Continuity

MA 101 (Mathematics-I)

Anupam Saikia and Rupam BarmanDepartment of Mathematics

IIT Guwahati

A. Saikia & R. Barman MA-101 (2019)

Continuity

Introduction

Let D ⊆ R and let x0 ∈ R be such that for some h > 0,(x0 − h, x0 + h) \ {x0} ⊆ D.

If f : D → R, then ` ∈ R is said to be the limit of f at x0 iffor each ε > 0, there exists δ > 0 such that

whenever x ∈ D and 0 < |x − x0| < δ ⇒ |f (x)− `| < ε.

We write: limx→x0

f (x) = `.

Result: If limit exists, then it is unique.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Introduction

Let D ⊆ R and let x0 ∈ R be such that for some h > 0,(x0 − h, x0 + h) \ {x0} ⊆ D.

If f : D → R, then ` ∈ R is said to be the limit of f at x0 iffor each ε > 0, there exists δ > 0 such that

whenever x ∈ D and 0 < |x − x0| < δ ⇒ |f (x)− `| < ε.

We write: limx→x0

f (x) = `.

Result: If limit exists, then it is unique.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Introduction

Let D ⊆ R and let x0 ∈ R be such that for some h > 0,(x0 − h, x0 + h) \ {x0} ⊆ D.

If f : D → R, then ` ∈ R is said to be the limit of f at x0 iffor each ε > 0, there exists δ > 0 such that

whenever x ∈ D and 0 < |x − x0| < δ ⇒ |f (x)− `| < ε.

We write: limx→x0

f (x) = `.

Result: If limit exists, then it is unique.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Introduction

Let D ⊆ R and let x0 ∈ R be such that for some h > 0,(x0 − h, x0 + h) \ {x0} ⊆ D.

If f : D → R, then ` ∈ R is said to be the limit of f at x0 iffor each ε > 0, there exists δ > 0 such that

whenever x ∈ D and 0 < |x − x0| < δ ⇒ |f (x)− `| < ε.

We write: limx→x0

f (x) = `.

Result: If limit exists, then it is unique.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Example: limx→1

(3x

2− 1) =

1

2. Let ε > 0. We have to find

δ > 0 such that 0 < |x − 1| < δ ⇒ |f (x)− `| < ε holds with` = 1/2. Working backwards,

3

2|x − 1| < ε whenever |x − 1| < δ :=

2

3ε.

Theorem (Sequential criterion)

Let D ⊆ R and let x0 ∈ R such that for some h > 0,(x0 − h, x0 + h) \ {x0} ⊆ D. Let f : D → R. Then thefollowing are equivalent.

(a) limx→x0

f (x) = `.

(b) For any sequence (xn) in D with xn 6= x0 for all n ≥ 1 andxn → x0, the sequence (f (xn)) converges to `.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Example: limx→1

(3x

2− 1) =

1

2. Let ε > 0. We have to find

δ > 0 such that 0 < |x − 1| < δ ⇒ |f (x)− `| < ε holds with` = 1/2. Working backwards,

3

2|x − 1| < ε whenever |x − 1| < δ :=

2

3ε.

Theorem (Sequential criterion)

Let D ⊆ R and let x0 ∈ R such that for some h > 0,(x0 − h, x0 + h) \ {x0} ⊆ D. Let f : D → R. Then thefollowing are equivalent.

(a) limx→x0

f (x) = `.

(b) For any sequence (xn) in D with xn 6= x0 for all n ≥ 1 andxn → x0, the sequence (f (xn)) converges to `.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Example: limx→0

sin 1x

does not exist.

Solution: Let xn = 2(4n+1)π

and yn = 1nπ

for all n ∈ N. Then

xn → 0 and yn → 0.

Since sin 1xn

= 1 and sin 1yn

= 0 for all n ∈ N, we get sin 1xn→ 1

and sin 1yn→ 0.

Therefore by the sequential criterion for limit, limx→0

sin 1x

does

not exist.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Example: limx→0

sin 1x

does not exist.

Solution: Let xn = 2(4n+1)π

and yn = 1nπ

for all n ∈ N. Then

xn → 0 and yn → 0.

Since sin 1xn

= 1 and sin 1yn

= 0 for all n ∈ N, we get sin 1xn→ 1

and sin 1yn→ 0.

Therefore by the sequential criterion for limit, limx→0

sin 1x

does

not exist.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Example: limx→0

sin 1x

does not exist.

Solution: Let xn = 2(4n+1)π

and yn = 1nπ

for all n ∈ N. Then

xn → 0 and yn → 0.

Since sin 1xn

= 1 and sin 1yn

= 0 for all n ∈ N, we get sin 1xn→ 1

and sin 1yn→ 0.

Therefore by the sequential criterion for limit, limx→0

sin 1x

does

not exist.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Example: limx→0

sin 1x

does not exist.

Solution: Let xn = 2(4n+1)π

and yn = 1nπ

for all n ∈ N. Then

xn → 0 and yn → 0.

Since sin 1xn

= 1 and sin 1yn

= 0 for all n ∈ N, we get sin 1xn→ 1

and sin 1yn→ 0.

Therefore by the sequential criterion for limit, limx→0

sin 1x

does

not exist.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Example: limx→0

sin 1x

does not exist.

Solution: Let xn = 2(4n+1)π

and yn = 1nπ

for all n ∈ N. Then

xn → 0 and yn → 0.

Since sin 1xn

= 1 and sin 1yn

= 0 for all n ∈ N, we get sin 1xn→ 1

and sin 1yn→ 0.

Therefore by the sequential criterion for limit, limx→0

sin 1x

does

not exist.

A. Saikia & R. Barman MA-101 (2019)

Continuity

A function f : D → R is called bounded if there exists M > 0such that |f (x)| < M for all x ∈ D.

Result: Let f : D → R. Suppose that limx→x0

f (x) = `. Then

there exists some δ > 0 such that f is bounded on(x0 − δ, x0 + δ) \ {x0}. That is, there exists M > 0 such that|f (x)| < M for all x ∈ (x0 − δ, x0 + δ) with x 6= x0.

Proof: Let ε > 0. Since limx→x0

f (x) = `, so there is some δ > 0

such that |f (x)− `| < ε whenever 0 < |x − x0| < δ.

Take M = ε + |`|.

Remark: If x0 ∈ D, then by taking M = max{ε + |`|, |f (x0)|}we have |f (x)| < M for all x ∈ (x0 − δ, x0 + δ).

A. Saikia & R. Barman MA-101 (2019)

Continuity

A function f : D → R is called bounded if there exists M > 0such that |f (x)| < M for all x ∈ D.

Result: Let f : D → R. Suppose that limx→x0

f (x) = `. Then

there exists some δ > 0 such that f is bounded on(x0 − δ, x0 + δ) \ {x0}. That is, there exists M > 0 such that|f (x)| < M for all x ∈ (x0 − δ, x0 + δ) with x 6= x0.

Proof: Let ε > 0. Since limx→x0

f (x) = `, so there is some δ > 0

such that |f (x)− `| < ε whenever 0 < |x − x0| < δ.

Take M = ε + |`|.

Remark: If x0 ∈ D, then by taking M = max{ε + |`|, |f (x0)|}we have |f (x)| < M for all x ∈ (x0 − δ, x0 + δ).

A. Saikia & R. Barman MA-101 (2019)

Continuity

A function f : D → R is called bounded if there exists M > 0such that |f (x)| < M for all x ∈ D.

Result: Let f : D → R. Suppose that limx→x0

f (x) = `. Then

there exists some δ > 0 such that f is bounded on(x0 − δ, x0 + δ) \ {x0}. That is, there exists M > 0 such that|f (x)| < M for all x ∈ (x0 − δ, x0 + δ) with x 6= x0.

Proof: Let ε > 0. Since limx→x0

f (x) = `, so there is some δ > 0

such that |f (x)− `| < ε whenever 0 < |x − x0| < δ.

Take M = ε + |`|.

Remark: If x0 ∈ D, then by taking M = max{ε + |`|, |f (x0)|}we have |f (x)| < M for all x ∈ (x0 − δ, x0 + δ).

A. Saikia & R. Barman MA-101 (2019)

Continuity

A function f : D → R is called bounded if there exists M > 0such that |f (x)| < M for all x ∈ D.

Result: Let f : D → R. Suppose that limx→x0

f (x) = `. Then

there exists some δ > 0 such that f is bounded on(x0 − δ, x0 + δ) \ {x0}. That is, there exists M > 0 such that|f (x)| < M for all x ∈ (x0 − δ, x0 + δ) with x 6= x0.

Proof: Let ε > 0. Since limx→x0

f (x) = `, so there is some δ > 0

such that |f (x)− `| < ε whenever 0 < |x − x0| < δ.

Take M = ε + |`|.

Remark: If x0 ∈ D, then by taking M = max{ε + |`|, |f (x0)|}we have |f (x)| < M for all x ∈ (x0 − δ, x0 + δ).

A. Saikia & R. Barman MA-101 (2019)

Continuity

A function f : D → R is called bounded if there exists M > 0such that |f (x)| < M for all x ∈ D.

Result: Let f : D → R. Suppose that limx→x0

f (x) = `. Then

there exists some δ > 0 such that f is bounded on(x0 − δ, x0 + δ) \ {x0}. That is, there exists M > 0 such that|f (x)| < M for all x ∈ (x0 − δ, x0 + δ) with x 6= x0.

Proof: Let ε > 0. Since limx→x0

f (x) = `, so there is some δ > 0

such that |f (x)− `| < ε whenever 0 < |x − x0| < δ.

Take M = ε + |`|.

Remark: If x0 ∈ D, then by taking M = max{ε + |`|, |f (x0)|}we have |f (x)| < M for all x ∈ (x0 − δ, x0 + δ).

A. Saikia & R. Barman MA-101 (2019)

Continuity

Theorem (Limit Theorems)

Let D ⊆ R and let x0 ∈ R such that for some h > 0,(x0 − h, x0 + h) \ {x0} ⊆ D. Let f , g , j : D → R. Supposethat lim

x→x0f (x) = ` and lim

x→x0g(x) = m. Then

(1) limx→x0

(f (x)± g(x)) = `±m.

(2) If f (x) ≤ g(x) for all x ∈ (x0 − h, x0 + h) \ {x0}, then` ≤ m.

(3) limx→x0

(fg)(x) = `m and if m 6= 0 and g(x) 6= 0 for all

x ∈ D, then limx→x0

1

g(x)=

1

m.

(4) If f (x) ≤ j(x) ≤ g(x) for all x ∈ (x0 − h, x0 + h) \ {x0}and ` = m, then lim

x→x0j(x) = `.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Theorem (Limit Theorems)

Let D ⊆ R and let x0 ∈ R such that for some h > 0,(x0 − h, x0 + h) \ {x0} ⊆ D. Let f , g , j : D → R. Supposethat lim

x→x0f (x) = ` and lim

x→x0g(x) = m. Then

(1) limx→x0

(f (x)± g(x)) = `±m.

(2) If f (x) ≤ g(x) for all x ∈ (x0 − h, x0 + h) \ {x0}, then` ≤ m.

(3) limx→x0

(fg)(x) = `m and if m 6= 0 and g(x) 6= 0 for all

x ∈ D, then limx→x0

1

g(x)=

1

m.

(4) If f (x) ≤ j(x) ≤ g(x) for all x ∈ (x0 − h, x0 + h) \ {x0}and ` = m, then lim

x→x0j(x) = `.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Theorem (Limit Theorems)

Let D ⊆ R and let x0 ∈ R such that for some h > 0,(x0 − h, x0 + h) \ {x0} ⊆ D. Let f , g , j : D → R. Supposethat lim

x→x0f (x) = ` and lim

x→x0g(x) = m. Then

(1) limx→x0

(f (x)± g(x)) = `±m.

(2) If f (x) ≤ g(x) for all x ∈ (x0 − h, x0 + h) \ {x0}, then` ≤ m.

(3) limx→x0

(fg)(x) = `m and if m 6= 0 and g(x) 6= 0 for all

x ∈ D, then limx→x0

1

g(x)=

1

m.

(4) If f (x) ≤ j(x) ≤ g(x) for all x ∈ (x0 − h, x0 + h) \ {x0}and ` = m, then lim

x→x0j(x) = `.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Theorem (Limit Theorems)

Let D ⊆ R and let x0 ∈ R such that for some h > 0,(x0 − h, x0 + h) \ {x0} ⊆ D. Let f , g , j : D → R. Supposethat lim

x→x0f (x) = ` and lim

x→x0g(x) = m. Then

(1) limx→x0

(f (x)± g(x)) = `±m.

(2) If f (x) ≤ g(x) for all x ∈ (x0 − h, x0 + h) \ {x0}, then` ≤ m.

(3) limx→x0

(fg)(x) = `m and if m 6= 0 and g(x) 6= 0 for all

x ∈ D, then limx→x0

1

g(x)=

1

m.

(4) If f (x) ≤ j(x) ≤ g(x) for all x ∈ (x0 − h, x0 + h) \ {x0}and ` = m, then lim

x→x0j(x) = `.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Theorem (Limit Theorems)

Let D ⊆ R and let x0 ∈ R such that for some h > 0,(x0 − h, x0 + h) \ {x0} ⊆ D. Let f , g , j : D → R. Supposethat lim

x→x0f (x) = ` and lim

x→x0g(x) = m. Then

(1) limx→x0

(f (x)± g(x)) = `±m.

(2) If f (x) ≤ g(x) for all x ∈ (x0 − h, x0 + h) \ {x0}, then` ≤ m.

(3) limx→x0

(fg)(x) = `m and if m 6= 0 and g(x) 6= 0 for all

x ∈ D, then limx→x0

1

g(x)=

1

m.

(4) If f (x) ≤ j(x) ≤ g(x) for all x ∈ (x0 − h, x0 + h) \ {x0}and ` = m, then lim

x→x0j(x) = `.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Result: Suppose that f (x) is bounded in(x0 − h, x0 + h) \ {x0} for some h > 0 and lim

x→x0g(x) = 0.

Then limx→x0

f (x)g(x) = 0.

Example: limx→0

x sin 1x

= 0.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Result: Suppose that f (x) is bounded in(x0 − h, x0 + h) \ {x0} for some h > 0 and lim

x→x0g(x) = 0.

Then limx→x0

f (x)g(x) = 0.

Example: limx→0

x sin 1x

= 0.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Result: Suppose that f (x) is bounded in(x0 − h, x0 + h) \ {x0} for some h > 0 and lim

x→x0g(x) = 0.

Then limx→x0

f (x)g(x) = 0.

Example: limx→0

x sin 1x

= 0.

A. Saikia & R. Barman MA-101 (2019)

Continuity

One sided limits

Let D ⊆ R and let x0 ∈ R such that for some h > 0,(x0, x0 + h) ⊆ D.

If f : D → R, then ` ∈ R is said to be the right hand limit off at x0 if for each ε > 0, there exists δ > 0 such that

whenever x ∈ D and 0 < x − x0 < δ ⇒ |f (x)− `| < ε.

We write: limx→x0+

f (x) = `.

Similarly one defines left hand limit of f at x0 and is denotedby lim

x→x0−f (x).

Result: limx→x0

f (x) = `⇔ limx→x0+

f (x) = limx→x0−

f (x) = `.

A. Saikia & R. Barman MA-101 (2019)

Continuity

One sided limits

Let D ⊆ R and let x0 ∈ R such that for some h > 0,(x0, x0 + h) ⊆ D.

If f : D → R, then ` ∈ R is said to be the right hand limit off at x0 if for each ε > 0, there exists δ > 0 such that

whenever x ∈ D and 0 < x − x0 < δ ⇒ |f (x)− `| < ε.

We write: limx→x0+

f (x) = `.

Similarly one defines left hand limit of f at x0 and is denotedby lim

x→x0−f (x).

Result: limx→x0

f (x) = `⇔ limx→x0+

f (x) = limx→x0−

f (x) = `.

A. Saikia & R. Barman MA-101 (2019)

Continuity

One sided limits

Let D ⊆ R and let x0 ∈ R such that for some h > 0,(x0, x0 + h) ⊆ D.

If f : D → R, then ` ∈ R is said to be the right hand limit off at x0 if for each ε > 0, there exists δ > 0 such that

whenever x ∈ D and 0 < x − x0 < δ ⇒ |f (x)− `| < ε.

We write: limx→x0+

f (x) = `.

Similarly one defines left hand limit of f at x0 and is denotedby lim

x→x0−f (x).

Result: limx→x0

f (x) = `⇔ limx→x0+

f (x) = limx→x0−

f (x) = `.

A. Saikia & R. Barman MA-101 (2019)

Continuity

One sided limits

Let D ⊆ R and let x0 ∈ R such that for some h > 0,(x0, x0 + h) ⊆ D.

If f : D → R, then ` ∈ R is said to be the right hand limit off at x0 if for each ε > 0, there exists δ > 0 such that

whenever x ∈ D and 0 < x − x0 < δ ⇒ |f (x)− `| < ε.

We write: limx→x0+

f (x) = `.

Similarly one defines left hand limit of f at x0 and is denotedby lim

x→x0−f (x).

Result: limx→x0

f (x) = `⇔ limx→x0+

f (x) = limx→x0−

f (x) = `.

A. Saikia & R. Barman MA-101 (2019)

Continuity

One sided limits

Let D ⊆ R and let x0 ∈ R such that for some h > 0,(x0, x0 + h) ⊆ D.

If f : D → R, then ` ∈ R is said to be the right hand limit off at x0 if for each ε > 0, there exists δ > 0 such that

whenever x ∈ D and 0 < x − x0 < δ ⇒ |f (x)− `| < ε.

We write: limx→x0+

f (x) = `.

Similarly one defines left hand limit of f at x0 and is denotedby lim

x→x0−f (x).

Result: limx→x0

f (x) = `⇔ limx→x0+

f (x) = limx→x0−

f (x) = `.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Example: Show that limθ→0

sin θθ

= 1.

T

O

B

A

P

Q

θ

(1, 0)(0, 0)

Unit circle

Area of ∆OAP < Area of the circular section OAPO < Area of ∆OAT

Area of ∆OAP= (sin θ)/2

θ

Area of the circular section OAPO = θ/2

Area of ∆OAT= (tan θ)/2

We will use Sandwich Thm for limit(Limit Thm (4))

A. Saikia & R. Barman MA-101 (2019)

Continuity

Example: Show that limθ→0

sin θθ

= 1.

T

O

B

A

P

Q

θ

(1, 0)(0, 0)

Unit circle

Area of ∆OAP < Area of the circular section OAPO < Area of ∆OAT

Area of ∆OAP= (sin θ)/2

θ

Area of the circular section OAPO = θ/2

Area of ∆OAT= (tan θ)/2

We will use Sandwich Thm for limit(Limit Thm (4))

A. Saikia & R. Barman MA-101 (2019)

Continuity

Limits at infinity and infinite limits

Definition: f (x) has limit ` as x approaches +∞, if for anygiven ε > 0, there exists M > 0 such that

x > M =⇒ |f (x)− `| < ε.

Similarly, one can define limit of f (x) as x approaches −∞.

Example: (i) limx→∞

1

x= 0, (ii) limx→−∞

1x

= 0,

(iii) limx→∞

sin x does not exist.

Definition: A function f (x) approaches ∞ (f (x)→∞) asx → x0 if, for every real M > 0, there exists δ > 0 such that

0 < |x − x0| < δ =⇒ f (x) > M .

Similarly, one can define limit of f (x) approaching −∞.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Limits at infinity and infinite limits

Definition: f (x) has limit ` as x approaches +∞, if for anygiven ε > 0, there exists M > 0 such that

x > M =⇒ |f (x)− `| < ε.

Similarly, one can define limit of f (x) as x approaches −∞.

Example: (i) limx→∞

1

x= 0, (ii) limx→−∞

1x

= 0,

(iii) limx→∞

sin x does not exist.

Definition: A function f (x) approaches ∞ (f (x)→∞) asx → x0 if, for every real M > 0, there exists δ > 0 such that

0 < |x − x0| < δ =⇒ f (x) > M .

Similarly, one can define limit of f (x) approaching −∞.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Limits at infinity and infinite limits

Definition: f (x) has limit ` as x approaches +∞, if for anygiven ε > 0, there exists M > 0 such that

x > M =⇒ |f (x)− `| < ε.

Similarly, one can define limit of f (x) as x approaches −∞.

Example: (i) limx→∞

1

x= 0, (ii) limx→−∞

1x

= 0,

(iii) limx→∞

sin x does not exist.

Definition: A function f (x) approaches ∞ (f (x)→∞) asx → x0 if, for every real M > 0, there exists δ > 0 such that

0 < |x − x0| < δ =⇒ f (x) > M .

Similarly, one can define limit of f (x) approaching −∞.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Example: (i) limx→0

1

x2=∞, (ii) lim

x→0

1

x2sin(1/x) does not exist.

For (ii), let xn = 1π2+2nπ

and yn = 1nπ

. Then xn, yn → 0 asn→∞.

But limn→∞

f (xn) =1

x2n→∞ and lim

n→∞f (yn) = 0.

TheoremSuppose that lim

x→x0f (x) = `. If ` 6= 0, then there exists some δ

such that f (x) 6= 0 for all x ∈ (x0 − δ, x0 + δ) \ {x0}.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Example: (i) limx→0

1

x2=∞, (ii) lim

x→0

1

x2sin(1/x) does not exist.

For (ii), let xn = 1π2+2nπ

and yn = 1nπ

. Then xn, yn → 0 asn→∞.

But limn→∞

f (xn) =1

x2n→∞ and lim

n→∞f (yn) = 0.

TheoremSuppose that lim

x→x0f (x) = `. If ` 6= 0, then there exists some δ

such that f (x) 6= 0 for all x ∈ (x0 − δ, x0 + δ) \ {x0}.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Example: (i) limx→0

1

x2=∞, (ii) lim

x→0

1

x2sin(1/x) does not exist.

For (ii), let xn = 1π2+2nπ

and yn = 1nπ

. Then xn, yn → 0 asn→∞.

But limn→∞

f (xn) =1

x2n→∞ and lim

n→∞f (yn) = 0.

TheoremSuppose that lim

x→x0f (x) = `. If ` 6= 0, then there exists some δ

such that f (x) 6= 0 for all x ∈ (x0 − δ, x0 + δ) \ {x0}.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Continuous functions

Let D be a nonempty subset of R and f : D → R. We saythat f is continuous at x0 ∈ D if for each ε > 0, there existsδ > 0 such that |f (x)− f (x0)| < ε for all x ∈ D satisfying|x − x0| < δ.

Let f : [a, b]→ R. Then f is continuous at c ∈ (a, b) iflimx→c

f (x) = f (c). f is continuous at a if limx→a+

f (x) = f (a).

Similarly, f is continuous at b if limx→b−

f (x) = f (b).

Sequential criterion of continuity: f : D → R iscontinuous at x0 ∈ D if and only if for every sequence (xn) inD such that xn → x0, we have f (xn)→ f (x0).

We say that f : D → R is continuous if f is continuous ateach x0 ∈ D.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Continuous functions

Let D be a nonempty subset of R and f : D → R. We saythat f is continuous at x0 ∈ D if for each ε > 0, there existsδ > 0 such that |f (x)− f (x0)| < ε for all x ∈ D satisfying|x − x0| < δ.

Let f : [a, b]→ R. Then f is continuous at c ∈ (a, b) iflimx→c

f (x) = f (c). f is continuous at a if limx→a+

f (x) = f (a).

Similarly, f is continuous at b if limx→b−

f (x) = f (b).

Sequential criterion of continuity: f : D → R iscontinuous at x0 ∈ D if and only if for every sequence (xn) inD such that xn → x0, we have f (xn)→ f (x0).

We say that f : D → R is continuous if f is continuous ateach x0 ∈ D.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Continuous functions

Let D be a nonempty subset of R and f : D → R. We saythat f is continuous at x0 ∈ D if for each ε > 0, there existsδ > 0 such that |f (x)− f (x0)| < ε for all x ∈ D satisfying|x − x0| < δ.

Let f : [a, b]→ R. Then f is continuous at c ∈ (a, b) iflimx→c

f (x) = f (c). f is continuous at a if limx→a+

f (x) = f (a).

Similarly, f is continuous at b if limx→b−

f (x) = f (b).

Sequential criterion of continuity: f : D → R iscontinuous at x0 ∈ D if and only if for every sequence (xn) inD such that xn → x0, we have f (xn)→ f (x0).

We say that f : D → R is continuous if f is continuous ateach x0 ∈ D.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Continuous functions

Let D be a nonempty subset of R and f : D → R. We saythat f is continuous at x0 ∈ D if for each ε > 0, there existsδ > 0 such that |f (x)− f (x0)| < ε for all x ∈ D satisfying|x − x0| < δ.

Let f : [a, b]→ R. Then f is continuous at c ∈ (a, b) iflimx→c

f (x) = f (c). f is continuous at a if limx→a+

f (x) = f (a).

Similarly, f is continuous at b if limx→b−

f (x) = f (b).

Sequential criterion of continuity: f : D → R iscontinuous at x0 ∈ D if and only if for every sequence (xn) inD such that xn → x0, we have f (xn)→ f (x0).

We say that f : D → R is continuous if f is continuous ateach x0 ∈ D.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Examples

(1) f (x) =

{3x + 2 if x < 1,

4x2 if x ≥ 1.

(2) f (x) =

{x sin 1

xif x 6= 0,

0 if x = 0.

(3) f (x) =

{sin 1

xif x 6= 0,

0 if x = 0.

(4) f (x) =

{1 if x ∈ Q,0 if x ∈ R \Q.

(5) f (x) =

{x if x ∈ Q,−x if x ∈ R \Q.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Examples

(1) f (x) =

{3x + 2 if x < 1,

4x2 if x ≥ 1.

(2) f (x) =

{x sin 1

xif x 6= 0,

0 if x = 0.

(3) f (x) =

{sin 1

xif x 6= 0,

0 if x = 0.

(4) f (x) =

{1 if x ∈ Q,0 if x ∈ R \Q.

(5) f (x) =

{x if x ∈ Q,−x if x ∈ R \Q.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Examples

(1) f (x) =

{3x + 2 if x < 1,

4x2 if x ≥ 1.

(2) f (x) =

{x sin 1

xif x 6= 0,

0 if x = 0.

(3) f (x) =

{sin 1

xif x 6= 0,

0 if x = 0.

(4) f (x) =

{1 if x ∈ Q,0 if x ∈ R \Q.

(5) f (x) =

{x if x ∈ Q,−x if x ∈ R \Q.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Examples

(1) f (x) =

{3x + 2 if x < 1,

4x2 if x ≥ 1.

(2) f (x) =

{x sin 1

xif x 6= 0,

0 if x = 0.

(3) f (x) =

{sin 1

xif x 6= 0,

0 if x = 0.

(4) f (x) =

{1 if x ∈ Q,0 if x ∈ R \Q.

(5) f (x) =

{x if x ∈ Q,−x if x ∈ R \Q.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Examples

(1) f (x) =

{3x + 2 if x < 1,

4x2 if x ≥ 1.

(2) f (x) =

{x sin 1

xif x 6= 0,

0 if x = 0.

(3) f (x) =

{sin 1

xif x 6= 0,

0 if x = 0.

(4) f (x) =

{1 if x ∈ Q,0 if x ∈ R \Q.

(5) f (x) =

{x if x ∈ Q,−x if x ∈ R \Q.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Result: Let f , g : D → R be continuous at x0 ∈ D. Then

(a) f ± g , fg and |f | are continuous at x0.

(b) f /g is continuous at x0 if g(x) 6= 0 for all x ∈ D.

Result: Composition of two continuous functions is continuous.

Further examples of continuous functions:

Polynomial function, Rational function, sine function, cosinefunction, exponential function, etc.

Result: If f : D → R is continuous at x0 ∈ D and f (x0) 6= 0,then there exists δ > 0 such that f (x) 6= 0 for all x ∈ Dsatisfying |x − x0| < δ.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Result: Let f , g : D → R be continuous at x0 ∈ D. Then

(a) f ± g , fg and |f | are continuous at x0.

(b) f /g is continuous at x0 if g(x) 6= 0 for all x ∈ D.

Result: Composition of two continuous functions is continuous.

Further examples of continuous functions:

Polynomial function, Rational function, sine function, cosinefunction, exponential function, etc.

Result: If f : D → R is continuous at x0 ∈ D and f (x0) 6= 0,then there exists δ > 0 such that f (x) 6= 0 for all x ∈ Dsatisfying |x − x0| < δ.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Result: Let f , g : D → R be continuous at x0 ∈ D. Then

(a) f ± g , fg and |f | are continuous at x0.

(b) f /g is continuous at x0 if g(x) 6= 0 for all x ∈ D.

Result: Composition of two continuous functions is continuous.

Further examples of continuous functions:

Polynomial function, Rational function, sine function, cosinefunction, exponential function, etc.

Result: If f : D → R is continuous at x0 ∈ D and f (x0) 6= 0,then there exists δ > 0 such that f (x) 6= 0 for all x ∈ Dsatisfying |x − x0| < δ.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Result: Let f , g : D → R be continuous at x0 ∈ D. Then

(a) f ± g , fg and |f | are continuous at x0.

(b) f /g is continuous at x0 if g(x) 6= 0 for all x ∈ D.

Result: Composition of two continuous functions is continuous.

Further examples of continuous functions:

Polynomial function, Rational function, sine function, cosinefunction, exponential function, etc.

Result: If f : D → R is continuous at x0 ∈ D and f (x0) 6= 0,then there exists δ > 0 such that f (x) 6= 0 for all x ∈ Dsatisfying |x − x0| < δ.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Result: Let f , g : D → R be continuous at x0 ∈ D. Then

(a) f ± g , fg and |f | are continuous at x0.

(b) f /g is continuous at x0 if g(x) 6= 0 for all x ∈ D.

Result: Composition of two continuous functions is continuous.

Further examples of continuous functions:

Polynomial function, Rational function, sine function, cosinefunction, exponential function, etc.

Result: If f : D → R is continuous at x0 ∈ D and f (x0) 6= 0,then there exists δ > 0 such that f (x) 6= 0 for all x ∈ Dsatisfying |x − x0| < δ.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Result: Let f , g : D → R be continuous at x0 ∈ D. Then

(a) f ± g , fg and |f | are continuous at x0.

(b) f /g is continuous at x0 if g(x) 6= 0 for all x ∈ D.

Result: Composition of two continuous functions is continuous.

Further examples of continuous functions:

Polynomial function, Rational function, sine function, cosinefunction, exponential function, etc.

Result: If f : D → R is continuous at x0 ∈ D and f (x0) 6= 0,then there exists δ > 0 such that f (x) 6= 0 for all x ∈ Dsatisfying |x − x0| < δ.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Result: If f : [a, b]→ R is continuous and if f (a) · f (b) < 0,then there exists c ∈ (a, b) such that f (c) = 0.

Intermediate value theorem: Let I be an interval of R and letf : I → R be continuous. If a, b ∈ I with a < b and iff (a) < k < f (b), then there exists c ∈ (a, b) such thatf (c) = k .Examples:

(a) The equation x2 = x sin x + cos x has at least two realroots.

(b) (Fixed point) If f : [a, b]→ [a, b] is continuous, then thereexists c ∈ [a, b] such that f (c) = c .

A. Saikia & R. Barman MA-101 (2019)

Continuity

Result: If f : [a, b]→ R is continuous and if f (a) · f (b) < 0,then there exists c ∈ (a, b) such that f (c) = 0.

Intermediate value theorem: Let I be an interval of R and letf : I → R be continuous. If a, b ∈ I with a < b and iff (a) < k < f (b), then there exists c ∈ (a, b) such thatf (c) = k .

Examples:

(a) The equation x2 = x sin x + cos x has at least two realroots.

(b) (Fixed point) If f : [a, b]→ [a, b] is continuous, then thereexists c ∈ [a, b] such that f (c) = c .

A. Saikia & R. Barman MA-101 (2019)

Continuity

Result: If f : [a, b]→ R is continuous and if f (a) · f (b) < 0,then there exists c ∈ (a, b) such that f (c) = 0.

Intermediate value theorem: Let I be an interval of R and letf : I → R be continuous. If a, b ∈ I with a < b and iff (a) < k < f (b), then there exists c ∈ (a, b) such thatf (c) = k .Examples:

(a) The equation x2 = x sin x + cos x has at least two realroots.

(b) (Fixed point) If f : [a, b]→ [a, b] is continuous, then thereexists c ∈ [a, b] such that f (c) = c .

A. Saikia & R. Barman MA-101 (2019)

Continuity

Result: If f : [a, b]→ R is continuous and if f (a) · f (b) < 0,then there exists c ∈ (a, b) such that f (c) = 0.

Intermediate value theorem: Let I be an interval of R and letf : I → R be continuous. If a, b ∈ I with a < b and iff (a) < k < f (b), then there exists c ∈ (a, b) such thatf (c) = k .Examples:

(a) The equation x2 = x sin x + cos x has at least two realroots.

(b) (Fixed point) If f : [a, b]→ [a, b] is continuous, then thereexists c ∈ [a, b] such that f (c) = c .

A. Saikia & R. Barman MA-101 (2019)

Continuity

Recall that a function f : D → R is called bounded if thereexists M > 0 such that |f (x)| < M for all x ∈ D.

Result: If f : [a, b]→ R is continuous, then f is bounded.

Example: There does not exist any continuous function from[0, 1] onto (0,∞).

Result: If f : [a, b]→ R is continuous, then the maximum andthe minimum of f (x) are attained in [a, b]. That is, there existx0, y0 ∈ [a, b] such that f (x0) ≤ f (x) ≤ f (y0) for all x ∈ [a, b].

A. Saikia & R. Barman MA-101 (2019)

Continuity

Recall that a function f : D → R is called bounded if thereexists M > 0 such that |f (x)| < M for all x ∈ D.

Result: If f : [a, b]→ R is continuous, then f is bounded.

Example: There does not exist any continuous function from[0, 1] onto (0,∞).

Result: If f : [a, b]→ R is continuous, then the maximum andthe minimum of f (x) are attained in [a, b]. That is, there existx0, y0 ∈ [a, b] such that f (x0) ≤ f (x) ≤ f (y0) for all x ∈ [a, b].

A. Saikia & R. Barman MA-101 (2019)

Continuity

Recall that a function f : D → R is called bounded if thereexists M > 0 such that |f (x)| < M for all x ∈ D.

Result: If f : [a, b]→ R is continuous, then f is bounded.

Example: There does not exist any continuous function from[0, 1] onto (0,∞).

Result: If f : [a, b]→ R is continuous, then the maximum andthe minimum of f (x) are attained in [a, b]. That is, there existx0, y0 ∈ [a, b] such that f (x0) ≤ f (x) ≤ f (y0) for all x ∈ [a, b].

A. Saikia & R. Barman MA-101 (2019)

Continuity

Recall that a function f : D → R is called bounded if thereexists M > 0 such that |f (x)| < M for all x ∈ D.

Result: If f : [a, b]→ R is continuous, then f is bounded.

Example: There does not exist any continuous function from[0, 1] onto (0,∞).

Result: If f : [a, b]→ R is continuous, then the maximum andthe minimum of f (x) are attained in [a, b]. That is, there existx0, y0 ∈ [a, b] such that f (x0) ≤ f (x) ≤ f (y0) for all x ∈ [a, b].

A. Saikia & R. Barman MA-101 (2019)

Continuity

Limit point of a set: Let A ⊆ R. A real number x is called alimit point of A if there exists a sequence (xn) in A convergingto x .

Let A ⊆ R. Then A is called a closed set if A contains all itslimit points. That is, if (xn) is a sequence in A converging tox , then x ∈ A.

For example, R, [a, b], {x1, x2, . . . , xn}, N are closed sets. But,(a, b), Q are not closed sets.

Result: Let A be a closed and bounded subset of R. Iff : A→ R is continuous, then f is bounded.

Remark: The above result is not true if A is bounded but notclosed. For example f (x) = 1/x on (0, 1). Also, the result isnot true if A is closed but not bounded. For example,f (x) = x on R.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Limit point of a set: Let A ⊆ R. A real number x is called alimit point of A if there exists a sequence (xn) in A convergingto x .

Let A ⊆ R. Then A is called a closed set if A contains all itslimit points. That is, if (xn) is a sequence in A converging tox , then x ∈ A.

For example, R, [a, b], {x1, x2, . . . , xn}, N are closed sets. But,(a, b), Q are not closed sets.

Result: Let A be a closed and bounded subset of R. Iff : A→ R is continuous, then f is bounded.

Remark: The above result is not true if A is bounded but notclosed. For example f (x) = 1/x on (0, 1). Also, the result isnot true if A is closed but not bounded. For example,f (x) = x on R.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Limit point of a set: Let A ⊆ R. A real number x is called alimit point of A if there exists a sequence (xn) in A convergingto x .

Let A ⊆ R. Then A is called a closed set if A contains all itslimit points. That is, if (xn) is a sequence in A converging tox , then x ∈ A.

For example, R, [a, b], {x1, x2, . . . , xn}, N are closed sets. But,(a, b), Q are not closed sets.

Result: Let A be a closed and bounded subset of R. Iff : A→ R is continuous, then f is bounded.

Remark: The above result is not true if A is bounded but notclosed. For example f (x) = 1/x on (0, 1). Also, the result isnot true if A is closed but not bounded. For example,f (x) = x on R.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Limit point of a set: Let A ⊆ R. A real number x is called alimit point of A if there exists a sequence (xn) in A convergingto x .

Let A ⊆ R. Then A is called a closed set if A contains all itslimit points. That is, if (xn) is a sequence in A converging tox , then x ∈ A.

For example, R, [a, b], {x1, x2, . . . , xn}, N are closed sets. But,(a, b), Q are not closed sets.

Result: Let A be a closed and bounded subset of R. Iff : A→ R is continuous, then f is bounded.

Remark: The above result is not true if A is bounded but notclosed. For example f (x) = 1/x on (0, 1). Also, the result isnot true if A is closed but not bounded. For example,f (x) = x on R.

A. Saikia & R. Barman MA-101 (2019)

Continuity

Limit point of a set: Let A ⊆ R. A real number x is called alimit point of A if there exists a sequence (xn) in A convergingto x .

Let A ⊆ R. Then A is called a closed set if A contains all itslimit points. That is, if (xn) is a sequence in A converging tox , then x ∈ A.

For example, R, [a, b], {x1, x2, . . . , xn}, N are closed sets. But,(a, b), Q are not closed sets.

Result: Let A be a closed and bounded subset of R. Iff : A→ R is continuous, then f is bounded.

Remark: The above result is not true if A is bounded but notclosed. For example f (x) = 1/x on (0, 1). Also, the result isnot true if A is closed but not bounded. For example,f (x) = x on R.

A. Saikia & R. Barman MA-101 (2019)