V
δ
a
, f (x) belongs to U
ε
A
:
∀U
ε
A
∃◦V
δ
a
∀x ∈ E ∩◦V
δ
a
f(x) ∈ U
ε
A
.
8. The limit of a function 67
Definition 3 of the function limit (in the language ε–δ).
The function f has a limit, as x → a, equal to A ∈ R if for any number
ε > 0 there exists a number δ > 0 such that for any x from E satisfying the
condition 0 < |x − a| < δ, the estimate |(f(x) − A| < ε holds:
∀ε > 0 ∃δ > 0 ∀x ∈ E, 0 < |x − a| < δ, |f(x) − A| < ε.
Theorem (on the equivalence of three definitions of the
limit of a function).
Definitions 1, 2, and 3 of the limit of a function are equivalent.
Proof4.
The equivalence of definitions 2 and 3 follows from the definition of symmetric neighborhood.
Let us prove the equivalence of definitions 1 and 2. It is clear that if A
is a limit in the sense of definition 1 then A is also a limit in the sense of
definition 2 (since any symmetric neighborhood of a point is also its ordinary
neighborhood). Let us prove that if A is a limit in the sense of definition 2
then A is also a limit in the sense of definition 1.
Let UAbe an arbitrary neighborhood. Then there exists a symmetric
neighborhood U
ε
A
⊂ UA. For a symmetric neighborhood U
ε
A
, according to
definition 2, there exists a neighborhood◦V
δ
a
such that for all x ∈ E ∩◦V
δ
a
, f (x)
belongs to the neighborhood U
ε
A
. By the embedding U
ε
A
⊂ UA, this means
that f (x) also belongs to the neighborhood UA. Thus, for any neighborhood
of the point A, definition 1 holds. The equivalence of definitions 1 and 2 is
proved.
Since each of definitions 1 and 3 is equivalent to definition 2, definitions 1
and 3 are also equivalent.
Remark.
In definition of the limit of a function, a punctured neighborhood of the
point a is used, since the value of the function at this point should not affect
the value of the limit. It should also be noted that the function may not be
defined at the point a: the definition does not require that a belongs to E,
but it requires that a be the limit point of this set.
The uniqueness theorem
for the limit of a function 8A/00:00 (12:21)
Theorem (on the uniqueness of the limit of a function).
If a function has a limit at a point, then this limit is unique.
4
In video lectures, there is no proof of this theorem.
68 M. E. Abramyan. Lectures on differential calculus
Proof.
Let f : E → R be a function, let a be the limit point of the set E. Let
lim
x→a
f(x) = A and lim
x→a
f(x) = B. We should prove that A = B.
Let us prove this fact by contradiction: suppose that A 6= B.
We will use the definition of the limit in the language of neighborhoods.
Let us choose neighborhoods UAand UBso that they do not intersect:
UA∩ UB= ∅. This can always be done, since A 6= B.
By definition 1 of the limit of a function, we obtain the following relations
for these neighborhoods:
∃◦V
0
a
∀x ∈ E ∩◦V
0
a
f(x) ∈ UA, (1)
∃◦V
00
a
∀x ∈ E ∩◦V
00
a
f(x) ∈ UB. (2)
Consider the point x0that belongs to the intersection of three sets:
x0∈ E ∩◦V
0
a
∩◦V
00
a
. Such a point exists, since the intersection a◦V
0
a
∩◦V
00
a
of the punctured neighborhoods of the point is itself a punctured neighborhood of this point, and the intersection of any punctured neighborhood of the
point a and the set E is not empty, since a is the limit point of E.
Then for this point x0, due to (1) and (2), we have: f (x0) ∈ UAand
f(x0) ∈ UB. So f (x0) ∈ UA∩ UB. But UA∩ UB= ∅, therefore f(x0) ∈ ∅,
which is impossible. The resulting contradiction means that A = B.
Criterion for the existence of the limit of a function
in terms of sequences
Formulation of the criterion 8A/12:21 (03:49)
Theorem (criterion for the existence of the limit of a func-
tion in terms of sequences).
Let f : E → R be a function, let a be the limit point of the set E.
For the function f to have the limit lim
x→a
f(x) = A, it is necessary and
sufficient that for any sequence {xn} satisfying the conditions xn∈ E \ {a}
and lim
n→∞xn
= a, the limit of the sequence {f(xn)} exists and is equal
to A:
∀ {xn}, xn∈ E \ {a}, lim
n→∞
xn= a, ∃ lim
n→∞
f(xn) = A. (3)
Thus, the statement of the theorem can be written as follows:
lim
x→a
f(x) = A⇔condition (3) holds.
8. The limit of a function 69
Proof of necessity 8A/16:10 (11:56)
Given: lim
x→a
f(x) = A. Prove: condition (3) holds.
Let us choose an arbitrary sequence {xn} satisfying all the required con-
ditions:
xn∈ E \ {a}, lim
n→∞
xn= a.
We should show that the limit of the sequence {f (xn)} exists and is equal
to A. Let us write the definition of the limit of a function in the language
ε–δ:
∀ε > 0 ∃δ > 0 ∀x ∈ E, 0 < |x − a| < δ, |f(x) − A| < ε. (4)
Let us choose an arbitrary value ε > 0, find the corresponding δ > 0 for
it, and write down the definition of the limit of the sequence {xn} for this δ:
∃N ∈ N ∀ n > N 0 < |xn− a| < δ. (5)
In the obtained double estimate (5), the right-hand side (|xn− a| < δ) is
satisfied because the sequence {xn} has the limit a, and the left-hand side
(0 < |xn− a|) is satisfied due to the condition xn∈ E \ {a}.
Since for all n > N, by virtue of (5), the condition 0 < |xn−a| < δ holds,
we obtain, by virtue of (4), that the estimate |f (xn) −A| < ε also holds for
these n.
Thus, we have shown that for an arbitrarily chosen ε > 0 one can, using
the auxiliary value δ, find a number N such that for all n > N the estimate
|f(xn) −A| < ε holds:
∀ε > 0 ∃N ∈ N ∀n > N |f (xn) −A| < ε.
The last relation means that lim
n→∞
f(xn) = A.
Proof of sufficiency 8B/00:00 (13:41)
Given: condition (3) holds. Prove: lim
x→a
f(x) = A.
We will prove this fact by contradiction. Suppose that the number A is
not the limit of the function f (x) as x → a. Let us write down what this
means by applying the logical negation operation to the definition of the limit
in the language of symmetrical neighbors:
∀ε > 0 ∃δ > 0 ∀x ∈◦V
δ
a
∩ E f(x) ∈ U
ε
A
,
∃ε > 0 ∀δ > 0 ∃x ∈◦V
δ
a
∩ E f(x) /∈ U
ε
A
. (6)
70 M. E. Abramyan. Lectures on differential calculus
Let us construct the sequence {xn} choosing the values δ equal to
1,
1
2
,
1
3
, . . . ,
1
n
, . . . in the relation (6). For δ = 1, there exists a number
x1∈◦V
1
a
∩ E for which the condition f (x1) /∈ U
ε
A
is fulfilled; for δ =
1
2
,
there exists a number x2∈◦V
1/2
a
∩ E for which the condition f(x2) /∈ U
ε
A
is
fulfilled; . . . ; for δ =
1
n
, there exists a number xn∈◦V
1/n
a
∩ E for which the
condition f (xn) /∈ U
ε
A
is fulfilled, and so on.
Since for any n ∈ N for the elements of the constructed sequence {xn}, the
condition xn∈◦V
1/n
a
∩E holds, we obtain that the elements xnbelong to the
set E \{a} and lim
n→∞xn
= a. At the same time, the sequence {f(xn)} does
not have a limit equal to A, because there exists a symmetric neighborhood
U
ε
A
of the point A that does not contain any element of the sequence {f (xn)}.
We have constructed the sequence {xn} for which condition (3) is violated, however, this condition must be satisfied for any sequence with the
specified properties. This contradiction means that our assumption that the
number A is not the limit of the function f(x), as x → a, is false, and
lim
x→a
f(x) = A.
Remarks.
1. The proved criterion means that condition (3) can be considered as
another definition of the limit of a function. A feature of such a definition is
that it defines the limit of a function through the limits of auxiliary sequences.
2. This criterion makes it easy to prove the absence of the limit of the
function f at the point a. To prove this, it is enough to give two sequences
{x
0
n
} and {x
00
n
} with elements from E \ {a} that converge to a and such that
the sequences {f(x
0
n
)} and {f(x
00
n
)} do not converge to the same limit.
Examples of functions
with and without limits 8B/13:41 (20:11)
1. Consider the function f(x) = x sin
1
x
, E = R \ {0}.
We will show that lim
x→0
x sin
1
x
= 0 using the definition of the limit of a
function in the language ε–δ.
For any x 6= 0, the estimate
x sin
1
x
≤ |x| holds. Let us choose an
arbitrary value ε > 0 and set δ = ε. Then for x such that 0 < |x| < δ, we
obtain
x sin
1
x
≤ |x| < δ = ε.
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