## Advanced Calculus (I)

WEN-CHING LIEN

Department of Mathematics National Cheng Kung University

WEN-CHINGLIEN **Advanced Calculus (I)**

## 3.3 Continuity

Definition

* Let E be a nonempty subset of R and f* :

*E*→

**R.**

(i)

*f is said to be continuous at a point a*∈*E if and only if*
given ǫ >0 there is aδ >0 (which in general depends on
ǫ, f, and a) such that

|*x*−*a*| < δ *and x* ∈*E imply* |*f*(x) −*f*(a)| < ǫ.

WEN-CHINGLIEN **Advanced Calculus (I)**

## 3.3 Continuity

Definition

* Let E be a nonempty subset of R and f* :

*E*→

**R.**

(i)

*f is said to be continuous at a point a*∈*E if and only if*
given ǫ >0 there is aδ >0 (which in general depends on
ǫ, f, and a) such that

|*x*−*a*| < δ *and x* ∈*E imply* |*f*(x) −*f*(a)| < ǫ.

WEN-CHINGLIEN **Advanced Calculus (I)**

## 3.3 Continuity

Definition

* Let E be a nonempty subset of R and f* :

*E*→

**R.**

(i)

*f is said to be continuous at a point a*∈*E if and only if*
given ǫ >0 there is aδ >0 (which in general depends on
ǫ, f, and a) such that

|*x*−*a*| < δ *and x* ∈*E imply* |*f*(x) −*f*(a)| < ǫ.

WEN-CHINGLIEN **Advanced Calculus (I)**

## 3.3 Continuity

Definition

* Let E be a nonempty subset of R and f* :

*E*→

**R.**

(i)

*f is said to be continuous at a point a*∈*E if and only if*
given ǫ >0 there is aδ >0 (which in general depends on
ǫ, f, and a) such that

|*x*−*a*| < δ *and x* ∈*E imply* |*f*(x) −*f*(a)| < ǫ.

WEN-CHINGLIEN **Advanced Calculus (I)**

Definition (ii)

*f is said to be continuous on E (notation: f* :*E* →**R is**
*continuous) if and only if f is continuous at every x* ∈*E.*

WEN-CHINGLIEN **Advanced Calculus (I)**

Definition (ii)

*f is said to be continuous on E (notation: f* :*E* →**R is**
*continuous) if and only if f is continuous at every x* ∈*E.*

WEN-CHINGLIEN **Advanced Calculus (I)**

Definition (ii)

*f is said to be continuous on E (notation: f* :*E* →**R is**
*continuous) if and only if f is continuous at every x* ∈*E.*

WEN-CHINGLIEN **Advanced Calculus (I)**

**Remark:**

Let I be an open interval that contains a point a and
*f* :*I* →*R. Then f is continuous at a*∈*I if and only if*

*f*(a) = lim

*x→a**f*(x).

WEN-CHINGLIEN **Advanced Calculus (I)**

**Remark:**

Let I be an open interval that contains a point a and
*f* :*I* →*R. Then f is continuous at a*∈*I if and only if*

*f*(a) = lim

*x→a**f*(x).

WEN-CHINGLIEN **Advanced Calculus (I)**

**Example:**

*1. f*(x) =*x*^{2}*, x* ∈ [0,1].

*2. f*(x) =*x*^{2}*, x* ∈*R.*

*3. f*(x) = 1

*x, x* ∈ (0,1].

*4. f*(x) =√

*x, x* ∈*R*^{+}.

WEN-CHINGLIEN **Advanced Calculus (I)**

Theorem

* Suppose that E is a nonempty subset of R, a*∈

*E, and*

*f*:

*E*→

**R. Then the following statements are equivalent:***(i) f is continuous at a* ∈*E*

*(ii) If x**n* *converges to a and x**n*∈*E, then f*(x*n*) →*f*(a)*as*
*n* → ∞*.*

WEN-CHINGLIEN **Advanced Calculus (I)**

Theorem

* Suppose that E is a nonempty subset of R, a*∈

*E, and*

*f*:

*E*→

**R. Then the following statements are equivalent:***(i) f is continuous at a* ∈*E*

*(ii) If x**n* *converges to a and x**n*∈*E, then f*(x*n*) →*f*(a)*as*
*n* → ∞*.*

WEN-CHINGLIEN **Advanced Calculus (I)**

Theorem

* Suppose that E is a nonempty subset of R, a*∈

*E, and*

*f*:

*E*→

**R. Then the following statements are equivalent:***(i) f is continuous at a* ∈*E*

*(ii) If x**n* *converges to a and x**n*∈*E, then f*(x*n*) →*f*(a)*as*
*n* → ∞*.*

WEN-CHINGLIEN **Advanced Calculus (I)**

Theorem

* Suppose that E is a nonempty subset of R, a*∈

*E, and*

*f*:

*E*→

**R. Then the following statements are equivalent:***(i) f is continuous at a* ∈*E*

*(ii) If x**n* *converges to a and x**n*∈*E, then f*(x*n*) →*f*(a)*as*
*n* → ∞*.*

WEN-CHINGLIEN **Advanced Calculus (I)**

Theorem

* Suppose that A and B are subsets of R and that f* :

*A*→

**R**

*and g*:

*B*→

*R with f*(A) ⊆

*B.*

WEN-CHINGLIEN **Advanced Calculus (I)**

Theorem

* Suppose that A and B are subsets of R and that f* :

*A*→

**R**

*and g*:

*B*→

*R with f*(A) ⊆

*B.*

WEN-CHINGLIEN **Advanced Calculus (I)**

Theorem
*(i)*

*If A*:=*I*\ {*a*}*, where I is a nondegenerate interval that*
*either contains a or has a as one of its endpoint if*

*L*:= lim

*x*→*a*
*x∈I*

*f*(x)

*exists and belongs to B, and if g is continuous at L*∈*B,*
*then*

*x*lim→*a*
*x∈I*

(g◦*f*)(x) =*g* lim

*x*→*a*
*x∈I*

*f*(x)

! .

WEN-CHINGLIEN **Advanced Calculus (I)**

Theorem
*(i)*

*If A*:=*I*\ {*a*}*, where I is a nondegenerate interval that*
*either contains a or has a as one of its endpoint if*

*L*:= lim

*x*→*a*
*x∈I*

*f*(x)

*exists and belongs to B, and if g is continuous at L*∈*B,*
*then*

*x*lim→*a*
*x∈I*

(g◦*f*)(x) =*g* lim

*x*→*a*
*x∈I*

*f*(x)

! .

WEN-CHINGLIEN **Advanced Calculus (I)**

Theorem
*(ii)*

*If f is continuous at a* ∈*A and g is continuous at f*(a) ∈*B,*
*then g*◦*f is continuous at a*∈*A.*

WEN-CHINGLIEN **Advanced Calculus (I)**

Theorem
*(ii)*

*If f is continuous at a* ∈*A and g is continuous at f*(a) ∈*B,*
*then g*◦*f is continuous at a*∈*A.*

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

*Suppose that x**n* ∈*I*\ {*a*}*and x**n*→*a as n*→ ∞. Since
*f*(A) ⊆*B,f*(x*n*) ∈*B.*Also, by the Sequential

*Characterization of Limits (Theorem 3.17), f*(x*n*) →*L as*
*n* → ∞*. Since g is continuous at L*∈*B, it follows from*
*Theorem 3.21 that g*◦*f*(x*n*) :=*g(f*(x*n*)) →*g(L)as n*→ ∞.
*Hence by Theorem 3.17, g*◦*f*(x) →*g(L)as x* →*a in I.*

This prove (i). A similar proof establishes part (ii). 2

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

*Suppose that x**n* ∈*I*\ {*a*}*and x**n*→*a as n*→ ∞. Since
*f*(A) ⊆*B, f*(x*n*) ∈*B.*Also, by the Sequential

*Characterization of Limits (Theorem 3.17), f*(x*n*) →*L as*
*n* → ∞*. Since g is continuous at L*∈*B, it follows from*
*Theorem 3.21 that g*◦*f*(x*n*) :=*g(f*(x*n*)) →*g(L)as n*→ ∞.
*Hence by Theorem 3.17, g*◦*f*(x) →*g(L)as x* →*a in I.*

This prove (i). A similar proof establishes part (ii). 2

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

*Suppose that x**n* ∈*I*\ {*a*}*and x**n*→*a as n*→ ∞. Since
*f*(A) ⊆*B,f*(x*n*) ∈*B.*Also, by the Sequential

*Characterization of Limits (Theorem 3.17), f*(x*n*) →*L as*
*n* → ∞*. Since g is continuous at L*∈*B, it follows from*
*Theorem 3.21 that g*◦*f*(x*n*) :=*g(f*(x*n*)) →*g(L)as n*→ ∞.
*Hence by Theorem 3.17, g*◦*f*(x) →*g(L)as x* →*a in I.*

This prove (i). A similar proof establishes part (ii). 2

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

*Suppose that x**n* ∈*I*\ {*a*}*and x**n*→*a as n*→ ∞. Since
*f*(A) ⊆*B, f*(x*n*) ∈*B.*Also, by the Sequential

*Characterization of Limits (Theorem 3.17), f*(x*n*) →*L as*
*n* → ∞. *Since g is continuous at L*∈*B, it follows from*
*Theorem 3.21 that g*◦*f*(x*n*) :=*g(f*(x*n*)) →*g(L)as n*→ ∞.
*Hence by Theorem 3.17, g*◦*f*(x) →*g(L)as x* →*a in I.*

This prove (i). A similar proof establishes part (ii). 2

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

*Suppose that x**n* ∈*I*\ {*a*}*and x**n*→*a as n*→ ∞. Since
*f*(A) ⊆*B, f*(x*n*) ∈*B.*Also, by the Sequential

*Characterization of Limits (Theorem 3.17), f*(x*n*) →*L as*
*n* → ∞*. Since g is continuous at L*∈*B,*it follows from
*Theorem 3.21 that g*◦*f*(x*n*) :=*g(f*(x*n*)) →*g(L)as n*→ ∞.
*Hence by Theorem 3.17, g*◦*f*(x) →*g(L)as x* →*a in I.*

This prove (i). A similar proof establishes part (ii). 2

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

*Suppose that x**n* ∈*I*\ {*a*}*and x**n*→*a as n*→ ∞. Since
*f*(A) ⊆*B, f*(x*n*) ∈*B.*Also, by the Sequential

*Characterization of Limits (Theorem 3.17), f*(x*n*) →*L as*
*n* → ∞. *Since g is continuous at L*∈*B, it follows from*
*Theorem 3.21 that g*◦*f*(x*n*) :=*g(f*(x*n*)) →*g(L)as n*→ ∞.
*Hence by Theorem 3.17, g*◦*f*(x) →*g(L)as x* →*a in I.*

This prove (i). A similar proof establishes part (ii). 2

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

*Suppose that x**n* ∈*I*\ {*a*}*and x**n*→*a as n*→ ∞. Since
*f*(A) ⊆*B, f*(x*n*) ∈*B.*Also, by the Sequential

*Characterization of Limits (Theorem 3.17), f*(x*n*) →*L as*
*n* → ∞*. Since g is continuous at L*∈*B,*it follows from
*Theorem 3.21 that g*◦*f*(x*n*) :=*g(f*(x*n*)) →*g(L)as n*→ ∞.
Hence by Theorem 3.17,*g*◦*f*(x) →*g(L)as x* →*a in I.*

This prove (i). A similar proof establishes part (ii). 2

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

*Suppose that x**n* ∈*I*\ {*a*}*and x**n*→*a as n*→ ∞. Since
*f*(A) ⊆*B, f*(x*n*) ∈*B.*Also, by the Sequential

*Characterization of Limits (Theorem 3.17), f*(x*n*) →*L as*
*n* → ∞*. Since g is continuous at L*∈*B, it follows from*
*Theorem 3.21 that g*◦*f*(x*n*) :=*g(f*(x*n*)) →*g(L)as n*→ ∞.
*Hence by Theorem 3.17, g*◦*f*(x) →*g(L)as x* →*a in I.*

This prove (i). A similar proof establishes part (ii). 2

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

*Suppose that x**n* ∈*I*\ {*a*}*and x**n*→*a as n*→ ∞. Since
*f*(A) ⊆*B, f*(x*n*) ∈*B.*Also, by the Sequential

*Characterization of Limits (Theorem 3.17), f*(x*n*) →*L as*
*n* → ∞*. Since g is continuous at L*∈*B, it follows from*
*Theorem 3.21 that g*◦*f*(x*n*) :=*g(f*(x*n*)) →*g(L)as n*→ ∞.
Hence by Theorem 3.17,*g*◦*f*(x) →*g(L)as x* →*a in I.*

This prove (i). A similar proof establishes part (ii). 2

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

*Suppose that x**n* ∈*I*\ {*a*}*and x**n*→*a as n*→ ∞. Since
*f*(A) ⊆*B, f*(x*n*) ∈*B.*Also, by the Sequential

*Characterization of Limits (Theorem 3.17), f*(x*n*) →*L as*
*n* → ∞*. Since g is continuous at L*∈*B, it follows from*
*Theorem 3.21 that g*◦*f*(x*n*) :=*g(f*(x*n*)) →*g(L)as n*→ ∞.
*Hence by Theorem 3.17, g*◦*f*(x) →*g(L)as x* →*a in I.*

This prove (i). A similar proof establishes part (ii). 2

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

*Suppose that x**n* ∈*I*\ {*a*}*and x**n*→*a as n*→ ∞. Since
*f*(A) ⊆*B, f*(x*n*) ∈*B.*Also, by the Sequential

*Characterization of Limits (Theorem 3.17), f*(x*n*) →*L as*
*n* → ∞*. Since g is continuous at L*∈*B, it follows from*
*Theorem 3.21 that g*◦*f*(x*n*) :=*g(f*(x*n*)) →*g(L)as n*→ ∞.
*Hence by Theorem 3.17, g*◦*f*(x) →*g(L)as x* →*a in I.*

This prove (i). A similar proof establishes part (ii). 2

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

*Suppose that x**n* ∈*I*\ {*a*}*and x**n*→*a as n*→ ∞. Since
*f*(A) ⊆*B, f*(x*n*) ∈*B.*Also, by the Sequential

*Characterization of Limits (Theorem 3.17), f*(x*n*) →*L as*
*n* → ∞*. Since g is continuous at L*∈*B, it follows from*
*Theorem 3.21 that g*◦*f*(x*n*) :=*g(f*(x*n*)) →*g(L)as n*→ ∞.
*Hence by Theorem 3.17, g*◦*f*(x) →*g(L)as x* →*a in I.*

This prove (i). A similar proof establishes part (ii). 2

WEN-CHINGLIEN **Advanced Calculus (I)**

Definition

* Let E be a nonempty subset of R. A function f* :

*E*→

**R is**

*said to be bounded on E if and only if there is an M*∈

*R*such that|

*f*(x)| ≤

*M for all x*∈

*E.*

WEN-CHINGLIEN **Advanced Calculus (I)**

Definition

* Let E be a nonempty subset of R. A function f* :

*E*→

**R is**

*said to be bounded on E if and only if there is an M*∈

*R*such that|

*f*(x)| ≤

*M for all x*∈

*E.*

WEN-CHINGLIEN **Advanced Calculus (I)**

Theorem (Extreme Value Theorem)

*If I is a closed, bounded interval and f* :*I* →**R is***continuous on I, then f is bounded on I. Moreover, if*

*M* =sup

*x∈I*

*f*(x) *and m*=inf

*x*∈I*f*(x),
*then there exist points x**m*,*x**M* ∈*I such that*
*(6)*

*f*(x*M*) =*M and f*(x*m*) =*m.*

WEN-CHINGLIEN **Advanced Calculus (I)**

Theorem (Extreme Value Theorem)

*If I is a closed, bounded interval and f* :*I* →**R is***continuous on I, then f is bounded on I. Moreover, if*

*M* =sup

*x∈I*

*f*(x) *and m*=inf

*x*∈I*f*(x),
*then there exist points x**m*,*x**M* ∈*I such that*
*(6)*

*f*(x*M*) =*M and f*(x*m*) =*m.*

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I. Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded, we know (by the Bolzano-Weierstrass*
Theorem) that{*x**n*}has a convergent subsequence, say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed, we also know (by the

*Comparison Theorem) that a*∈

*I. In particular, f*(a) ∈

*R.*

*On the other hand, substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞, we have|*f*(a)| = ∞, a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I.Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded,* we know (by the Bolzano-Weierstrass
Theorem) that{*x**n*}has a convergent subsequence, say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed, we also know (by the

*Comparison Theorem) that a*∈

*I. In particular, f*(a) ∈

*R.*

*On the other hand, substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞, we have|*f*(a)| = ∞, a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I. Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded, we know (by the Bolzano-Weierstrass*
Theorem) that{*x**n*}has a convergent subsequence,say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed, we also know (by the

*Comparison Theorem) that a*∈

*I. In particular, f*(a) ∈

*R.*

*On the other hand, substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞, we have|*f*(a)| = ∞, a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I. Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded,* we know (by the Bolzano-Weierstrass
Theorem) that{*x**n*}has a convergent subsequence, say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed, we also know (by the

*Comparison Theorem) that a*∈

*I. In particular, f*(a) ∈

*R.*

*On the other hand, substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞, we have|*f*(a)| = ∞, a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I. Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded, we know (by the Bolzano-Weierstrass*
Theorem) that{*x**n*}has a convergent subsequence,say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed,we also know (by the

*Comparison Theorem) that a*∈

*I. In particular, f*(a) ∈

*R.*

*On the other hand, substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞, we have|*f*(a)| = ∞, a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I. Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded, we know (by the Bolzano-Weierstrass*
Theorem) that{*x**n*}has a convergent subsequence, say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed, we also know (by the

*Comparison Theorem) that a*∈

*I.*

*In particular, f*(a) ∈

*R.*

*On the other hand, substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞, we have|*f*(a)| = ∞, a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I. Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded, we know (by the Bolzano-Weierstrass*
Theorem) that{*x**n*}has a convergent subsequence, say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed,we also know (by the

*Comparison Theorem) that a*∈

*I. In particular, f*(a) ∈

*R.*

*On the other hand, substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞, we have|*f*(a)| = ∞, a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I. Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded, we know (by the Bolzano-Weierstrass*
Theorem) that{*x**n*}has a convergent subsequence, say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed, we also know (by the

*Comparison Theorem) that a*∈

*I.*

*In particular, f*(a) ∈

*R.*

On the other hand, *substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞, we have|*f*(a)| = ∞, a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I. Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded, we know (by the Bolzano-Weierstrass*
Theorem) that{*x**n*}has a convergent subsequence, say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed, we also know (by the

*Comparison Theorem) that a*∈

*I. In particular, f*(a) ∈

*R.*

*On the other hand, substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞,we have|*f*(a)| = ∞, a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I. Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded, we know (by the Bolzano-Weierstrass*
Theorem) that{*x**n*}has a convergent subsequence, say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed, we also know (by the

*Comparison Theorem) that a*∈

*I. In particular, f*(a) ∈

*R.*

On the other hand, *substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞, we have|*f*(a)| = ∞,a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I. Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded, we know (by the Bolzano-Weierstrass*
Theorem) that{*x**n*}has a convergent subsequence, say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed, we also know (by the

*Comparison Theorem) that a*∈

*I. In particular, f*(a) ∈

*R.*

*On the other hand, substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞,we have|*f*(a)| = ∞, a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I. Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded, we know (by the Bolzano-Weierstrass*
Theorem) that{*x**n*}has a convergent subsequence, say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed, we also know (by the

*Comparison Theorem) that a*∈

*I. In particular, f*(a) ∈

*R.*

*On the other hand, substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞, we have|*f*(a)| = ∞,a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I. Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded, we know (by the Bolzano-Weierstrass*
Theorem) that{*x**n*}has a convergent subsequence, say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed, we also know (by the

*Comparison Theorem) that a*∈

*I. In particular, f*(a) ∈

*R.*

*On the other hand, substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞, we have|*f*(a)| = ∞, a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I. Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded, we know (by the Bolzano-Weierstrass*
Theorem) that{*x**n*}has a convergent subsequence, say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed, we also know (by the

*Comparison Theorem) that a*∈

*I. In particular, f*(a) ∈

*R.*

*On the other hand, substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞, we have|*f*(a)| = ∞, a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Proof:**

Suppose first that f is not bounded on I. Then there exist
*x**n* ∈*I such that*

(7) |*f*(x*n*)| > *n,* *n*∈**N.**

*Since I is bounded, we know (by the Bolzano-Weierstrass*
Theorem) that{*x**n*}has a convergent subsequence, say
*x**n** _{k}* →

*a as k*→ ∞. Since I is closed, we also know (by the

*Comparison Theorem) that a*∈

*I. In particular, f*(a) ∈

*R.*

*On the other hand, substituting n**k* for n in (7) and taking
*the limit of this inequility as k* → ∞, we have|*f*(a)| = ∞, a
contradiction. Hence, the function f is bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

We have proved that both M and m are finite real
*numbers. To show that there is an x**M* ∈*I such that*
*f*(x*M*) =*M,suppose to the contrary that f*(x) <*M for all*
*x* ∈*I. Then the function*

*g(x) =* 1

*M*−*f*(x)
is continuous, hence, bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

We have proved that both M and m are finite real
numbers. *To show that there is an x**M* ∈*I such that*
*f*(x*M*) =*M, suppose to the contrary that f*(x) <*M for all*
*x* ∈*I.* Then the function

*g(x) =* 1

*M*−*f*(x)
is continuous, hence, bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

We have proved that both M and m are finite real
*numbers. To show that there is an x**M* ∈*I such that*
*f*(x*M*) =*M,suppose to the contrary that f*(x) <*M for all*
*x* ∈*I. Then the function*

*g(x) =* 1
*M*−*f*(x)
is continuous, hence, bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

We have proved that both M and m are finite real
*numbers. To show that there is an x**M* ∈*I such that*
*f*(x*M*) =*M, suppose to the contrary that f*(x) <*M for all*
*x* ∈*I.* Then the function

*g(x) =* 1
*M*−*f*(x)
is continuous, hence,bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

We have proved that both M and m are finite real
*numbers. To show that there is an x**M* ∈*I such that*
*f*(x*M*) =*M, suppose to the contrary that f*(x) <*M for all*
*x* ∈*I. Then the function*

*g(x) =* 1
*M*−*f*(x)
is continuous, hence, bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

*numbers. To show that there is an x**M* ∈*I such that*
*f*(x*M*) =*M, suppose to the contrary that f*(x) <*M for all*
*x* ∈*I. Then the function*

*g(x) =* 1
*M*−*f*(x)
is continuous, hence,bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

*numbers. To show that there is an x**M* ∈*I such that*
*f*(x*M*) =*M, suppose to the contrary that f*(x) <*M for all*
*x* ∈*I. Then the function*

*g(x) =* 1
*M*−*f*(x)
is continuous, hence, bounded on I.

WEN-CHINGLIEN **Advanced Calculus (I)**

*In particular, there is a C* >0 such that|*g(x*)| =*g(x*) ≤*C.*

It follows that

(8) *f*(x) ≤*M*− 1

*c*

*for all x* ∈*I. Taking the supremum of (8) over all x* ∈*I, we*
*obtain M* ≤*M*−1/C <*M, a contradiction. Hence, there*
*is an x**M* ∈*I, such that f*(x*M*) =*M.*A similar argument
*proves that there is an x**m* ∈*I such that f*(x*m*) = *m*. 2

WEN-CHINGLIEN **Advanced Calculus (I)**

In particular, *there is a C* >0 such that|*g(x*)| =*g(x*) ≤*C.*

It follows that

(8) *f*(x) ≤*M*− 1

*c*

*for all x* ∈*I.* *Taking the supremum of (8) over all x* ∈*I, we*
*obtain M* ≤*M*−1/C <*M, a contradiction. Hence, there*
*is an x**M* ∈*I, such that f*(x*M*) =*M.*A similar argument
*proves that there is an x**m* ∈*I such that f*(x*m*) = *m*. 2

WEN-CHINGLIEN **Advanced Calculus (I)**

*In particular, there is a C* >0 such that|*g(x*)| =*g(x*) ≤*C.*

It follows that

(8) *f*(x) ≤*M*− 1

*c*

*for all x* ∈*I. Taking the supremum of (8) over all x* ∈*I,*we
*obtain M* ≤*M*−1/C <*M, a contradiction. Hence, there*
*is an x**M* ∈*I, such that f*(x*M*) =*M.*A similar argument
*proves that there is an x**m* ∈*I such that f*(x*m*) = *m*. 2

WEN-CHINGLIEN **Advanced Calculus (I)**

*In particular, there is a C* >0 such that|*g(x*)| =*g(x*) ≤*C.*

It follows that

(8) *f*(x) ≤*M*− 1

*c*

*for all x* ∈*I.* *Taking the supremum of (8) over all x* ∈*I, we*
*obtain M* ≤*M*−1/C <*M,* a contradiction. Hence, there
*is an x**M* ∈*I, such that f*(x*M*) =*M.*A similar argument
*proves that there is an x**m* ∈*I such that f*(x*m*) = *m*. 2

WEN-CHINGLIEN **Advanced Calculus (I)**

*In particular, there is a C* >0 such that|*g(x*)| =*g(x*) ≤*C.*

It follows that

(8) *f*(x) ≤*M*− 1

*c*

*for all x* ∈*I. Taking the supremum of (8) over all x* ∈*I,*we
*obtain M* ≤*M*−1/C <*M, a contradiction.* Hence, there
*is an x**M* ∈*I, such that f*(x*M*) =*M.*A similar argument
*proves that there is an x**m* ∈*I such that f*(x*m*) = *m*. 2

WEN-CHINGLIEN **Advanced Calculus (I)**

*In particular, there is a C* >0 such that|*g(x*)| =*g(x*) ≤*C.*

It follows that

(8) *f*(x) ≤*M*− 1

*c*

*for all x* ∈*I. Taking the supremum of (8) over all x* ∈*I, we*
*obtain M* ≤*M*−1/C <*M,* a contradiction. Hence,there
*is an x**M* ∈*I, such that f*(x*M*) =*M.*A similar argument
*proves that there is an x**m* ∈*I such that f*(x*m*) = *m*. 2

WEN-CHINGLIEN **Advanced Calculus (I)**

*In particular, there is a C* >0 such that|*g(x*)| =*g(x*) ≤*C.*

It follows that

(8) *f*(x) ≤*M*− 1

*c*

*for all x* ∈*I. Taking the supremum of (8) over all x* ∈*I, we*
*obtain M* ≤*M*−1/C <*M, a contradiction.* Hence, there
*is an x**M* ∈*I, such that f*(x*M*) =*M.*A similar argument
*proves that there is an x**m* ∈*I such that f*(x*m*) = *m*. 2

WEN-CHINGLIEN **Advanced Calculus (I)**

*In particular, there is a C* >0 such that|*g(x*)| =*g(x*) ≤*C.*

It follows that

(8) *f*(x) ≤*M*− 1

*c*

*for all x* ∈*I. Taking the supremum of (8) over all x* ∈*I, we*
*obtain M* ≤*M*−1/C <*M, a contradiction. Hence,*there
*is an x**M* ∈*I, such that f*(x*M*) =*M.*A similar argument
*proves that there is an x**m* ∈*I such that f*(x*m*) = *m*. 2

WEN-CHINGLIEN **Advanced Calculus (I)**

*In particular, there is a C* >0 such that|*g(x*)| =*g(x*) ≤*C.*

It follows that

(8) *f*(x) ≤*M*− 1

*c*

*for all x* ∈*I. Taking the supremum of (8) over all x* ∈*I, we*
*obtain M* ≤*M*−1/C <*M, a contradiction. Hence, there*
*is an x**M* ∈*I, such that f*(x*M*) =*M.*A similar argument
*proves that there is an x**m* ∈*I such that f*(x*m*) = *m*. 2

WEN-CHINGLIEN **Advanced Calculus (I)**

*In particular, there is a C* >0 such that|*g(x*)| =*g(x*) ≤*C.*

It follows that

(8) *f*(x) ≤*M*− 1

*c*

*for all x* ∈*I. Taking the supremum of (8) over all x* ∈*I, we*
*obtain M* ≤*M*−1/C <*M, a contradiction. Hence, there*
*is an x**M* ∈*I, such that f*(x*M*) =*M.*A similar argument
*proves that there is an x**m* ∈*I such that f*(x*m*) = *m*. 2

WEN-CHINGLIEN **Advanced Calculus (I)**

**Lemma:**

[Sign-Preserving Property].
*Let f* :*I* →**R where I is an open, nondegenerate interval.**

*If f is continuous at a point x*0∈*I and f*(x0) >0, then there
are positive numberǫandδ such that|*x*−*x*0| < δ implies
*f*(x) > ǫ.

WEN-CHINGLIEN **Advanced Calculus (I)**

**Lemma:**

[Sign-Preserving Property].
*Let f* :*I* →**R where I is an open, nondegenerate interval.**

*If f is continuous at a point x*0∈*I and f*(x0) >0, then there
are positive numberǫandδ such that|*x*−*x*0| < δ implies
*f*(x) > ǫ.

WEN-CHINGLIEN **Advanced Calculus (I)**

Theorem (Intermediate Value Theorem)

*Let I be a nondegenerate interval and f* :*I* →**R be***continuous. If a,b*∈*I with a*<*b, and if y*0*lies between*
*f*(a)*and f*(b), then there is an x0∈ (*a,b)such that*
*f*(x0) =*y*0*.*

WEN-CHINGLIEN **Advanced Calculus (I)**

Theorem (Intermediate Value Theorem)

*Let I be a nondegenerate interval and f* :*I* →**R be***continuous. If a,b*∈*I with a*<*b, and if y*0*lies between*
*f*(a)*and f*(b), then there is an x0∈ (*a,b)such that*
*f*(x0) =*y*0*.*

WEN-CHINGLIEN **Advanced Calculus (I)**

**Example:**

**Assuming that sin x is continuous on R, prove that**

*f*(x) =
(

sin 1

*x* *x* 6=0

1 *x* =0

is continuous on (−∞,0)and(0, ∞), discontinuous at 0,
*and neither f*(0+)*nor f*(0−)exists. (see Figure 3.1 on
p.61.)

WEN-CHINGLIEN **Advanced Calculus (I)**

**Example:**

**Assuming that sin x is continuous on R, prove that**

*f*(x) =
(

sin 1

*x* *x* 6=0

1 *x* =0

is continuous on (−∞,0)and(0, ∞), discontinuous at 0,
*and neither f*(0+)*nor f*(0−)exists. (see Figure 3.1 on
p.61.)

WEN-CHINGLIEN **Advanced Calculus (I)**

**Example:**

**The Dirchlet function is defined on R by**

*f*(x) :=

1 *x* ∈**Q**
0 *x* ∈**Q**

*Prove that every point x* ∈**R is a point of discontinuity of f.**

*(Such functions are called nowhere continuous.)*

WEN-CHINGLIEN **Advanced Calculus (I)**

**Example:**

**The Dirchlet function is defined on R by***f*(x) :=

1 *x* ∈**Q**
0 *x* ∈**Q**

*Prove that every point x* ∈**R is a point of discontinuity of f.**

*(Such functions are called nowhere continuous.)*

WEN-CHINGLIEN **Advanced Calculus (I)**

**Example:**

Prove that the function

*f*(x) =

1

*q* *x* = *p*

*q* ∈**Q** (in reduced form)
0 *x* ∈**Q.**

is continuous at every irrational in the interval (0,1)but discontinuous at every rational in (0,1).

WEN-CHINGLIEN **Advanced Calculus (I)**

**Example:**

Prove that the function

*f*(x) =

1

*q* *x* = *p*

*q* ∈**Q** (in reduced form)
0 *x* ∈**Q.**

is continuous at every irrational in the interval (0,1)but discontinuous at every rational in (0,1).

WEN-CHINGLIEN **Advanced Calculus (I)**

*Thank you.*

WEN-CHINGLIEN **Advanced Calculus (I)**