Calculus 1 Topic 1
Calculus 1 Topic 1
Calculus 1 Topic 1
Bounds
Lemma. Let A be a set of real numbers. A real number M ¿(m ¿ ¿ is the supremum
¿ ¿
(infimum) of A if and only if a ≤ M (a ≥ m ) for all a ∈ A and for every positive number ε there is
¿ ¿
an element a ε ∈ A such that M −ε <a ε ( m + ε >a ε ¿.
Sequences
Definition1. A Sequence of real numbers is a function defined on the set N= { 1,2,3, … }
of natural numbers whose range is contained in the set R of real numbers. By a sequence of real
numbers we mean a function f : IN → R. We shall denote the value of f at n by
a n=f ( n ) , n ∈ N or { a n } or { a n }n=∞
n=1
.
1
Sequence Examples . 1. a n= ; 2. an =n ; 3.e n= 1+
n n ( )
1 n
;
It is clear that
If for arbitrary positive number ε > 0 ( however large) there is a positive number
N ( ε) such that the inequality |an|>ε holds for all n> N (ε ¿ , we say that the
sequence {a n } is infinitely large.
Similarly, the definition of the limit of the sequence is paraphrased for the
any case when this limit is infinite with a definite sign.
2n−1 n
an=(−(2 n−1 )) ; an =(−n ) .
Infinitesimals
1
lim α n =0 an =
{ α n } is an infinitesimal ⇔ n→∞ . Examples of infinitesimals are n ,
().
n
(−1)n 1
an = 2 a n =
1 1 a= 1
an = an = n
n , n , √n , n! , 2
Definition. 1.{xn}={yn} ⇔ x n= y n ∀ n ∈ N ;
5. λ {x n }={λxn } , ∀ n ∈ N .
Theorem. A sum, a difference and a product of two infinitesimals is again an
infinitesimal.
sin n
{ }
Example. Let us consider the sequence n which can be written as
sin n 1
=sin n⋅
n n
1
x =sin n α n=
It is clear that n is bounded since −1≤sin n≤1 , and n is an
sin n
α n x n=
infinitesimal. Therefore the product n is an infinitesimal.
lim α n =0 ⇒ lim x n =a
x n=a+α n , where n→∞ n→∞ .
Properties of limits
lim x n =C
1. If x n=C , C - constant ∀n then n→∞ . This means that the limit of a constant
equals the constant.
2. If a sequence has a limit then this limit is unique, i.e., there no other limits of
the sequences.
lim x n =a
Proof. Let {x n } be convergent and n→∞ . Take ε=1 . According to definition
there exist a number N such that the inequality |x n −a|<1 holds for all n>N . Let
d=max {1 ,|x 1 −a|, ... ,|x N −a|}
. Then the inequality |x n −a|≤d holds for all n .
−d≤x n −a≤d
or a−d≤x n ≤a+d
4. The limit of a sum of sequences is equal to the sum of the limits of these
sequences.
lim x n =a , lim y n =b
Proof. Suppose { x n }and { y n } are convergent and n→∞ n→∞
It follows that
x n + y n =a+ b+α n + β n or
lim γ n =0
x n + y n =a+ b+ γ n where γ n =α n + β n and n→∞ .
lim x n =a lim y n =b
Proof. Let {x n } and {y n } be convergent and n→∞ n →∞ .
It follows that
x n− y n =a−b+α n −β n lim γ n =0
or x n− y n =a−b+γ n , where γ n =α n− β n and n→∞ .
6. The limit of a product of sequences is equal to the product of the limits of these
sequences.
lim x n =a lim y n =b
Proof. Let { x n } and { y n } be convergent and n→∞ and n→∞
It follows that
where γ n =α n b+ β n a+α n β n
lim γ n =0
According to the properties of infinitesimals it is clear that n→∞ . Hence
lim Cx n=C⋅lim x n
n→∞ n→ ∞ , where C is constant.
8. The limit of a ratio of sequences is equal to the ratio of the limits of these
sequences.
lim x n
xn n→ ∞
lim =
yn lim y n lim y n ≠0
( y n ≠0 and n→∞
n→∞
n→ ∞ )
Proof. We have
a−ε < x n ≤ y n ≤z n < a+ε or a−ε < y n <a+ ε or |y n −a|<ε for all n> N .
Hence we get
lim y n =a
By definition, this means n→∞ and we get what we had to prove.
10. Theorem
{
p p−1
a0 n +a1 n +...+ ak a0
lim q q−1 =¿ for p=q, ¿ {∞ , for p>q,¿¿¿¿
n→∞ b0 n +b1 n +.. .+bs b0
Monotone sequences
1 1 1
x n= , x n =−n , x n = n , x n =
n 3 √n
Note. Not every sequence is monotone. For example, the sequences {(−1 ) } and
n
Theorem. An increasing and bounded above sequence has a finite limit, i.e.
Number e
( )
n
1
x n= 1+ , n=1,2 ,,. ..
Consider the sequence n
( )
x n= 1+
1 n
n
n 1 n( n−1) 1 n(n−1)(n−2) 1
=1+ ⋅ +
1! n 2!
⋅ 2+
n 3!
⋅ 3 +. . .+
n
n(n−1).. .(n−n+1) 1
n!
⋅ n=
n
1+1+
1
2!( ) ( )( )
1 1
1− +
n 3!
1 2
1− 1− +. . .+
n n
1
n!
1 2
1− 1− 1−
n n
n−1
n ( )( )( )
Replacing n by n+1 we obtain an analogous expression for x n+1 .
( )
n+1
1 n+1 1 (n+1 )⋅n 1 (n+1 )⋅n⋅(n−1) 1
x n+1 = 1+ =1+ ⋅ + ⋅ + ⋅ +.. .
n+1 1 ! n+1 2! (n+1) 3 !
2
(n+1)3
.. .+
(n+1)⋅n( n−1 ). ..(n+1−n ) 1
(n+1)!
⋅
(n+1)n+1
=1+1+
1
2!
1−
1
+
n+1 3 !
1
1−
1
n+1 (
1−
2
n+1
+. ..) ( )( )
.. .+
1
(n+1)!
1−
1
(
n+1
1−
2
)(
n+1
. .. 1−
n
n+1 ) ( )
Now, comparing these expressions for x n and x n+1 we conclude that the first two
terms in both sums coincide and that every subsequent term in x n is less than the
k k
1− <1− , k=1,2 , .. . , n−1
n n+1
1 1 1 11 1
2! 3! n! 2 2 2
11 1 1
22 2 2 ( 11 1
xn<1+1+ + +.. .+ <1+1+ + 2 +...+ n−1 <¿¿<1+1+ + 2 +.. .+ n−1 + n +.. .<1+ 1+ + 2 +...+ n +...
22 2 )
The sum of the progression in the parentheses being equal to 2 we obtain x n <3 .
( ) possesses a limit as
n
1
x n= 1+
Theorem. The sequence n n→ ∞ .
This limit plays a very important role in mathematics and is traditionally denoted
by e.
Definition. The number e is the limit
( )
n
1
lim 1+ =e
n→∞ n .
e=2,718281828459045 … . ≈ 2,72.
I 1 ⊇ I 2 ⊇ … ⊇ I n ⊇ I n 1 ⊇… .
Proof.
Theorem. If I n=[ a n , b n ] , n ∈ N , is a nested sequence of closed, bounded intervals such that the
lengths b n−a n of I nsatisfy
∃! ξ=¿ n=1 ¿ ∞ I n .
( 1n ) for n ∈ N , then this intervals is nested, but the intervals have no point
Example1. If I n= 0 ,
Example2. The sequence of intervals I n=( n , ∞ ) , n∈ N , is nested but has no common point.
TEST
3
lim √
3
n +2 n−1
1. n→∞ n+2
2
( √ n2+1+ n )
lim
2. n→∞ √3 n6 +1
1 1 1
1+ + +.. .+ n
2 4 2
lim
n→∞ 1 1 1
1+ + +. .. n
3. 3 9 3
5 4 C)2 D)1 1
A) 3 B) 3 E) 3
4.
lim
n→∞
( 1+2+3+. . .+n n
n−2
−
2 )
A) 2 B) -1 1 D)3 1
−
C) 2 E) 4
lim √
9 n2−5
5. n→∞ 6 n+11
1 1 1 1 1
A) 6 B) 5 C) 4 D) 3 E) 2
6.
lim
n→∞
( 1 n2
+
n n+3 )
A) ∞ B) 0 C)1 D)2 E)3
n−1
lim
7. n→∞ √ n +3
2
2
16 n −3 n+7
lim 2
8. n→∞ (2 n+1)
n
n−(−1 )
lim n
9. n→∞ n+(−1 )
lim √ n+1− √ n
10. n→∞