MATH F111- Mathematics I
Saranya G. Nair
Department of Mathematics
BITS Pilani
May 25, 2026
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Mathematics I
May 25, 2026
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Notations
N- Set of Natural numbers
Q- Set of rational numbers
R- Set of real numbers
∀ - For all
∃- There exists
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Mathematics I
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Intervals
Definition
A subset I of R is said to be an interval if
a, b ∈ I and a < x < b =⇒ x ∈ I .
Let a, b ∈ R and a < b.
• (a, b) := {x ∈ R : a < x < b} (open interval)
• [a, b] := {x ∈ R : a ≤ x ≤ b} (closed interval)
• [a, b) := {x ∈ R : a ≤ x < b} and (a, b] := {x ∈ R : a < x ≤ b} are
half-open (or half-closed) intervals.
• (a, ∞) := {x ∈ R : x > a} and (−∞, a) := {x ∈ R : x < a} are
infinite open intervals.
• [a, ∞) := {x ∈ R : x ≥ a} and (−∞, a] := {x ∈ R : x ≤ a} are
infinite closed intervals.
Let a ∈ R and ϵ > 0. Then (a − ϵ, a + ϵ) is called the ϵ-neighborhood of
a.
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Sequences
Definition
A sequence of real numbers (or a sequence in R) is a function
x : N → R.
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Sequences
Definition
A sequence of real numbers (or a sequence in R) is a function
x : N → R.
• If x : N → R is a sequence, we will usually denote the value of x(n)
by the symbol xn .
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Mathematics I
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Sequences
Definition
A sequence of real numbers (or a sequence in R) is a function
x : N → R.
• If x : N → R is a sequence, we will usually denote the value of x(n)
by the symbol xn .
• The values xn are also called the terms or the elements of the
sequence and xn (that is, the value of x at n) is called the n-th term
of the sequence.
We will denote this sequence by the notations
(xn ), or
(xn : n ∈ N).
In this course, we will consider only Real sequences.
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Example
• (n : n ∈ N) = (1, 2, 3, 4, . . .)
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Example
• (n : n ∈ N) = (1, 2, 3, 4, . . .)
• (1/n : n ∈ N) = (1, 1/2, 1/3, 1/4, . . .)
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Example
• (n : n ∈ N) = (1, 2, 3, 4, . . .)
• (1/n : n ∈ N) = (1, 1/2, 1/3, 1/4, . . .)
• (n2 : n ∈ N) = (1, 4, 9, 16, . . .)
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Example
• (n : n ∈ N) = (1, 2, 3, 4, . . .)
• (1/n : n ∈ N) = (1, 1/2, 1/3, 1/4, . . .)
• (n2 : n ∈ N) = (1, 4, 9, 16, . . .)
• If b ∈ R, the sequence (b, b, b, . . .), all of whose terms equal b, is
called the constant sequence b.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
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Example
• (n : n ∈ N) = (1, 2, 3, 4, . . .)
• (1/n : n ∈ N) = (1, 1/2, 1/3, 1/4, . . .)
• (n2 : n ∈ N) = (1, 4, 9, 16, . . .)
• If b ∈ R, the sequence (b, b, b, . . .), all of whose terms equal b, is
called the constant sequence b.
• (2n : n ∈ N) = (2, 4, 8, 16, . . .)
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Example
• (n : n ∈ N) = (1, 2, 3, 4, . . .)
• (1/n : n ∈ N) = (1, 1/2, 1/3, 1/4, . . .)
• (n2 : n ∈ N) = (1, 4, 9, 16, . . .)
• If b ∈ R, the sequence (b, b, b, . . .), all of whose terms equal b, is
called the constant sequence b.
• (2n : n ∈ N) = (2, 4, 8, 16, . . .)
• (−1)n : n ∈ N = (−1, 1, −1, 1, . . .)
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Example
• (n : n ∈ N) = (1, 2, 3, 4, . . .)
• (1/n : n ∈ N) = (1, 1/2, 1/3, 1/4, . . .)
• (n2 : n ∈ N) = (1, 4, 9, 16, . . .)
• If b ∈ R, the sequence (b, b, b, . . .), all of whose terms equal b, is
called the constant sequence b.
• (2n : n ∈ N) = (2, 4, 8, 16, . . .)
• (−1)n : n ∈ N = (−1, 1, −1, 1, . . .)
• x1 := 1, x2 := 1 and xn := xn−1 + xn−2 for n ≥ 3:
(1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .) This sequence is known as the
Fibonacci sequence.
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Bounded Sequences
A sequence (an ) of real numbers is said to be bounded above if there is a
real number α such that an ≤ α for every (∀) n ∈ N.
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Bounded Sequences
A sequence (an ) of real numbers is said to be bounded above if there is a
real number α such that an ≤ α for every (∀) n ∈ N. eg. (an ) = −n.
A sequence (an ) of real numbers is said to be bounded below if there is a
real number β such that β ≤ an for every n ∈ N.
Saranya G. Nair (BITS Pilani)
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Bounded Sequences
A sequence (an ) of real numbers is said to be bounded above if there is a
real number α such that an ≤ α for every (∀) n ∈ N. eg. (an ) = −n.
A sequence (an ) of real numbers is said to be bounded below if there is a
real number β such that β ≤ an for every n ∈ N. eg. (an ) = n2
A sequence (an ) of real numbers is said to be bounded if there are real
numbers α, β such that β ≤ an ≤ α for every n ∈ N.
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Bounded Sequences
A sequence (an ) of real numbers is said to be bounded above if there is a
real number α such that an ≤ α for every (∀) n ∈ N. eg. (an ) = −n.
A sequence (an ) of real numbers is said to be bounded below if there is a
real number β such that β ≤ an for every n ∈ N. eg. (an ) = n2
A sequence (an ) of real numbers is said to be bounded if there are real
numbers α, β such that β ≤ an ≤ α for every n ∈ N. eg. (an ) = n1 ,
(an ) = (−1)n
If a sequence is not bounded, it is said to be unbounded. eg.
(an ) = (−1)n n
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• (an ) = {1, 12 , 13 , · · · , n1 · · · }
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• (an ) = {1, 12 , 13 , · · · , n1 · · · } approaches 0 as n gets large.
• (an ) = {0, 12 , 23 , · · · , 1 − n1 · · · }
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• (an ) = {1, 12 , 13 , · · · , n1 · · · } approaches 0 as n gets large.
• (an ) = {0, 12 , 23 , · · · , 1 − n1 · · · } approaches 1 as n gets large.
√ √ √
√
• (an ) = { 1, 2, 3, · · · , n · · · }
Saranya G. Nair (BITS Pilani)
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• (an ) = {1, 12 , 13 , · · · , n1 · · · } approaches 0 as n gets large.
• (an ) = {0, 12 , 23 , · · · , 1 − n1 · · · } approaches 1 as n gets large.
√ √ √
√
• (an ) = { 1, 2, 3, · · · , n · · · } have terms that get larger than
any number as n increases.
Saranya G. Nair (BITS Pilani)
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• (an ) = {1, 12 , 13 , · · · , n1 · · · } approaches 0 as n gets large.
• (an ) = {0, 12 , 23 , · · · , 1 − n1 · · · } approaches 1 as n gets large.
√ √ √
√
• (an ) = { 1, 2, 3, · · · , n · · · } have terms that get larger than
any number as n increases.
• (an ) = {1, −1, 1, −1, · · · }
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• (an ) = {1, 12 , 13 , · · · , n1 · · · } approaches 0 as n gets large.
• (an ) = {0, 12 , 23 , · · · , 1 − n1 · · · } approaches 1 as n gets large.
√ √ √
√
• (an ) = { 1, 2, 3, · · · , n · · · } have terms that get larger than
any number as n increases.
• (an ) = {1, −1, 1, −1, · · · } bounce back and forth between 1 and −1,
never approaching to a single value.
Saranya G. Nair (BITS Pilani)
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May 25, 2026
7 / 39
• (an ) = {1, 12 , 13 , · · · , n1 · · · } approaches 0 as n gets large.
• (an ) = {0, 12 , 23 , · · · , 1 − n1 · · · } approaches 1 as n gets large.
√ √ √
√
• (an ) = { 1, 2, 3, · · · , n · · · } have terms that get larger than
any number as n increases.
• (an ) = {1, −1, 1, −1, · · · } bounce back and forth between 1 and −1,
never approaching to a single value.
Remark
Question: What do we mean by a sequence converges?
It says that if we go far enough out in the sequence, the difference
between an and the limit of the sequence becomes less than any
preselected number ϵ > 0.
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Let us see this with (an ) = n1 .
• Can you find an integer N such that |an − 0| < 12 , ∀n ≥ N?
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Let us see this with (an ) = n1 .
• Can you find an integer N such that |an − 0| < 12 , ∀n ≥ N? Yes,
choose N = 3.
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Let us see this with (an ) = n1 .
• Can you find an integer N such that |an − 0| < 12 , ∀n ≥ N? Yes,
choose N = 3.
1
• Can you find an integer N such that |an − 0| < 10000
, ∀n ≥ N?
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Let us see this with (an ) = n1 .
• Can you find an integer N such that |an − 0| < 12 , ∀n ≥ N? Yes,
choose N = 3.
1
• Can you find an integer N such that |an − 0| < 10000
, ∀n ≥ N? Yes,
choose N = 10001.
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Let us see this with (an ) = n1 .
• Can you find an integer N such that |an − 0| < 12 , ∀n ≥ N? Yes,
choose N = 3.
1
• Can you find an integer N such that |an − 0| < 10000
, ∀n ≥ N? Yes,
choose N = 10001.
• For any preselected positive number, say ϵ > 0, can you find an
integer N such that |an − 0| < ϵ, ∀n ≥ N?
Saranya G. Nair (BITS Pilani)
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Let us see this with (an ) = n1 .
• Can you find an integer N such that |an − 0| < 12 , ∀n ≥ N? Yes,
choose N = 3.
1
• Can you find an integer N such that |an − 0| < 10000
, ∀n ≥ N? Yes,
choose N = 10001.
• For any preselected positive number, say ϵ > 0, can you find an
integer N such that |an − 0| < ϵ, ∀n ≥ N? Yes, choose N > 1ϵ . In
particular choose N = ⌊ 1ϵ ⌋ + 1.
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The Limit of a Sequence
Definition
A sequence (an ) in R is said to converge to ℓ ∈ R, or ℓ is said to be a limit
of (an ),
if for every ϵ > 0, there exists an integer N ∈ N such that
|an − ℓ| < ϵ, for all n ≥ N.
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The Limit of a Sequence
Definition
A sequence (an ) in R is said to converge to ℓ ∈ R, or ℓ is said to be a limit
of (an ),
if for every ϵ > 0, there exists an integer N ∈ N such that
|an − ℓ| < ϵ, for all n ≥ N.
ie,
an ∈ (ℓ − ϵ, ℓ + ϵ), ∀n ≥ N.
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The Limit of a Sequence
Definition
A sequence (an ) in R is said to converge to ℓ ∈ R, or ℓ is said to be a limit
of (an ),
if for every ϵ > 0, there exists an integer N ∈ N such that
|an − ℓ| < ϵ, for all n ≥ N.
ie,
an ∈ (ℓ − ϵ, ℓ + ϵ), ∀n ≥ N.
Remark
The choice of N depends on the value of ϵ.
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When a sequence (an ) has limit ℓ, we will use the notation
lim an = ℓ.
We will sometimes use the symbolism an → ℓ, which indicates the intuitive
idea that the values an “approach” the number ℓ as n → ∞.
• If a sequence has a limit, we say that the sequence is convergent
• If a sequence has no limit, we say that the sequence is divergent.
There is also a notion of divergence if the sequence is not bounded.
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Definition
The sequence (an ) diverges to infinity or
lim an = ∞
n→∞
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Definition
The sequence (an ) diverges to infinity or
lim an = ∞
n→∞
if for every number M there is an integer N such that ∀n > N, an > M.
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Definition
The sequence (an ) diverges to infinity or
lim an = ∞
n→∞
if for every number M there is an integer N such that ∀n > N, an > M.
Similarly we say (an ) diverges to negative infinity or
lim an = −∞
n→∞
if for every number m, there is an integer N such that ∀n > N, we have
an < m.
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The convergence of a sequence is unaltered if a finite number of its terms
are replaced by some other terms.
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The convergence of a sequence is unaltered if a finite number of its terms
are replaced by some other terms.
Examples:
(i) Let a ∈ R and an := a for all n ∈ N. Then an → a. We can let
N := 1 for any choice of ϵ.
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The convergence of a sequence is unaltered if a finite number of its terms
are replaced by some other terms.
Examples:
(i) Let a ∈ R and an := a for all n ∈ N. Then an → a. We can let
N := 1 for any choice of ϵ.
(ii) an := 1/n for all n ∈ N. Then an → 0.
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(iii) an := 2/(n2 + 1)
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(iii) an := 2/(n2 + 1)
2
n2 + 1
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−0 =
2
n2 + 1
<
Mathematics I
2
n2
for all n ∈ N.
May 25, 2026
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(iii) an := 2/(n2 + 1)
2
n2 + 1
−0 =
2
n2 + 1
Choose N ∈ N such that
<
r
N>
Then
|an − 0| <
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2
n2
for all n ∈ N.
2
.
ϵ
2
< ϵ, ∀n ≥ N.
n2
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(iii) an := 2/(n2 + 1)
2
n2 + 1
−0 =
2
n2 + 1
Choose N ∈ N such that
<
r
N>
Then
|an − 0| <
2
n2
for all n ∈ N.
2
.
ϵ
2
< ϵ, ∀n ≥ N.
n2
Then an → 0.
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(iii) an := 2/(n2 + 1)
2
n2 + 1
−0 =
2
n2 + 1
Choose N ∈ N such that
<
r
N>
Then
|an − 0| <
2
n2
for all n ∈ N.
2
.
ϵ
2
< ϵ, ∀n ≥ N.
n2
Then an → 0.
(iv) an := 5/(3n + 1)
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(iii) an := 2/(n2 + 1)
2
n2 + 1
−0 =
2
n2 + 1
Choose N ∈ N such that
<
r
N>
Then
|an − 0| <
2
n2
for all n ∈ N.
2
.
ϵ
2
< ϵ, ∀n ≥ N.
n2
Then an → 0.
(iv) an := 5/(3n + 1)
5
5
<
3n + 1
3n
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for all n ∈ N.
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(iii) an := 2/(n2 + 1)
2
n2 + 1
−0 =
2
n2 + 1
Choose N ∈ N such that
<
r
N>
Then
|an − 0| <
2
n2
for all n ∈ N.
2
.
ϵ
2
< ϵ, ∀n ≥ N.
n2
Then an → 0.
(iv) an := 5/(3n + 1)
5
5
<
for all n ∈ N.
3n + 1
3n
5
Choose N ∈ N such that N > 5/3ϵ. Then |an − 0| < 3n
< ϵ for all
n ≥ N. Then an → 0.
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Examples
(v). lim r n = 0 for |r | < 1.
n→∞
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Examples
(v). lim r n = 0 for |r | < 1.
n→∞
Case 1. r = 0
In this case the sequence is {0, 0, 0, . . .} which converges to 0.
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Examples
(v). lim r n = 0 for |r | < 1.
n→∞
Case 1. r = 0
In this case the sequence is {0, 0, 0, . . .} which converges to 0.
Case 2. r ̸= 0 and |r | < 1.
Since |r | < 1, |r1| > 1. Let |r1| = 1 + a where a > 0.
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Examples
(v). lim r n = 0 for |r | < 1.
n→∞
Case 1. r = 0
In this case the sequence is {0, 0, 0, . . .} which converges to 0.
Case 2. r ̸= 0 and |r | < 1.
Since |r | < 1, |r1| > 1. Let |r1| = 1 + a where a > 0. Then
|r n − 0| = |r |n =
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1
.
(1 + a)n
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Examples
(v). lim r n = 0 for |r | < 1.
n→∞
Case 1. r = 0
In this case the sequence is {0, 0, 0, . . .} which converges to 0.
Case 2. r ̸= 0 and |r | < 1.
Since |r | < 1, |r1| > 1. Let |r1| = 1 + a where a > 0. Then
|r n − 0| = |r |n =
1
.
(1 + a)n
We have (1 + a)n > na for all n ∈ N and hence,
|r n − 0| <
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1
for all n ∈ N.
na
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Let ϵ > 0 be given. Then
|r n − 0| < ϵ holds if n >
1
.
aϵ
1
Choose any N ∈ N such that N > aϵ
. Then
∀n ≥ N, |r n − 0| < ϵ.
Since ϵ > 0 is arbitrary,
lim r n = 0.
n→∞
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Properties of limits
Theorem
Uniqueness of Limits. A sequence in R can have at most one limit.
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Properties of limits
Theorem
Uniqueness of Limits. A sequence in R can have at most one limit.
Proof. Let (an ) be a real sequence and suppose that ℓ1 and ℓ2 are both
limits for (an ) and let ℓ1 ̸= ℓ2 .
• Let ϵ := |ℓ1 − ℓ2 |/2. Since ℓ1 ̸= ℓ2 , ϵ > 0.
• Since ℓ1 is a limit of the sequence, for the chosen ϵ, ∃ N1 ∈ N such
that
|an − ℓ1 | < ϵ, for all n ≥ N1 .
• Since ℓ2 is a limit of the sequence, for the chosen ϵ, ∃ N2 ∈ N such
that
|an − ℓ2 | < ϵ, for all n ≥ N2 .
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Let N = max{N1 , N2 }. Then:
|an − ℓ1 | < ϵ and |an − ℓ2 | < ϵ for all n ≥ N,
and hence,
|ℓ1 − ℓ2 | = |(aN − ℓ2 ) − (aN − ℓ1 )| ≤ |aN − ℓ1 | + |aN − ℓ2 | < ϵ + ϵ = |ℓ1 − ℓ2 |
which is a contradiction. Hence, ℓ1 = ℓ2 .
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■
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Theorem
A convergent sequence is bounded.
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Theorem
A convergent sequence is bounded.
Suppose an → ℓ. Let ϵ := 1. There is N ∈ N such that
|an − ℓ| < 1 for all n ≥ N.
Hence
|an | ≤ |an − ℓ| + |ℓ| < 1 + |ℓ| for all n ≥ N.
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Theorem
A convergent sequence is bounded.
Suppose an → ℓ. Let ϵ := 1. There is N ∈ N such that
|an − ℓ| < 1 for all n ≥ N.
Hence
|an | ≤ |an − ℓ| + |ℓ| < 1 + |ℓ| for all n ≥ N.
• Thus it remains to find a bound for a1 , a2 · · · , aN−1 . Choose
β = max{|a1 |, |a2 |, · · · , |aN−1 |}. Then |an | ≤ β, for all
1 ≤ n ≤ N − 1.
• Define α := max |a1 |, . . . , |aN−1 |, |ℓ| + 1 . Then
|an | ≤ α for all n ∈ N. Hence (an ) is bounded.
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If an is convergent, then an is bounded. Equivalently, if an is not bounded,
then an is not convergent. This result can be used to show if a sequence is
not bounded.
• The sequence {(−1)n n : n ∈ N} divergent since it is not bounded.
• A bounded sequence need not be convergent. For example, the
sequence {(−1)n : n ∈ N} is bounded but not convergent.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
19 / 39
Theorem
Limit theorems. Let {an } and {bn } be two convergent sequences that
converge to A and B respectively. Then:
• lim(an ± bn ) = A ± B.
• lim(an bn ) = AB.
In particular, lim(can ) = cA for c ∈ R.
• lim ban = BA , provided (bn ) is a sequence of non-zero real numbers and
n
B ̸= 0.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
20 / 39
Theorem
Limit theorems. Let {an } and {bn } be two convergent sequences that
converge to A and B respectively. Then:
• lim(an ± bn ) = A ± B.
• lim(an bn ) = AB.
In particular, lim(can ) = cA for c ∈ R.
• lim ban = BA , provided (bn ) is a sequence of non-zero real numbers and
n
B ̸= 0.
1
• limn→∞ −1
n = − limn→∞ n = 0.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
20 / 39
Theorem
Limit theorems. Let {an } and {bn } be two convergent sequences that
converge to A and B respectively. Then:
• lim(an ± bn ) = A ± B.
• lim(an bn ) = AB.
In particular, lim(can ) = cA for c ∈ R.
• lim ban = BA , provided (bn ) is a sequence of non-zero real numbers and
n
B ̸= 0.
1
• limn→∞ −1
n = − limn→∞ n = 0.
1
• limn→∞ n+1
n = limn→∞ 1 + n = 1.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
20 / 39
Theorem
Limit theorems. Let {an } and {bn } be two convergent sequences that
converge to A and B respectively. Then:
• lim(an ± bn ) = A ± B.
• lim(an bn ) = AB.
In particular, lim(can ) = cA for c ∈ R.
• lim ban = BA , provided (bn ) is a sequence of non-zero real numbers and
n
B ̸= 0.
1
• limn→∞ −1
n = − limn→∞ n = 0.
1
• limn→∞ n+1
n = limn→∞ 1 + n = 1.
6
• limn→∞ 4−7n
= limn→∞
n6 +3
Saranya G. Nair (BITS Pilani)
4
−7
n6
1+ 36
n
= 0−7
1+0 = −7.
Mathematics I
May 25, 2026
20 / 39
Theorem
Let (xn ) be a convergent sequence of real numbers and there exists a
positive integer m such that xn ≥ 0 for all n ≥ m. Then lim xn ≥ 0.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
21 / 39
Theorem
Let (xn ) be a convergent sequence of real numbers and there exists a
positive integer m such that xn ≥ 0 for all n ≥ m. Then lim xn ≥ 0.
Corollary
If (xn ) and (yn ) are convergent sequences of real numbers and if there is a
positive integer m such that xn ≤ yn for all n ≥ m, then lim xn ≤ lim yn .
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
21 / 39
Theorem
Let (xn ) be a convergent sequence of real numbers and there exists a
positive integer m such that xn ≥ 0 for all n ≥ m. Then lim xn ≥ 0.
Corollary
If (xn ) and (yn ) are convergent sequences of real numbers and if there is a
positive integer m such that xn ≤ yn for all n ≥ m, then lim xn ≤ lim yn .
Proof. Let zn := yn − xn .
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
21 / 39
Theorem
Let (xn ) be a convergent sequence of real numbers and there exists a
positive integer m such that xn ≥ 0 for all n ≥ m. Then lim xn ≥ 0.
Corollary
If (xn ) and (yn ) are convergent sequences of real numbers and if there is a
positive integer m such that xn ≤ yn for all n ≥ m, then lim xn ≤ lim yn .
Proof. Let zn := yn − xn . Then (zn ) is convergent sequence of real
numbers such that zn ≥ 0 for all n ≥ m. It then follows from the
preceding theorem that
lim zn = lim(yn − xn ) = lim yn − lim xn ≥ 0.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
21 / 39
Theorem
Sandwich Theorem. Let (an ), (bn ), (cn ) be three sequences of real
numbers and there is a natural number m such that
an ≤ bn ≤ cn for all n ≥ m.
If lim an = lim cn = ℓ, then (bn ) is convergent and lim bn = ℓ.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
22 / 39
Theorem
Sandwich Theorem. Let (an ), (bn ), (cn ) be three sequences of real
numbers and there is a natural number m such that
an ≤ bn ≤ cn for all n ≥ m.
If lim an = lim cn = ℓ, then (bn ) is convergent and lim bn = ℓ.
Examples:
(i) lim
n→∞
cos n
n
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
22 / 39
Theorem
Sandwich Theorem. Let (an ), (bn ), (cn ) be three sequences of real
numbers and there is a natural number m such that
an ≤ bn ≤ cn for all n ≥ m.
If lim an = lim cn = ℓ, then (bn ) is convergent and lim bn = ℓ.
Examples:
cos n
n
cos n
= 0.
−1 ≤ cos n ≤ 1. Therefore − n1 ≤ cosn n ≤ n1 and lim
n→∞ n
(i) lim
n→∞
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
22 / 39
Theorem
Sandwich Theorem. Let (an ), (bn ), (cn ) be three sequences of real
numbers and there is a natural number m such that
an ≤ bn ≤ cn for all n ≥ m.
If lim an = lim cn = ℓ, then (bn ) is convergent and lim bn = ℓ.
Examples:
cos n
n
cos n
= 0.
−1 ≤ cos n ≤ 1. Therefore − n1 ≤ cosn n ≤ n1 and lim
n→∞ n
1
(ii) lim n = 0
n→∞ 2
as 0 ≤ 21n ≤ n1
(i) lim
n→∞
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
22 / 39
Theorem
Sandwich Theorem. Let (an ), (bn ), (cn ) be three sequences of real
numbers and there is a natural number m such that
an ≤ bn ≤ cn for all n ≥ m.
If lim an = lim cn = ℓ, then (bn ) is convergent and lim bn = ℓ.
Examples:
cos n
n
cos n
= 0.
−1 ≤ cos n ≤ 1. Therefore − n1 ≤ cosn n ≤ n1 and lim
n→∞ n
1
(ii) lim n = 0
n→∞ 2
as 0 ≤ 21n ≤ n1
1
(iii) lim (−1)n = 0.
n→∞
n
(i) lim
n→∞
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
22 / 39
Examples (continued)
(iv) Let an :=
n3 + 3n2 + 1
n4 + 8n2 + 2
(iii) Let an :=
1
1
sin
for n ∈ N. Then an → 0,
n
n
Saranya G. Nair (BITS Pilani)
for n ∈ N. Then an → 0,
Mathematics I
May 25, 2026
23 / 39
Examples (continued)
(iv) Let an :=
n3 + 3n2 + 1
n4 + 8n2 + 2
since 0 ≤ an ≤
(iii) Let an :=
for n ∈ N. Then an → 0,
n3 + 3n2 + 1
3
1
1
≤ + 2 + 4 → 0.
n4
n n
n
1
1
sin
for n ∈ N. Then an → 0,
n
n
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
23 / 39
Examples (continued)
(iv) Let an :=
n3 + 3n2 + 1
n4 + 8n2 + 2
since 0 ≤ an ≤
(iii) Let an :=
for n ∈ N. Then an → 0,
n3 + 3n2 + 1
3
1
1
≤ + 2 + 4 → 0.
n4
n n
n
1
1
sin
for n ∈ N. Then an → 0,
n
n
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
23 / 39
Examples (continued)
(iv) Let an :=
n3 + 3n2 + 1
n4 + 8n2 + 2
since 0 ≤ an ≤
(iii) Let an :=
for n ∈ N. Then an → 0,
n3 + 3n2 + 1
3
1
1
≤ + 2 + 4 → 0.
n4
n n
n
1
1
sin
for n ∈ N. Then an → 0,
n
n
since |an | ≤
1
→ 0.
n
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
23 / 39
Understanding limit of a function
Definition
Let f (x) be defined on an open interval about x0 , except possibly at x0
itself. We say that the limit of f (x) as x approaches x0 is the number L,
and write
lim f (x) = L
x→x0
if,
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
24 / 39
Understanding limit of a function
Definition
Let f (x) be defined on an open interval about x0 , except possibly at x0
itself. We say that the limit of f (x) as x approaches x0 is the number L,
and write
lim f (x) = L
x→x0
if, for every number ϵ > 0, there exists a corresponding number δ > 0 such
that for all x,
0 < |x − x0 | < δ =⇒ |f (x) − L| < ϵ.
Refer Chapter 2 in Thomas Calculus.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
24 / 39
Definition: Continuity
Definition
Let f : D → R be a function where D ⊆ R. For x0 , we say that the
function is continuous at x0 if the following conditions hold:
1
x0 ∈ D.
2
limx→x0 f (x) exists.
3
limx→x0 f (x) = f (x0 ).
A function is continuous if it is continuous at all points of it’s domain.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
25 / 39
Continuous function theorem for Sequences
Remark
• If (an ) is a sequence and if f is any function from R → R, is f (an ) a
sequence?
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
26 / 39
Continuous function theorem for Sequences
Remark
• If (an ) is a sequence and if f is any function from R → R, is f (an ) a
sequence? Yes
• What can we say about convergence of f (an ) if we know about
convergence of an ?
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
26 / 39
Continuous function theorem for Sequences
Remark
• If (an ) is a sequence and if f is any function from R → R, is f (an ) a
sequence? Yes
• What can we say about convergence of f (an ) if we know about
convergence of an ?
Theorem
Theorem 3: Let (an ) be a sequence of real numbers. If
• an → ℓ and
• if f is a function that is continuous at ℓ and defined at all an , then
f (an ) → f (ℓ).
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
26 / 39
• Show that
q
(n+1)
→ 1.
n
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
27 / 39
√
(n+1)
→ 1. We know that n+1
→ 1 and f (x) = x is
n
n
q
continuous at ℓ = 1. So by Theorem 3, ( (n+1)
n ) → 1.
• Show that
q
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
27 / 39
√
(n+1)
→ 1. We know that n+1
→ 1 and f (x) = x is
n
n
q
continuous at ℓ = 1. So by Theorem 3, ( (n+1)
n ) → 1.
• Show that
q
1
• Show that (2 n ) → 1.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
27 / 39
√
(n+1)
→ 1. We know that n+1
→ 1 and f (x) = x is
n
n
q
continuous at ℓ = 1. So by Theorem 3, ( (n+1)
n ) → 1.
• Show that
q
1
• Show that (2 n ) → 1. n1 → 0 and f (x) = 2x is continuous at x = 0.
1
Thus 2 n → 1.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
27 / 39
• Given a real number x0 , can you construct sequences that converge to
x0 ?
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
28 / 39
• Given a real number x0 , can you construct sequences that converge to
x0 ?
• Can you construct rational sequences that converge to x0 ? How
about irrational sequences converging to x0 ?
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
28 / 39
• Given a real number x0 , can you construct sequences that converge to
x0 ?
• Can you construct rational sequences that converge to x0 ? How
about irrational sequences converging to x0 ?
There are infinite sequences that converge to a point x0 . How these
sequences can be used to check continuity at the point x0 ?
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
28 / 39
• Given a real number x0 , can you construct sequences that converge to
x0 ?
• Can you construct rational sequences that converge to x0 ? How
about irrational sequences converging to x0 ?
There are infinite sequences that converge to a point x0 . How these
sequences can be used to check continuity at the point x0 ?
Theorem
Sequential criteria for continuity A function f : D → R is continuous at
x0 ∈ D iff for every sequence (xn ) in D such that xn → x0 , we have
f (xn ) → f (x0 ).
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
28 / 39
• Given a real number x0 , can you construct sequences that converge to
x0 ?
• Can you construct rational sequences that converge to x0 ? How
about irrational sequences converging to x0 ?
There are infinite sequences that converge to a point x0 . How these
sequences can be used to check continuity at the point x0 ?
Theorem
Sequential criteria for continuity A function f : D → R is continuous at
x0 ∈ D iff for every sequence (xn ) in D such that xn → x0 , we have
f (xn ) → f (x0 ).
This theorem is particularly useful if you want to show that a function is
not continuous at x0 . We only need to construct two sequences xn and yn
both converging to x0 , but f (xn ) ̸= f (yn ).
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
28 / 39
Let
(
1
f (x) =
0
if x ∈ Q
if x ∈ R \ Q.
Show that f is discontinuous at every real number x ∈ R.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
29 / 39
Let
(
1
f (x) =
0
if x ∈ Q
if x ∈ R \ Q.
Show that f is discontinuous at every real number x ∈ R.
• Let x0 be any real number. Let xn be a rational sequence converging
to x0 and yn be an irrational sequence converging to x0 .
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
29 / 39
Let
(
1
f (x) =
0
if x ∈ Q
if x ∈ R \ Q.
Show that f is discontinuous at every real number x ∈ R.
• Let x0 be any real number. Let xn be a rational sequence converging
to x0 and yn be an irrational sequence converging to x0 .
• Thus f (xn ) = 1 for every n and f (yn ) = 0 for every n.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
29 / 39
Let
(
1
f (x) =
0
if x ∈ Q
if x ∈ R \ Q.
Show that f is discontinuous at every real number x ∈ R.
• Let x0 be any real number. Let xn be a rational sequence converging
to x0 and yn be an irrational sequence converging to x0 .
• Thus f (xn ) = 1 for every n and f (yn ) = 0 for every n.
• If x0 ∈ Q, then f (x0 ) = 1. Thus yn → x0 , but f (yn ) = 0 ↛ f (x0 ) = 1.
• If x0 ∈ R \ Q, then f (x0 ) = 0. Thus xn → x0 , but
f (xn ) = 1 ↛ f (x0 ) = 0.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
29 / 39
Let
(
1
f (x) =
0
if x ∈ Q
if x ∈ R \ Q.
Show that f is discontinuous at every real number x ∈ R.
• Let x0 be any real number. Let xn be a rational sequence converging
to x0 and yn be an irrational sequence converging to x0 .
• Thus f (xn ) = 1 for every n and f (yn ) = 0 for every n.
• If x0 ∈ Q, then f (x0 ) = 1. Thus yn → x0 , but f (yn ) = 0 ↛ f (x0 ) = 1.
• If x0 ∈ R \ Q, then f (x0 ) = 0. Thus xn → x0 , but
f (xn ) = 1 ↛ f (x0 ) = 0.
Therefore by sequential criteria for continuity f is not continuous at x0 .
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
29 / 39
Let
(
1
f (x) =
0
if x ∈ Q
if x ∈ R \ Q.
Show that f is discontinuous at every real number x ∈ R.
• Let x0 be any real number. Let xn be a rational sequence converging
to x0 and yn be an irrational sequence converging to x0 .
• Thus f (xn ) = 1 for every n and f (yn ) = 0 for every n.
• If x0 ∈ Q, then f (x0 ) = 1. Thus yn → x0 , but f (yn ) = 0 ↛ f (x0 ) = 1.
• If x0 ∈ R \ Q, then f (x0 ) = 0. Thus xn → x0 , but
f (xn ) = 1 ↛ f (x0 ) = 0.
Therefore by sequential criteria for continuity f is not continuous at x0 .
Since x0 is arbitrary, f (x) is discontinuous at every real number.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
29 / 39
Subsequences
Definition
A subsequence of a sequence is a sequence that can be derived from the
given sequence by deleting some elements without changing the order of
the remaining elements.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
30 / 39
Subsequences
Definition
A subsequence of a sequence is a sequence that can be derived from the
given sequence by deleting some elements without changing the order of
the remaining elements.
Let (an ) = {a1 , a2 , a3 , . . .}. Then
{a1 , a5 , a6 , a13 , . . .}
{a1 , a3 , a5 , a7 , . . .}
{a1001 , a100001 , a200001 . . .} are subsequences
{a5 , a4 , a6 , a7 , . . .} is not a subsequence.
Remark
Why are we interested in subsequences?
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
30 / 39
Theorem
A sequence converges to a limit ℓ if and only if every subsequence also
converges to the same limit ℓ.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
31 / 39
Theorem
A sequence converges to a limit ℓ if and only if every subsequence also
converges to the same limit ℓ.
{1, −1, 1, −1, · · · } doesn’t converge as it has a subsequence {1, 1, 1, · · · }
that converges to 1 and another subsequence {−1, −1, −1, · · · } that
converges to −1. Since the limits of both subsequences are different we
can conclude that original sequence doesn’t converge.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
31 / 39
Bounded Sequences
A sequence (an ) of real numbers is said to be bounded above if there is a
real number α such that an ≤ α for every (∀) n ∈ N.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
32 / 39
Bounded Sequences
A sequence (an ) of real numbers is said to be bounded above if there is a
real number α such that an ≤ α for every (∀) n ∈ N. The number α is an
upper bound for (an ). If α is an upper bound for an but no number less
than α is an upper bound for an , then α is the least upper bound for
an .
A sequence (an ) of real numbers is said to be bounded below if there is a
real number β such that β ≤ an for every n ∈ N.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
32 / 39
Bounded Sequences
A sequence (an ) of real numbers is said to be bounded above if there is a
real number α such that an ≤ α for every (∀) n ∈ N. The number α is an
upper bound for (an ). If α is an upper bound for an but no number less
than α is an upper bound for an , then α is the least upper bound for
an .
A sequence (an ) of real numbers is said to be bounded below if there is a
real number β such that β ≤ an for every n ∈ N. The number β is a lower
bound for an . If β is a lower bound for an but no number greater than β is
a lower bound for an , then β is the greatest lower bound for an .
A sequence (an ) of real numbers is said to be bounded if there are real
numbers α, β such that β ≤ an ≤ α for every n ∈ N.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
32 / 39
Bounded Sequences
A sequence (an ) of real numbers is said to be bounded above if there is a
real number α such that an ≤ α for every (∀) n ∈ N. The number α is an
upper bound for (an ). If α is an upper bound for an but no number less
than α is an upper bound for an , then α is the least upper bound for
an .
A sequence (an ) of real numbers is said to be bounded below if there is a
real number β such that β ≤ an for every n ∈ N. The number β is a lower
bound for an . If β is a lower bound for an but no number greater than β is
a lower bound for an , then β is the greatest lower bound for an .
A sequence (an ) of real numbers is said to be bounded if there are real
numbers α, β such that β ≤ an ≤ α for every n ∈ N.
If a sequence is not bounded, it is said to be unbounded.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
32 / 39
Monotone sequences and convergence
• A sequence (xn ) is said to be monotone increasing or nondecreasing
if xn ≤ xn+1 for all n ∈ N, that is, x1 ≤ x2 ≤ x3 ≤ · · · .
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
33 / 39
Monotone sequences and convergence
• A sequence (xn ) is said to be monotone increasing or nondecreasing
if xn ≤ xn+1 for all n ∈ N, that is, x1 ≤ x2 ≤ x3 ≤ · · · .
n
• 12 , 23 , . . . , n+1
, . . . is monotone increasing
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
33 / 39
Monotone sequences and convergence
• A sequence (xn ) is said to be monotone increasing or nondecreasing
if xn ≤ xn+1 for all n ∈ N, that is, x1 ≤ x2 ≤ x3 ≤ · · · .
n
• 12 , 23 , . . . , n+1
, . . . is monotone increasing
• A sequence (xn ) is said to be monotone decreasing or nonincreasing
if xn ≥ xn+1 for all n ∈ N, that is, x1 ≥ x2 ≥ x3 ≥ · · · .
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
33 / 39
Monotone sequences and convergence
• A sequence (xn ) is said to be monotone increasing or nondecreasing
if xn ≤ xn+1 for all n ∈ N, that is, x1 ≤ x2 ≤ x3 ≤ · · · .
n
• 12 , 23 , . . . , n+1
, . . . is monotone increasing
• A sequence (xn ) is said to be monotone decreasing or nonincreasing
if xn ≥ xn+1 for all n ∈ N, that is, x1 ≥ x2 ≥ x3 ≥ · · · .
• 1, 12 , 14 , . . . , 21n , . . . is monotone decreasing.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
33 / 39
Monotone sequences and convergence
• A sequence (xn ) is said to be monotone increasing or nondecreasing
if xn ≤ xn+1 for all n ∈ N, that is, x1 ≤ x2 ≤ x3 ≤ · · · .
n
• 12 , 23 , . . . , n+1
, . . . is monotone increasing
• A sequence (xn ) is said to be monotone decreasing or nonincreasing
if xn ≥ xn+1 for all n ∈ N, that is, x1 ≥ x2 ≥ x3 ≥ · · · .
• 1, 12 , 14 , . . . , 21n , . . . is monotone decreasing.
• A sequence is monotone if it is either monotone increasing or
monotone decreasing.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
33 / 39
Monotone sequences and convergence
• A sequence (xn ) is said to be monotone increasing or nondecreasing
if xn ≤ xn+1 for all n ∈ N, that is, x1 ≤ x2 ≤ x3 ≤ · · · .
n
• 12 , 23 , . . . , n+1
, . . . is monotone increasing
• A sequence (xn ) is said to be monotone decreasing or nonincreasing
if xn ≥ xn+1 for all n ∈ N, that is, x1 ≥ x2 ≥ x3 ≥ · · · .
• 1, 12 , 14 , . . . , 21n , . . . is monotone decreasing.
• A sequence is monotone if it is either monotone increasing or
monotone decreasing.
A nondecreasing sequence that is bounded from above always has a least
upper bound. Likewise, a nonincreasing sequence bounded from below
always has a greatest lower bound. These results are based on the
completeness property of the real numbers.
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
33 / 39
Monotone convergence theorem
Theorem
• A monotone increasing sequence that is bounded above, is convergent
and it converges to the least upper bound.
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Monotone convergence theorem
Theorem
• A monotone increasing sequence that is bounded above, is convergent
and it converges to the least upper bound.
• A monotone decreasing sequence that is bounded below, is
convergent and it converges to the greatest lower bound.
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Monotone convergence theorem
Theorem
• A monotone increasing sequence that is bounded above, is convergent
and it converges to the least upper bound.
• A monotone decreasing sequence that is bounded below, is
convergent and it converges to the greatest lower bound.
• Let an be a monotone increasing sequence that is bounded above.
• Let L be its least upperbound. By definition of least upper bound,
L − ϵ is not an upperbound. i.e some term ak from the sequence
satisfies ak > L − ϵ.
• Since (an ) is increasing an ≥ ak , ∀n ≥ k.
• Thus we have L − ϵ < ak ≤ an ≤ L < L + ϵ, ∀n ≥ k. Thus an → L.
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Example:
3
Let a1 :=
2
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and
1
an+1 :=
2
Mathematics I
2
an +
an
for n ∈ N.
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Example:
3
Let a1 :=
2
and
1
an+1 :=
2
2
an +
an
for n ∈ N.
Then an > 0 for all n ∈ N. Hence (an ) is bounded below by 0.
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Mathematics I
May 25, 2026
35 / 39
Example:
3
Let a1 :=
2
and
1
an+1 :=
2
2
an +
an
for n ∈ N.
Then an > 0 for all n ∈ N. Hence (an ) is bounded below by 0.
Let us check whether the sequence (an ) is decreasing. Since
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May 25, 2026
35 / 39
Example:
3
Let a1 :=
2
and
1
an+1 :=
2
2
an +
an
for n ∈ N.
Then an > 0 for all n ∈ N. Hence (an ) is bounded below by 0.
Let us check whether the sequence (an ) is decreasing. Since
an − an+1 = an −
an2 + 2
a2 − 2
= n
2an
2 an
for all n ∈ N,
(an ) is decreasing if and only if an2 − 2 ≥ 0 for all n ∈ N. But
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May 25, 2026
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Example:
3
Let a1 :=
2
and
1
an+1 :=
2
2
an +
an
for n ∈ N.
Then an > 0 for all n ∈ N. Hence (an ) is bounded below by 0.
Let us check whether the sequence (an ) is decreasing. Since
an − an+1 = an −
an2 + 2
a2 − 2
= n
2an
2 an
for all n ∈ N,
(an ) is decreasing if and only if an2 − 2 ≥ 0 for all n ∈ N. But
a12 ≥ 2
and
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2
an+1
−2=
an2 − 2
4an2
Mathematics I
2
≥ 0 for all n ∈ N.
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Example:
3
Let a1 :=
2
and
1
an+1 :=
2
2
an +
an
for n ∈ N.
Then an > 0 for all n ∈ N. Hence (an ) is bounded below by 0.
Let us check whether the sequence (an ) is decreasing. Since
an − an+1 = an −
an2 + 2
a2 − 2
= n
2an
2 an
for all n ∈ N,
(an ) is decreasing if and only if an2 − 2 ≥ 0 for all n ∈ N. But
a12 ≥ 2
and
2
an+1
−2=
an2 − 2
4an2
2
≥ 0 for all n ∈ N.
Hence the sequence (an ) is decreasing.
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It follows that (an ) is convergent. Let an → a. Then an+1 → a also. But
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36 / 39
It follows that (an ) is convergent. Let an → a. Then an+1 → a also. But
1
2
1
2
an+1 =
an +
→
a+
.
2
an
2
a
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May 25, 2026
36 / 39
It follows that (an ) is convergent. Let an → a. Then an+1 → a also. But
1
2
1
2
an+1 =
an +
→
a+
.
2
an
2
a
Since the limit of a sequence is unique, we see that 12 a + 2a = a, that is,
a2 = 2.
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Mathematics I
May 25, 2026
36 / 39
It follows that (an ) is convergent. Let an → a. Then an+1 → a also. But
1
2
1
2
an+1 =
an +
→
a+
.
2
an
2
a
Since the limit of a sequence is unique, we see that 12 a + 2a = a, that is,
a2 = 2.
Also, an > 0 for all n ∈ N and an → a,√so that a ≥ 0. Thus a is the
positive square root of 2, that is, a = 2.
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Mathematics I
May 25, 2026
36 / 39
Exercises
Determine if the sequences is monotonic and bounded.
n
• an = n+1
• an = 3n+1
n+1
• an = (2n+3)!
(n+1)!
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May 25, 2026
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Functions and sequences
Theorem
Suppose that f (x) is a function defined for all x ≥ n0 and that (an ) is a
sequence of real numbers such that an = f (n) for n ≥ n0 . Then
lim an = ℓ whenever lim f (x) = ℓ.
n→∞
x→∞
(i) Show that lim
n→∞
log n
= 0.
n
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Mathematics I
May 25, 2026
38 / 39
Functions and sequences
Theorem
Suppose that f (x) is a function defined for all x ≥ n0 and that (an ) is a
sequence of real numbers such that an = f (n) for n ≥ n0 . Then
lim an = ℓ whenever lim f (x) = ℓ.
n→∞
x→∞
(i) Show that lim
log n
= 0.
n→∞ n
We take f (x) = logx x and f (x) is defined for x ≥ 1. Therefore
lim
n→∞
log n
log x
= lim
= 0.
x→∞ x
n
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Mathematics I
May 25, 2026
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n
(ii) Let an = ( n+1
n−1 ) . Does an converge? Where?
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Mathematics I
May 25, 2026
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n
(ii) Let an = ( n+1
n−1 ) . Does an converge? Where?
The limit leads to the indeterminate form 1∞ . We can apply l’Hopital’s
rule if we first change the form by taking the natural logarithm of an .
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Mathematics I
May 25, 2026
39 / 39
n
(ii) Let an = ( n+1
n−1 ) . Does an converge? Where?
The limit leads to the indeterminate form 1∞ . We can apply l’Hopital’s
rule if we first change the form by taking the natural logarithm of an .
n+1
)
n−1
ln( n+1
n−1 )
= lim
n→∞ 1/n
−2/(n2 − 1)
= lim
n→∞
−1/n2
2n2
= lim 2
= 2.
n→∞ n − 1
lim ln(an ) = lim n ln(
n→∞
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n→∞
Mathematics I
May 25, 2026
39 / 39
n
(ii) Let an = ( n+1
n−1 ) . Does an converge? Where?
The limit leads to the indeterminate form 1∞ . We can apply l’Hopital’s
rule if we first change the form by taking the natural logarithm of an .
n+1
)
n−1
ln( n+1
n−1 )
= lim
n→∞ 1/n
−2/(n2 − 1)
= lim
n→∞
−1/n2
2n2
= lim 2
= 2.
n→∞ n − 1
lim ln(an ) = lim n ln(
n→∞
n→∞
Since ln(an ) → 2 and f (x) = e x is continuous, an → e 2 .
Saranya G. Nair (BITS Pilani)
Mathematics I
May 25, 2026
39 / 39