Chapter 1: Functions, Limit and Continuity

AM
PH
Duong T. PHAM

T.
g. CALCULUS I
on
Du

Duong T. PHAM 1 / 123

Outline

1 Lines and Planes

AM
2 Functions

PH
3 Limit of functions

4 Left-hand and right-hand limits

T.
5 Limits at infinity g.
on
6 Continuity
Du

7 Bounded Functions

8 Parametric Equations and polar coordinates

Duong T. PHAM 2 / 123

Lines

Definition 1.

AM
Let L be a line which passes through a point P0 (x0 , y0 , z0 ) and parallel to
a vector v = (a, b, c). Then the parametric equation of L is

PH
     
x x0 a
y  = y0  + t b  , t ∈ R;

T.
z z0 c
or
g.
on

x = x0 + ta

Du

y = y0 + tb , t ∈ R.

z = z0 + tc


Duong T. PHAM 3 / 123

Lines

Ex.
1 Find a parametric equation of the line L which passes through the

AM
point P0 (5, 1, 3) and is parallel to vector v = (1, 4, −2),
Find two other points on L.

PH
2

Sol.

T.
1 The parametric equation of L is
g.
   
x 5
 
1
on
y  = 1 + t  4 
z 3 −2
Du

2 Choose t1 = 1, we have P1 (6, 5, 1) ∈ L

Choose t2 = 2, we have P2 (7, 9, −1) ∈ L

Duong T. PHAM 4 / 123

Lines

Parametric equation of L
 
x = x0 + ta
 x − x0 = ta


AM
y = y0 + tb ⇐⇒ y − y0 = tb
 
z = z0 + tc z − z0 = tc
 

PH
T.
If none of a, b, c is 0, we obtain
x − x0
g. y − y0 z − z0
= =
a b c
on
Du

It is called symmetric equations of L

If e.g. a = 0, then
y − y0 z − z0
x = x0 , =
b c

Duong T. PHAM 5 / 123

Lines
Ex. Line L passes through points A(2, 4, −3) and B(3, −1, 1).
1 Find parametric and symmetric equations of L.
2 Find the intersection between L and xy -plane.

AM
Ans.
−→

PH
1 AB = (1, −5, 4): directional vector of L.

Parametric equ.

T.
symmetric equ.
     
x 2 1
g.
y  =  4  + t −5 x −2 y −4 z +3
= =
on
1 −5 4
z −3 4
Du

2 xy -plane is z = 0. Intersection between L and xy -plane is

x −2 y −4 3
= =
1 −5 4

Duong T. PHAM 6 / 123

Line Segment
Proposition 1.
Given two points A(a1 , a2 , a3 ) and B(b1 , b2 , b3 ). The line segment from A
to B is

AM
     
x1 a1 b1
x2  = (1 − t)a2  + t b2  0 ≤ t ≤ 1

PH
x3 a3 b3
−→

T.
Proof. AB = (b1 − a1 , b2 − a2 , b3 − a3 ). Let X (x1 , x2 , x3 ) be a point in
−→ −→
line segment AB. Then AX = t AB for 0 ≤ t ≤ 1. We have
g.
−→
AX = (x1 − a1 , x2 − a2 , x3 − a3 ). Thus
on
         
Du

x1 − a1 b1 − a1 x1 a1 b1
x2 − a2  = t b2 − a2  =⇒ x2  = (1 − t) a2  + t b2 
x3 − a3 b3 − a3 x3 a3 b3


Duong T. PHAM 7 / 123
Circles and Disks
y

P(x, y ) Let C (a, b). Let P(x, y ) belong to the

r circle centered at C (a, b) and radius r .
C
b Then

AM
CP 2 = r 2

PH
⇐⇒ (x − a)2 + (y − b)2 = r 2
a x
0

T.
The equation of the circle with center at C (a, b) and radius r is
g.(x − a)2 + (y − b)2 = r 2
on
The set of all points lie inside a circle is called the interior of the circle (an
open disk) .
Du

The set of all points inside a circle together with the circle itself is said to be
a closed disk (or a disk) and is represented by
(x − a)2 + (y − b)2 ≤ r 2

Duong T. PHAM 8 / 123

Equations of Parabolas
We consider the set of points that
y
are equidistant from the point
F (0, p) and the straight line y = L.

AM
If a point P(x, y ) satisfies the
F (0, p) above condition then PF = PH.
P

PH
x 2 + (y − p)2 = (y − L)2
0 x
x2 p 2 − L2

T.
H y =L ⇐⇒ y = +
2(p − L) 2(p − L)
g.
on
Definition 2.
Du

A parabola is a plane curve whose points are equidistant from a point F

and a straight line L which does not pass the point F .
The point F is the focus of the parabola. The line L is the directrix of
the parabola.
The line through F and perpendicular to L is the parabola’s axis .
Duong T. PHAM 9 / 123
Ellipses
y

minor axis

AM
major axis a x
0

PH
T.
Definition 3. g.
on
If a, b > 0, the equation x2 y2
+ 2 =1
Du

a 2 b
represents a curve (called ellipse ) that lies inside the rectangle
[−a, a] × [−b, b].
The line segments connecting (−a, 0) with (a, 0) and (0, −b) with (0, b) are
called principal axes of the ellipse.
Duong T. PHAM 10 / 123
Hyperbolas
y
x
a
+ yb = 0

AM
−a a

PH
0 x

T.
x
a
− yb = 0
g.
on
Definition 4.
Du

If a, b > 0, the equation x2 y2

− =1
a2 b2
represents a curve (called hyperbola ) that has center at the origin and
passes through (a, 0) and (−a, 0).
x y x y
The two asymptotes have equations a + b = 0 and a − b =0
Duong T. PHAM 11 / 123
Functions

Definition 5.
Let A and B be sets. A function f : A → B is an assignment of exactly one

AM
element of B to each element of A.

PH
Ex: Let A = {Tom, Hai, Thao, David, Nam} and B = {1, 2, 3, 4, 5}. The grades
of students in A are given by

T.
Tom 1
Hai
g.
2
The assigment grade of the students is
on
a function.
3
Du

Thao
grade(Tom) = 2; grade(Hai) = 2;
grade(Thao) = 4; grade(David) = 3;
David 4
grade(Nam) = 1.
Nam 5

Duong T. PHAM 12 / 123

Domain, Range, Image and Pre-image

Definition 6.
Let f : A → B be a function. We call f maps A to B and

AM
The set A is called the domain of f . The set B is called the codomain
If f (a) = b, then b is called the image of a and a is called the pre–image

PH
of b.
The set f (A) = {f (a)| a ∈ A} is called the range of f .

T.
grade
g.
on
Tom 1 {Tom, Hai, Thao, David, Nam} = domain
Du

Hai 2 {1, 2, 3, 4, 5} = codomain

Thao 3 range(grade) = {grade(Tom), grade(Hai),
grade(Thao), grade(David), grade(Nam)}
David 4
= {2, 2, 4, 3, 1} = {1, 2, 3, 4}
Nam 5

Duong T. PHAM 13 / 123

Domain of a function

Remark: If a function is given by a formula and the domain is not stated

explicitly, the convention is that the domain is the set of all numbers for

AM
which the formula makes sense and defines a real number .

PH
√
Ex: Find the domain of the function f (x) = x + 1.

T.
√
Ans: The formula x + 1 is well-defined when x + 1 ≥ 0, which is
equivalent to x ≥ −1. g.
on
The domain of the above function is
Du

D = [−1, +∞)

Duong T. PHAM 14 / 123

Example

Ex: We need a rectangular storage container with an open top which has
a volume of 10m3 . The length of its base is required to be twice its width.

AM
Material for the base costs $10/m2 ; material for the sides costs $6/m2 .
Express the cost of materials as a function of the width of the base.

PH
T.
g.
on
h
Du

w
2w

Duong T. PHAM 15 / 123

Example

Area of the base = w ∗ (2w ) = 2w 2

⇒ Cost = 10 ∗ (2w 2 ) = 20w 2 $

AM
Area of the front and back sides
h = 2 ∗ (2hw ) = 4hw

PH
w Area of the left and right sides
2w = 2 ∗ (hw ) = 2hw

T.
⇒ Area of 4 sides = 6hw ⇒ Cost = 6 ∗ (6hw ) = 36hw $
g.
⇒ Total cost = 20w 2 + 36hw ($)
on
Volume = 10 ⇒ 2hw 2 = 10 ⇒ h = 5/w 2
Du

180
⇒ Total cost: C (w ) = 20w 2 + w

Duong T. PHAM 16 / 123

The Vertical Line Test
The graph of a function is a curve in xy -plane.

Question : Which curves in the xy -plane are graphs of functions?

AM
Vertical Line Test: A curve in the xy -plane is the graph of a function of
x if and only if no vertical line intersects the curve more than once.

PH
y

T.
y

g. b
on
a x
Du

a
x c

f (a) = b or f (a) = c?
Duong T. PHAM 17 / 123
Piecewise defined functions
(
1−x if x ≤ 1
Ex: A function f is defined by: f (x) =
x2 if x > 1.
f (−1) = ? 1 − (−1) = 2

AM
f (2) =? 22 = 4
f (1) =? 1 − 1 = 0

PH
y
4

T.
g.
on
2
Du

−1 1 2 x
Duong T. PHAM 18 / 123
Piecewise defined functions

Ex: Sketch the graph of the absolute value function f (x) = |x|

Ans: We have

AM
(
x if x ≥ 0
f (x) = |x| =

PH
−x if x < 0

T.
y
g.
on
Du

Duong T. PHAM 19 / 123

Symmetry: Even Functions
y

4
f (−1) =? (−1)2 = 1;
f (1) =? 12 = 1;

AM
⇒ f (−1) = f (1)
f (−2) = (−2)2 = 4 and

PH
f (2) = 22 = 4
⇒ f (−2) = f (2)

T.
1
f (−x) = (−x 2 ) = x 2
g. f (x) = x 2 ⇒ f (−x) = f (x)
on
x
−2 −1 1 2
f is an even function
Du

Graph of function f (x) = x2

Definition 2.1.
A function f : D → R is said to be even if f (−x) = f (x) ∀x ∈ D

Duong T. PHAM 20 / 123

Symmetry: Odd Functions

f (x)

AM
−x
x

PH
f (−x)

T.
g.
on
f (−x) = −f (x) ∀x ∈ D and f is said to be an odd function
Du

Definition 2.2.
A function f : D → R is said to be odd if f (−x) = −f (x) ∀x ∈ D

Duong T. PHAM 21 / 123

Symmetry: Examples
Ex: Determine whether each of the following functions is even, odd, or
neither even nor odd

f (x) = x 3 − x; g (x) = 1 + x 2 ; h(x) = x + 1.

AM
Ans:

PH
(i) f (−x) = (−x)3 − (−x) = −x 3 + x = −(x 3 − x) = −f (x)
⇒ f is an odd function.

T.
g.
on
f (x)
Du

−x
x
f (−x)

Duong T. PHAM 22 / 123

Symmetry: Examples
Ans: g (x) = 1 + x 2 ;
g (−x) = 1 + (−x)2 = 1 + x 2 = g (x)
⇒ g is an even function.

AM
PH
T.
g.
on
g (−x) g (x)
Du

−x x

Duong T. PHAM 23 / 123

Symmetry: Examples
Ans: h(x) = 1 + x;
h(−x) = 1 + (−x) = 1 − x ⇒ h(−x) 6= h(x) and h(−x) 6= −h(x)
⇒ h is NOT either even or odd.

AM
y

PH
T.
g.
on
Du

Duong T. PHAM 24 / 123

Functions

Definition 7.
Let f , g : A → R. Then f + g and fg are functions from A to R defined by

AM
(f + g )(x) = f (x) + g (x)

PH
(fg )(x) = f (x)g (x)

T.
Ex: Let f , g : R → R be given by f (x) = x 2 and g (x) = x − x 2 . What are the
functions f + g and fg ? g.
on
Ans: f + g and fg are functions from R to R and
Du

(f + g )(x) = f (x) + g (x) = x 2 + (x − x 2 ) = x

(fg )(x) = f (x)g (x) = x 2 (x − x 2 ) = x 3 − x 4 .

Duong T. PHAM 25 / 123

Functions
Definition 8.
Let f : A → B be a function and S ⊂ A. The image of S, f (S), is a subset of B
given by

AM
f (S) = {f (x)| x ∈ S}

PH
Ex: Let A = {a, b, c, d, e} and B = {1, 2, 3, 4} and a function f given by
f (a) = 2, f (b) = 1, f (c) = 4, f (d) = 1 and f (e) = 1. Find the image f (S) of

T.
the subset S = {b, c, d}.
Ans: Exercise g.
on
Du

Duong T. PHAM 26 / 123

Composition
Definition 9.
Let f : A → B and g : B → C . The composition g ◦ f : A → C is defined by

AM
g ◦ f (a) = g (f (a)) , a ∈ A.

Ex: Let f : {a, b, c, d, e} → {1, 2, 3, 4, 5} be given by f (a) = 4, f (b) = 5,

PH
f (c) = 2, f (d) = 1, f (e) = 3, and the function
g : {1, 2, 3, 4, 5} → {Tom, Jerry, Donald, Hugo} be given by g (1) = Tom,

T.
g (2) = Hugo, g (3) = Jerry, g (4) = Tom, g (5) = Donald. Find g ◦ f ?
Ans: g.
on
f g
a 1 Tom
Du

b 2
Jerry
c 3
Donald
d 4
e 5 Hugo

Duong T. PHAM 27 / 123

Composition

f g
a 1 Tom
b 2
Jerry

AM
c 3
Donald
d 4

PH
e 5 Hugo

T.
The composition g ◦ f : {a, b, c, d, e} → {Tom, Jerry, Donald, Hugo} is defined
by g.
on



g ◦ f (a) = g f (a) = g 4 = Tom
Du

g ◦ f (b) = g f (b) = g 5 = Donald



g ◦ f (c) = g f (c) = g 2 = Hugo



g ◦ f (d) = g f (d) = g 1 = Tom



g ◦ f (e) = g f (e) = g 3 = Jerry

Duong T. PHAM 28 / 123

Composition

Ex: Let f : R → R and g : R → R be defined by

AM
f (x) = 2x + 3 and g (x) = 3x + 2.

PH
Find g ◦ f and f ◦ g .

T.
Ans: We have g ◦ f : R → R and

g.


g ◦ f (x) = g f (x) = g 2x + 3 = 3 2x + 3 + 2 = 6x + 11.
on
The composition f ◦ g : R → R is given by
Du

f ◦ g (x) = f g (x) = f 3x + 2 = 2 3x + 2 + 3 = 6x + 7.

Duong T. PHAM 29 / 123

One–to–one Functions

Definition 10.
A function f : A → B is said to be one-to-one (or injective) if and only if

AM
f (x) = f (y ) for any x, y ∈ A implies x = y

PH
Ex: Let f : {a, b, c, d} → {1, 2, 3, 4, 5} be a function given by f (a) = 4,

T.
f (b) = 5, f (c) = 1 and f (d) = 3. Determine if f is one-to-one.
Ans:
a
f
g.
1
on
2 The function f is one-to-one since there are no
Du

b
3 two different elements in {a, b, c, d} whose have
c the same image.
4
d 5

Duong T. PHAM 30 / 123

One-to-one functions

a 1
a 1 b 2
c

AM
b 2 3
c 3 d 4
d 4

PH
g
D E

T.
f
D E
g (b) = g (c)
g.
f is one-to-one and g is NOT one-to-one
on
Du

Remark: A function f is one-to-one if

f (x1 ) = f (x2 ) =⇒ x1 = x2

Duong T. PHAM 31 / 123

Horizontal line Test

Ex:
y f is not one-to-one

AM
PH
y = f (x)
f (x1 )
||

T.
f (x2 )
g.
on
Du

x1 x2 x

Horizontal line test: A function is one-to-one if and only if NO

horizontal line intersects its graph more than once.

Duong T. PHAM 32 / 123

One-to-one Functions
Ex: Consider the function f : R → R defined by f (x) = x 2 . Determine if f is
one-to-one.
Ans: The function f is not one-to-one since we have 1 6= −1 but

AM
f (1) = f (−1) = 12 .

PH
y = x2

T.
g.
on
Du

−1 1 x
Duong T. PHAM 33 / 123
One-to-one Functions
Ex: Determine whether the function f (x) = x + 1 is one-to-one or not.
Ans: Suppose that x1 , x2 ∈ R satisfy f (x1 ) = f (x2 ). Then we have
f (x1 ) = f (x2 ) ⇐⇒ x1 + 1 = x2 + 1

AM
=⇒ x1 = x2 .

PH
Hence, f (x) = x + 1 is one-to-one.

y =x +1

T.
y
g.
on
Du

Duong T. PHAM 34 / 123

Increasing and decreasing Functions

Definition 11.
Let A ⊂ R and f : A → R. Then

AM
If for any x, y ∈ A and x < y , there holds f (x) < f (y ), then f is said to be
strictly increasing

PH
If for any x, y ∈ A and x < y , there holds f (x) > f (y ), then f is said to be
strictly decreasing

T.
Ex: Consider the function f (x) = x 3 . Prove that f is strictly increasing.
g.
Ans: Let x, y ∈ R and x < y . We consider the difference f (x) − f (y ) as follows
on
f (x) − f (y ) = x 3 − y 3 = (x − y )(x 2 + xy + y 2 )
Du

y 2 3y 2
 
= (x − y ) (x + ) + < 0.
| {z } 2 4
<0 | {z }
>0

Hence, f (x) < f (y ) whenever x < y . The function f (x) = x 3 is increasing.

Duong T. PHAM 35 / 123
Onto Functions (surjective functions)

Definition 12.
Let f : A → B be a function. If for any b ∈ B, there is an a ∈ A such that

AM
f (a) = b, then f is said to be onto (or surjective). In this case, f is called a
surjection.

PH
Remark: f : A → B is surjective iff ∀(b ∈ B) ∃(a ∈ A) (f (a) = b)

T.
Ex:
a
f
1
g.
on
b 2
Du

This function is NOT surjective since there is

3 NO element in {a, b, c, d} whose image is 2.
c
4
d 5

Duong T. PHAM 36 / 123

Surjective Functions

AM
Ex: Prove that the function f : R → R defined by f (x) = x 3 + 1 is surjective.
√

PH
Ans: For any y ∈ R, there is x = 3 y − 1 ∈ R satisfying
p  p 3

T.
3
f (x) = f y −1 = 3 y −1 +1
= y − 1 + 1 = y.
g.
on
Hence, f is surjective.
Du

Duong T. PHAM 37 / 123

Bijective functions

Definition 13.

AM
A function f : A → B is said to be bijective if it is injective and surjective.

PH
Ex:
The function f is injective since for all

T.
f
a 1 x1 , x2 ∈ {a, b, c, d, e} such that x1 6= x2 ,
b 2 g. there holds f (x1 ) 6= f (x2 ).
on
c 3 The function f is surjective since for all
y ∈ {1, 2, 3, 4, 5}, there is a
Du

d 4
x ∈ {a, b, c, d, e} such that f (x) = y .
e 5
Hence, f is bijective.

Duong T. PHAM 38 / 123

Identity function

AM
PH
Definition 14.
Let A be a set. The identity function defined on A is given by

T.
iA (x) = x ∀x ∈ A.
g.
on
Du

Duong T. PHAM 39 / 123

Inverse function
Definition 15.
Let f : A → B be a bijective function. The inverse function of f , denoted by
f −1 , is defined by

AM
f −1 (b) := a ∀b ∈ B if f (a) = b
A function f is said to be invertible if it has an inverse

PH
Ex: Consider the bijective function f : {a, b, c, d, e} → {1, 2, 3, 4, 5} given by

T.
f (a) = 4, f (b) = 5, f (c) = 2, f (d) = 1, and f (e) = 3. Find f −1 ?
Ans: The inverse f −1 : {1, 2, 3, 4, 5} → {a, b, c, d, e} is given by f −1 (1) = d,
g.
f −1 (2) = c, f −1 (3) = e, f −1 (4) = a, and f −1 (5) = b.
on
f f −1
a a
Du

1 1
b 2 2 b
c 3 3 c
d 4 4 d
e 5 5 e
Duong T. PHAM 40 / 123
Inverse function

Remark: Let f be a one-to-one function with domain A and range B.

Then its inverse function f −1 has domain B and range A and is defined
by

AM
f −1 (y ) = x ⇐⇒ f (x) = y ∀y ∈ B.

PH
a 1 1 a
b 2 2 b

T.
c 3 3 c
Ex: d 4 4 d
g.
on
D
f
E E f −1 D
Du

f −1 ◦ f (a) = ? f −1 (f (a)) = f −1 (1) = a

f ◦ f −1 (3) = ? f (f −1 (3)) = f (c) = 3.

Duong T. PHAM 41 / 123

Cancellation equations

Let f be a one-to-one function with domain A and range B.

AM
f −1 (f (x)) = x ∀x ∈ A

PH
−1
f (f (y )) = y ∀y ∈ B

T.
g.
on
x f f (x) f −1 x
Du

Duong T. PHAM 42 / 123

Find an inverse

To find the inverse of a one-to-one function f :

Step 1: Write y = f (x)

AM
Step 2: Solve this equation for x in terms of y
Step 3: To obtain f −1 as a function of x, interchange x and y .

PH
The resulting equation is y = f −1 (x).

T.
Ex: Find the inverse of y = 3x 3 + 5
Ans: g.
on
y = 3x 3 + 5
Du

q
y −5
=⇒ 3x 3 = y − 5 =⇒ x = 3
3
q
x−5
y= 3
3 is the inverse of y = 3x 3 + 5.

Duong T. PHAM 43 / 123

Graphs
Definition 16.
Let f : A → B be a function. The graph of f is the set

AM
{(a, b)| a ∈ A and b = f (a)}

PH
Ex: Find the graph of the function f : R → R defined by f (x) = 2x + 1.
Ans: The graph of f is {(x, 2x + 1) | x ∈ R}

T.
y
g.
on
Du

1
-1
x

Duong T. PHAM 44 / 123

Graphs

Ex: Find the graph of the following function

AM
f
a 1
2

PH
b
3
c

T.
4
d 5
g.
on
Ans: The graph of f is
Du

{(x, f (x)) | x ∈ {a, b, c, d}} = {(a, f (a)), (b, f (b)), (c, f (c)), (d, f (d))}
= {(a, 4), (b, 5), (c, 1), (d, 3)}

Duong T. PHAM 45 / 123

Graphs
Ex: Consider f : Z → Z given by f (x) = x 2 . Find the graph of f .
Ans: The graph of f is
{(x, f (x)) | x ∈ Z} = {(x, x 2 ) | x ∈ Z}

AM
y

PH
9

T.
g.
on
Du

−3 −2 −1 1 2 3 x
Duong T. PHAM 46 / 123
Transformation of functions

Vertical and horizontal shifts: Suppose c > 0. To obtain the graph

AM
of
y = f (x) + c , shift the graph of y = f (x) a distance c units

PH
upward
y = f (x) − c , shift the graph of y = f (x) a distance c units

T.
downward
g.
y = f (x + c) , shift the graph of y = f (x) a distance c units to
on
the left
Du

y = f (x − c) , shift the graph of y = f (x) a distance c units to

the right

Duong T. PHAM 47 / 123

Vertical and horizontal shifts

y = f (x) + c

AM
PH
c y = f (x)
y = f (x + c) y = f (x − c)

T.
c
g. c
on
c
Du

y = f (x) − c

Duong T. PHAM 48 / 123

Vertical and horizontal stretching and reflecting

Vertical and horizontal Stretching and Reflecting: Suppose c > 1.

To obtain the graph of

AM
y = cf (x) , stretch the graph of y = f (x) vertically by a factor of c

PH
y = (1/c)f (x) , compress the graph of y = f (x) vertically by a
factor of c

T.
y = f (cx) , compress the graph of y = f (x) horizontally by a
factor of c g.
y = f (x/c) , stretch the graph of y = f (x) horizontally by a factor
on
of c
Du

y = −f (x) , reflect the graph of y = f (x) about the x-axis

y = f (−x) , reflect the graph of y = f (x) about the y -axis

Duong T. PHAM 49 / 123

Vertical and horizontal stretching and reflecting

Ex:

y y = 2 cos x

AM
2 y = cos x

PH
1
y = (0.5) ∗ cos x

T.
1/2

g. x
on
Du

Duong T. PHAM 50 / 123

Vertical and horizontal stretching and reflecting

Ex:
y

AM
y = cos 12 x

PH
1

T.
g. x
on
Du

y = cos x
y = cos 2x

Duong T. PHAM 51 / 123

Graphs of inverse functions

y y = f −1 (x)

AM
Suppose f (a) = b ⇒(a, b) ∈
graph of f a

PH
⇒f −1 (b) = a ⇒ (b, a) ∈ graph y = f (x)

of f −1 b

T.
The graph of f −1 is symmetric
to that of f g.
on
b a x
Du

The graph of f −1 is obtained by reflecting the graph of f about the line

y =x

Duong T. PHAM 52 / 123

Exercises

AM
PH
I.6:

T.
3–12; 15–18; 21–24
g.
on
Du

Duong T. PHAM 53 / 123

Floor and ceiling functions

Definition 17.

AM
The floor function assigns each real number to the largest integer that is
smaller than the real number itself.

PH
The ceiling function assigns each real number to the smallest integer that
is larger than the real number itself.

T.
The values of the floor and ceiling functions at x are denoted by bxc and
dxe, resp. g.
on
Du

Ex:
b5.2c = 5, d5.2e = 6,
b−1.2c = −2, d−1.2e = −1.

Duong T. PHAM 54 / 123

Floor and ceiling functions

Some properties of floor and ceiling functions (n is integer)

AM
bxc = n iff n ≤ x < n + 1
dxe = n iff n − 1 < x ≤ n

PH
bxc = n iff x − 1 < n ≤ x
dxe = n iff x ≤ n < x + 1

T.
x − 1 ≤ bxc ≤ x ≤ dxe < x + 1
g. b−xc = −dxe
on
d−xe = −bxc
Du

bx + nc = bxc + n
dx + ne = dxe + n

Duong T. PHAM 55 / 123

Linear Models
Def: A function y = f (x) is linear if its graph is a straight line. The
formula of a linear function has the following formula

y = ax + b,

AM
where a is the slope of the line and b is the y -intercept.

PH
Ex: y = − 21 x + 2

T.
y
g.
on
2
Du

0 4 x

Duong T. PHAM 56 / 123

Polynomials
Def: A function P is called a polynomial if

P(x) = an x n + an−1 x n−1 + . . . + a1 x + a0

AM
where

PH
n: nonnegative integer,
a0 , a1 , . . . , an are coefficients

T.
The domain of P is R = (−∞, ∞). If an 6= 0, the degree of P is n.
Ex:
g.
on
A linear function y = ax + b is a polynomial of degree 1,
Du

A polynomial of degree 2 has the form y = ax 2 + bx + c and is

called a quadratic function,
A polynomial of degree 3 has the form y = ax 3 + bx 2 + cx + d and
is called a cubic function.
Duong T. PHAM 57 / 123
Power Functions

A function of the form y = x α where α is a constant is called a power

function.

AM
Remark: Note here that α is a real number.
If α = n where n is a positive integer, then y = x n is a polynomial

PH
of degree n;
The domain of y = x n is R.

T.
If α = 1/n where n is a positive integer, then y = x 1/n is called a
root function.
√
g.
on
Note: y = n x ⇒ y n = x and
Du

(
√n
R+ = [0, ∞) if n is even
domain of y = x is
R if n is odd

Duong T. PHAM 58 / 123

Power Functions
√
n
y= x
y
y √
y = 3x

AM
√
y = x

PH
x

T.
x
y = x −1 = x1 ; domain is R\{0}
g.
on
Du

Duong T. PHAM 59 / 123

Rational Functions

Def: A rational function f is a ratio of two polynomials:

P(x)
f (x) =

AM
Q(x)

PH
where P and Q are polynomials.

Domain of f is

T.
D = {x ∈ R : Q(x) 6= 0}
g.
on
x
Ex: f (x) = is a rational function and the domain
x 2 − 3x + 2
Du

D = {x ∈ R : x 2 − 3x + 2 6= 0}
= {x ∈ R : x 6= 1 and x 6= 2}

Duong T. PHAM 60 / 123

Algebraic Functions

Def: A function f is called an algebraic function if it can be constructed

AM
using algebraic operations (such as addition, subtraction, multiplication,
division, and taking roots) starting with polynomials

PH
Ex:

T.
f (x) = an x n + . . . + a1 x + a0 and g (x) = bm x m + . . . + b1 x + b0 are
algebraic functions g. √
on
f + g , f − g , f ∗ g , f /g and k f are algebraic functions
Du

1
h(x) = √
k n
x +1
is an algebraic function

Duong T. PHAM 61 / 123

Trigonometric Functions

y = sin x
1

AM
−π π 2π

PH
−2π 0

T.
−1

y = cos x g.
on
1
Du

−π π 2π
−2π 0

−1

Duong T. PHAM 62 / 123

Exponential Functions

Def: Exponential functions are the functions of the form f (x) = ax

where a > 0.
The domain = R

AM
The range = {positive numbers}

PH
y y

T.
g.
on
Du

y = 2x y = ( 12 )x

x x

Duong T. PHAM 63 / 123

Logarithmic Functions
Def: The logarithmic functions f (x) = loga x, where the base a is a posi-
tive constant, are the inverse functions of the exponential functions.
Domain = (0, ∞)

AM
Range = (−∞, ∞)

PH
y

T.
g.
y = log2 x
on
Du

Duong T. PHAM 64 / 123

Transcendental Functions

Def: Transcendental functions are functions that are NOT algbraic

functions.

AM
Ex:
Trigonometric functions and their inverses are transcendental

PH
functions,
Exponential and logarithmic functions are transcendental functions.

T.
Ex: Classify the following functions as one of the types of functions that
g.
we have discussed.
on
1 3x −→ exponential function,
Du

2 x 5 −→ power function (or polynomial of degree 5)

1+x
3 √
1− x
−→ algebraic function,
4 1 − x − x 4 −→ polynomial of degree 4.

Duong T. PHAM 65 / 123

Limit of a function
(
2x − 1 if x 6= 3
Consider f (x) =
6 if x = 3

AM
y
x f (x) |5 − f (x)|

PH
2.99 4.98 0.02 5+
5
2.999 4.998 0.002 5−

T.
2.9999 4.9998 0.0002
2.99999 4.99998 0.00002
2.999999 4.999998
g.
0.000002
on
2.9999999 4.9999998 0.0000002
Du

0 3−δ 3 3+δ x
We say: f converges to 5 as x goes to 3, and we write
lim f (x) = 5
x→3

Duong T. PHAM 66 / 123

Limit of a function

AM
Definition 18.

PH
Let f be a function defined on some open intervals that contains a, except
possibly at a itself. Then
lim f (x) = L

T.
x→a

if for every  > 0, there is a number δ > 0 such that

g.
on
if 0 < |x − a| < δ then |f (x) − L| < .
Du

Duong T. PHAM 67 / 123

Limit of a function

Ex: Prove that lim (4x − 5) = 7

x→3
Ans:

AM
Guessing δ: given an arbitrary  > 0, we need to find δ > 0 s.t.
if 0 < |x − 3| < δ then |(4x − 5) − 7| < .

PH
We have

T.
|(4x − 5) − 7| <  ⇔ 4 |x − 3| < .
We choose δ = /4. g.
on
Proof: For any  > 0, there is δ = /4 satisfying that if
0 < |x − 3| < δ , then
Du

0 < |x − 3| < /4 ⇔ |(4x − 5) − 7| < .

This conclude that lim (4x − 5) = 7

x→3

Duong T. PHAM 68 / 123

Limit of a function
Ex: Prove that lim (x 2 − 1) = 0
x→1
Ans:
Guessing δ: given an arbitrary  > 0, we need to find δ > 0 s.t.

AM
if 0 < |x − 1| < δ then (x 2 − 1) − 0 < .
 

PH
 
We have (x 2 − 1) − 0 = |x − 1| |x + 1| <  (we want) .
We may choose δ = min{1/4, /3}.

T.
Proof: For any  > 0, there is δ = min{1/4, /3} satisfying that if
0 < |x − 1| < δ , then |x − 1| < δ ≤ 1/4 ⇒ 3/4 < x < 5/4 ⇒
g.
7/4 < x + 1 < 9/4 ⇒ 7/4 < |x + 1| < 9/4 .
on
Du

Hence, if 0 < |x − 1| < δ = min{1/4, /3}, we have

(x − 1) − 0 = |x − 1| |x + 1| < δ |x + 1| <  9 < .

 2 
34
This conclude that lim (x 2 − 1) = 0
x→1
Duong T. PHAM 69 / 123
Limit Laws

Limit Laws: Suppose that lim f (x) and lim g (x) exist and let c ∈ R. Then
x→a x→a

AM
(i) lim [f (x) + g (x)] = lim f (x) + lim g (x)
x→a x→a x→a

PH
(ii) lim [f (x) − g (x)] = lim f (x) − lim g (x)
x→a x→a x→a

T.
(iii) lim [cf (x)] = c lim f (x)
x→a x→a
g.
(iv) lim [f (x)g (x)] = lim f (x) · lim g (x)
on
x→a x→a x→a
Du

f (x) lim f (x)

(v) lim = x→a if lim g (x) 6= 0
x→a g (x) lim g (x) x→a
x→a

Duong T. PHAM 70 / 123

Limit Laws
Proof:
(i) Denote L := lim f (x) and M := lim g (x).
x→a x→a
Let  > 0 be an arbitrarily small positive number .

AM
Since L := lim f (x), there is a δ1 > 0 s.t.
x→a

PH
if 0 < |x − a| < δ1 then |f (x) − L| < /2. (1)

T.
Since M := lim g (x), there is a δ2 > 0 s.t.
x→a

if 0 < |x − a| < δ2 then |g (x) − M| < /2.

g. (2)
on
We choose δ = min{δ1 , δ2 }. From (1) and (2), there holds:
Du

if 0 < |x − a| < δ then

|(f (x) + g (x)) − (L + M)| ≤ |f (x) − L| + |g (x) − M|
</2 + /2 = .
Hence, lim [f (x) + g (x)] = L + M.
x→a

Duong T. PHAM 71 / 123

Limit Laws
Proof:
(ii) Let  > 0 be an arbitrarily small positive number .
Since L := lim f (x), there is a δ1 > 0 s.t.

AM
x→a

if 0 < |x − a| < δ1 then |f (x) − L| < /2. (3)

PH
Since M := lim g (x), there is a δ2 > 0 s.t.
x→a

T.
if 0 < |x − a| < δ2 then |g (x) − M| < /2. (4)
g.
We choose δ = min{δ1 , δ2 }. From (3) and (4), there holds:
on
if 0 < |x − a| < δ then
Du

|(f (x) − g (x)) − (L − M)| ≤ |(f (x) − L) − (g (x) − M)|

≤ |f (x) − L| + |g (x) − M|
</2 + /2 = .

Hence, lim [f (x) − g (x)] = L − M.

x→a

Duong T. PHAM 72 / 123

Limit Laws

Proof:
(iii) Recall L := lim f (x). Prove that lim cf (x) = cL.
x→a x→a

AM
If c = 0 then it is not hard to prove (iii).
Consider the case c 6= 0. Let  > 0 be an arbitrarily small positive number .

PH
Since L := lim f (x), there is a δ1 > 0 s.t.
x→a

T.

if 0 < |x − a| < δ1 then |f (x) − L| < . (5)
|c|
g.
on
We choose δ = δ1 > 0. From (5), there holds: if 0 < |x − a| < δ then
Du


|cf (x) − cL| = |c| |f (x) − L| < |c| = .
|c|

Hence, lim [cf (x)] = cL.

x→a

Duong T. PHAM 73 / 123

Limit Laws
Proof:
(iv) Recall: L := lim f (x) and M := lim g (x).
x→a x→a
Let  > 0 be an arbitrarily small positive number .

AM
Then there are δ1 > 0 and δ2 > 0 s.t.
if 0 < |x − a| < δ1 then |f (x) − L| < . (6)

PH
if 0 < |x − a| < δ2 then |g (x) − M| < . (7)

T.
We choose δ = min{δ1 , δ2 }. Suppose that
0 < |x − a| < δ = min{δ1 , δ2 }
g.
so that (6) and (7) hold. Then
on
|(f (x)g (x)) − (LM)| ≤ |f (x)(g (x) − M) + M(f (x) − L)|
Du

≤ |f (x)| |g (x) − M| + |M| |f (x) − L|

≤ (|f (x)| + |M|) ≤ (|f (x) − L| + |L| + |M|)
≤ ( + |L| + |M|)≤ (1 + |L| + |M|)
This proves (iv).
Duong T. PHAM 74 / 123
Limit Laws
Proof:
(v) Recall: L := lim f (x) and M := lim g (x), and M 6= 0.
x→a x→a
Let  > 0 be an arbitrarily small positive number .

AM
Then there are δ1 > 0 and δ2 > 0 s.t.

PH
if 0 < |x − a| < δ1 then |f (x) − L| < . (8)
M
if 0 < |x − a| < δ2 then |g (x) − M| <  so that |g (x)| > . (9)

T.
2
Moreover, |M| /2 > 0, there is δ3 > 0 s.t.
g.
on
|M|
if 0 < |x − a| < δ3 then |g (x) − M| < . (10)
2
Du

We choose δ = min{δ1 , δ2 , δ3 }. Suppose that

0 < |x − a| < δ = min{δ1 , δ2 , δ3 }

so that (8), (9) and (10) hold.

Duong T. PHAM 75 / 123
Limit Laws

Then
 
 f (x) L  |Mf (x) − Lg (x)|

AM
−
 g (x) M  =

|g (x)M|
2

PH
≤ 2 |Mf (x) − Lg (x)|
|M|
2

T.
≤ 2 |M(f (x) − L) + L(M − g (x))|
|M|
≤
g.2
2 (|M| |f (x) − L| + |L| |M − g (x)|)
on
|M|
Du

2
≤ 2 (|M| + |L|) 
|M|

This proves (v).

Duong T. PHAM 76 / 123

Some corollaries

In the rest of this course, except when being asked to use the definition of
limit to prove, we can use the six Limit Laws and the following simple
limits without proving:

AM
1 lim c = c
x→a

PH
2 lim x = a
x→a
sin x

T.
3 lim =1
x→a x
4 lim x n = an g.
x→a
on
√ √
5 lim n x = n a
Du

x→a
p q
6 lim n f (x) = n lim f (x) if the second limit exists and if n is even
x→a x→a
we assume further that lim f (x) ≥ 0
x→a
Proof:

Duong T. PHAM 77 / 123

Limit Laws
x +1
Ex: Evaluate lim (3x 3 − 2x 2 + 10) and lim
x→2 x→2 x 2 − 2x + 3
Ans:
lim (3x 3 − 2x 2 + 10) = lim (3x 3 ) − lim (2x 2 ) + lim 10

AM
x→2 x→2 x→2 x→2
= 3 lim x 3 − 2 lim x 2 + 10

PH
x→2 x→2
3 2
= 3 · 2 − 2 · 2 + 10 = 26

T.
and lim (x + 1)
x +1 x→2
lim g.
x→2 x 2 − 2x + 3
=
lim (x 2 − 2x + 3)
on
x→2
lim x + lim 1
Du

x→2 x→2
= 2
lim x − 2 lim x + lim 3
x→2 x→2 x→2
2+1
= 2 =1
2 −2·2+3

Duong T. PHAM 78 / 123

Limit Laws

Proposition 2.
If f (x) = g (x) for all x 6= a, then lim f (x) = lim g (x)
x→a x→a

AM
PH
Ex: Find lim f (x) where
x→2

T.
(
x2 + 1 if x 6= 2
f (x) =
g. 10 if x = 2
on
Du

Ans: Since f (x) = x 2 + 1 for all x 6= 2, we have

lim f (x) = lim (x 2 + 1)

x→2 x→2
=5

Duong T. PHAM 79 / 123

Limit Laws

(x − 2)2 − 4

AM
Ex: Evaluate lim
x→0 x

PH
Ans: We have
(x − 2)2 − 4 (x − 4)x

T.
= = x − 4 ∀x 6= 0.
x x
Hence,
g.
on
(x − 2)2 − 4
lim = lim (x − 4) = −4.
Du

x→0 x x→0

Duong T. PHAM 80 / 123

Left-hand and right-hand limits

Definition 19.
We say that

AM
lim f (x) = L
x→a−

PH
if for every  > 0, there is a δ > 0 such that

if a−δ <x <a then |f (x) − L| < 

T.
Definition 20. g.
on
We say that
Du

lim f (x) = L
x→a+

if for every  > 0, there is a δ > 0 such that

if a<x <a+δ then |f (x) − L| < 

Duong T. PHAM 81 / 123

Left-hand and right-hand limits

√
Ex: Prove that lim+ x =0
x→0

AM
Ans:
Guessing δ: Given  > 0, we need to find a δ > 0 satisfying

PH
√  √
if 0 < x < δ then  x − 0 = x < .

T.
We may choose δ = 2 ?
g.
For every  > 0, there is δ = 2 > 0 such that if 0 < x < δ = 2 then
on
√  √ √ √
 x − 0 = x < δ = 2 = .
Du

√
Hence, lim+ x = 0
x→0

Duong T. PHAM 82 / 123

Left-hand and right-hand limits

Theorem 1.

 lim f (x) = L
x→a−

AM
lim f (x) = L ⇐⇒
x→a  lim f (x) = L
x→a +

PH
Ex: Prove that lim |x| = 0

T.
x→0
(
x if x ≥ 0
g.
Ans: We have |x| =
on
−x if x < 0.
Thus,
Du


 lim |x| = lim x = 0
+
x→0 + x→0 ⇐⇒ lim |x| = 0
 lim |x| = lim (−x) = 0 x→0
x→0− x→0−

Duong T. PHAM 83 / 123

Left-hand and right-hand limits

|x|
Ex: Prove that lim does not exist
x→0 x

AM
x
|x|  x = 1 if x > 0
Ans: We have = −x

PH
x  = −1 if x < 0
x
Thus,

T.
|x|

lim = lim+ 1 = 1
g. 

x→0+ x x→0

 |x| |x|
on
⇔ lim 6= lim+
|x|  x→0− x x→0 x
Du


lim = lim (−1) = −1 

x→0− x x→0−

|x|
=⇒ lim does not exist.
x→0 x

Duong T. PHAM 84 / 123

Limit of functions
Theorem 2.
Let a ∈ (b, c). There holds
f (x) ≤ g (x) ∀x ∈ (b, c)\{a}

AM
)
=⇒ lim f (x) ≤ lim g (x)
lim f (x) and lim g (x) exist x→a x→a
x→a x→a

PH
T.
Proof: Denote lim f (x) = L and lim g (x) = M. Suppose further that M < L
x→a x→a
=⇒ 0 := (L − M)/3 > 0. g.
on
lim f (x)=L ⇒ ∃δ1 > 0 s.t. if 0 < |x − a| < δ1 then |f (x) − L| < 0
x→a
Du

lim g (x)=M ⇒ ∃δ2 > 0 s.t. if 0 < |x − a| < δ2 then |g (x) − M| < 0
x→a
Choose δ0 = min{δ1 , δ2 }. Then (
|f (x) − L| < 0
if 0 < |x − a| < δ0 then =⇒f (x) > g (x)
|g (x) − M| < 0

Duong T. PHAM 85 / 123

Squeeze Theorem

Theorem 3.
Let a ∈ (b, c). There holds

AM
f (x) ≤ g (x) ≤ h(x) ∀x ∈ (b, c)\{a}
)
=⇒ lim g (x) = L

PH
lim f (x) = lim h(x) = L x→a
x→a x→a

T.
1
Ex: Evaluate lim x 2 cos g.
x→0 x
on
Ans: We have −1 ≤ cos x1 ≤ 1 =⇒ −x 2 ≤ x 2 cos x1 ≤ x 2 .
Du

Moreover, lim (−x 2 ) = lim x 2 = 0

x→0 x→0

1
by Squeeze Theorem, lim x 2 cos = 0.
x→0 x
Duong T. PHAM 86 / 123
Infinite Limits
Definition 21.
Let a ∈ (b, c). Then lim f (x) = ∞ means that for arbitrary positive
x→a
number M, there exists δ > 0 satisfying

AM
if 0 < |x − a| < δ then f (x) > M

PH
1
Ex: Prove that lim = ∞.

T.
x→0 |x|

g.
Ans: Let M be an arbitrary positive number . We need to find δ > 0 s.t.
on
1
if 0 < |x| < δ then >M
|x|
Du

1 1
But |x| > M ⇐⇒ |x| < 1/M. Therefore, we choose δ = M , then clearly
we have
1 1
if 0 < |x| < δ = then > M,
M |x|
finishing the proof.
Duong T. PHAM 87 / 123
Infinite Limits

AM
Definition 22.

PH
Let a ∈ (b, c). Then lim f (x) = −∞ means that for arbitrary negative
x→a

T.
number N, there exists δ > 0 satisfying
g.
if 0 < |x − a| < δ then f (x) < N
on
Du

Duong T. PHAM 88 / 123

Limits at infinity
y

AM
1 lim
1
= ?

PH
y= x→∞ x
x

T.
g.
on
Du

x
Duong T. PHAM 89 / 123
Limit at Infinity
Definition 23.
Let f be a function defined on some interval (a, ∞). Then

AM
lim f (x) = L
x→∞

PH
if for all  > 0, there exists a number N such that
if x > N then |f (x) − L| < 

T.
1 g.
Ex: Prove that lim =0
x→∞ x
on
Ans:
Du

Let  > 0 be an arbitrary positive number.

We choose N = 1 > 0 . Then
 
1 1 1 1
if x > N = then   = < = .
x x x N
Hence, lim 1/x = 0.
x→∞
Duong T. PHAM 90 / 123
Limit at Infinity

AM
Definition 24.
Let f be a function defined on some interval (−∞, a). Then

PH
lim f (x) = L
x→−∞

T.
if for all  > 0, there exists a number N such that
g.
if x < N then |f (x) − L| < 
on
Du

Duong T. PHAM 91 / 123

Limit at Infinity

Definition 25.

AM
Let f be a function defined on some interval (a, ∞). Then

PH
lim f (x) = ∞
x→∞

if for all positive number M, there exists a number N such that

T.
if x > N
g. then f (x) > M
on
Du

Remark: Note that similar definitions apply when we replace ∞ by −∞.

Duong T. PHAM 92 / 123

Exercises

AM
2.3:

PH
1–20; 34–38; 46 – 48;

T.
2.4:
g.
1, 2; 19–25; 36, 37, 44.
on
Du

Duong T. PHAM 93 / 123

Continuity

Definition 26.
Let a ∈ (b, c) and let f be a function defined on (b, c). Function f is

AM
continuous at a if
lim f (x) = f (a)
x→a

PH
T.
f (x)
approaches
f (a)
g.
f (a)
on
Du

a
As x approaches a

Duong T. PHAM 94 / 123

Continuous functions

(
1−x if x ≤ 1
Ex: Given f (x) =
x2

AM
if x > 1.

PH
4 f (1) = 1 − 1 = 0

T.
lim f (x) = lim+ x 2 = 1
x→1+ x→1
2 g. lim f (x) = lim+ (1 − x) = 0
x→1− x→1
on
=⇒ lim f (x) does not exist
Du

1 x→1
f is NOT continuous at x = 1
−1 1 2 x

Duong T. PHAM 95 / 123

Continuous functions

Definition 27.

AM
Let f be a function defined on [a, c). Function f is continuous from
the right at a if

PH
lim+ f (x) = f (a)
x→a

T.
Definition 28.
g.
on
Let f be a function defined on (b, a]. Function f is continuous from
Du

the left at a if
lim f (x) = f (a)
x→a−

Duong T. PHAM 96 / 123

Continuous functions
(
1−x if x ≤ 1
Ex: Consider again f (x) =
x2 if x > 1.

AM
Known: f is NOT continuous at

PH
y x =1
4 lim f (x) = 1; lim f (x) = 0;
x→1+ x→1−

T.
and f (1) = 0
g. lim f (x) = f (1)
2 x→1−
on
⇒ f is continuous from the left
Du

1 at x = 1
lim f (x) 6= f (1)
x→1+
−1 1 2 x
⇒ f is NOT continuous from
the right at x = 1

Duong T. PHAM 97 / 123

Continuous functions

Theorem 4.

AM
A function f is continuous at x0 if and only if f is defined in an interval
(a, b) containing x0 and for each  > 0, there is a δ > 0 such that

PH
|f (x) − f (x0 )| <  whenever |x − x0 | < δ.

T.
A function f is continuous from the right at x0 if and only if f is defined
on an interval [x0 , b) and for each  > 0, there is a δ > 0 such that
g.
on
|f (x) − f (x0 )| <  whenever x0 ≤ x < x0 + δ.
Du

A function f is continuous from the left at x0 if and only if f is defined on

an interval (a, x0 ] and for each  > 0, there is a δ > 0 such that

|f (x) − f (x0 )| <  whenever x0 − δ < x ≤ x0 .

Duong T. PHAM 98 / 123

Continuous on intervals

Definition 29.

AM
A function f is said to be continuous on an open interval (a, b) if it is
continuous at any point x ∈ (a, b),

PH
A function f is said to be continuous on an closed interval [a, b] if it is
continuous at any point x ∈ (a, b), and continuous from the right at a and

T.
from the left at b.
g.
A function f is said to be continuous on a half-open interval [a, b) if it is
on
continuous at any point x ∈ (a, b), and continuous from the right at a.
Du

A function f is said to be continuous on a half-open interval (a, b] if it is

continuous at any point x ∈ (a, b), and continuous from the left at b.

Duong T. PHAM 99 / 123

Continuous on intervals
√
Ex: Show that function f (x) = 1 − 1 − x 2 is continuous on [−1, 1].
Ans:

AM
Let −1 < a < 1. Then
p q
lim f (x) = lim (1 − 1 − x 2 ) = 1 − 1 − lim x 2

PH
x→a x→a x→a
p
2
= 1 − 1 − a = f (a)

T.
=⇒ f is continuous at a.
Besides,
g.
on
p q
lim + f (x) = lim + (1− 1−x 2 ) = 1− 1− lim + x 2 = 1= f (−1)
x→−1 x→−1 x→−1
Du

p q
lim f (x) = lim (1− 1−x ) = 1− 1− lim x 2 = 1= f (1)
2
x→1− x→1− x→1−

=⇒ f is continuous at −1 from the right and at 1 from the left.

f is continuous on [−1, 1].

Duong T. PHAM 100 / 123

Continuous functions

Theorem 5.
If f and g are continuous at x = a. Then the following functions are continuous
at x = a:

AM
(i) f + g (iii) fg (v) cf where c is a

PH
f constant
(ii) f − g (iv) if g (a) 6= 0
g

T.
Proof:
(i) We have
g.
on
lim (f + g )(x) = lim [f (x) + g (x)]
Du

x→a x→a
= lim f (x) + lim g (x)
x→a x→a
= f (a) + g (a)
= (f + g )(a).

Duong T. PHAM 101 / 123

Continuous functions

AM
Theorem 6.
Any polynomial f (x) = an x n + . . . + a1 x + a0 is continuous on

PH
1

(−∞, ∞)

T.
g (x)
2 Any rational function f (x) = , where g and h are polynomials, is
h(x)
g.
continuous at wherever it is defined.
on
Du

Proof:

Duong T. PHAM 102 / 123

Continuous functions

x 2 + 2x − 3
Ex: Find lim
x→−2 (x − 3)(x + 3)

AM
Ans:

PH
x 2 + 2x − 3
Function f (x) is continuous at any point x 6= −3 and
(x − 3)(x + 3)

T.
x 6= 3
=⇒ It is continuous at x = −2
g.
x 2 + 2x − 3 (−2)2 + 2 · (−2) − 3
on
=⇒ lim = f (−2) =
x→−2 (x − 3)(x + 3) (−2 − 3)(−2 + 3)
Du

3
=
5

Duong T. PHAM 103 / 123

Continuous functions

Theorem 7.

AM
The following types of functions are continuous at every number in their
domains:

PH
polynomials
rational functions

T.
root functions
trigonometric functions
g.
on
inverse trigonometric functions
Du

exponential functions
logarithmic functions

Duong T. PHAM 104 / 123

Composition of continuous functions

Theorem 8.
Let f be continuous at a and let g be continuous at f (a). Then, the composition

AM
g ◦ f is continuous at a.

Proof: Let  > 0 be an arbitrarily small real number. Since g is continuous at

PH
f (a), there is a δ1 > 0 such that

T.
|g (y ) − g [f (a)]|<  whenever |y − f (a)| < δ1 . (11)

Since f is continuous at a, there is a δ > 0 such that

g.
on
|f (x) − f (a)| < δ1 whenever |x − a| < δ. (12)
Du

Combining (11) and (12) we derive

g [f (x)] − g [f (a)] <  whenever |x − a| < δ.

By Theorem 4, the function g ◦ f is continuous at a.

Duong T. PHAM 105 / 123

Bounded Functions

Definition 30.

AM
A function f is bounded below on a set S if there is a real number m such that

PH
f (x) ≥ m ∀x ∈ S.

In this case, the set V = {f (x) | x ∈ S} has an infimum α and we write

T.
g. α = inf f (x).
x∈S
on
If there is a x1 ∈ S such that f (x1 ) = α, then we say that α is the minimum of f
Du

on S and we write
α = min f (x).
x∈S

Duong T. PHAM 106 / 123

Bounded Functions

Definition 31.
A function f is bounded above on a set S if there is a real number M such that

AM
f (x) ≤ M ∀x ∈ S.

PH
In this case, the set V = {f (x) | x ∈ S} has a supremum β and we write

T.
β = sup f (x).
x∈S
g.
If there is a x1 ∈ S such that f (x1 ) = β, then we say that β is the maximum of f
on
on S and we write
β = max f (x).
Du

x∈S

Remark.
If f is bounded above and below on S, f is said to be bounded on S .

Duong T. PHAM 107 / 123

Bounded Functions

Theorem 9.
If f is continuous on a finite closed interval [a, b], then f is bounded on [a, b].

AM
Proof. Let t ∈ [a, b]. Since f is continuous at t, there is a δt > 0 such that

PH
|f (x) − f (t)| < 1 whenever x ∈ (t − δt , t + δt ) ∩ [a, b]. (13)

Denote It = (t − δt , t + δt ). Then the collection H = {It | t ∈ [a, b]} is an open

T.
covering of [a, b]. Since [a, b] is compact, Heine–Borel Theorem implies that
there are finitely many points t1 , t2 , . . . , tn such that the intervals It1 , It2 , . . . , Itn
g.
cover [a, b]. By (13), if x ∈ Iti then
on
|f (x)| − |f (ti )| ≤ |f (x) − f (ti )| < 1.
Du

This implies that |f (x)| < 1 + |f (ti )| for all x ∈ Iti and i = 1, . . . , n. Denote
M = 1 + max {|f (ti )| , i = 1, . . . , n}. We then have

|f (x)| ≤ M ∀x ∈ [a, b].

Duong T. PHAM 108 / 123

Continuous Functions
Theorem 10.
Let f : [a, b] → R be continuous on [a, b]. There are x1 , x2 ∈ [a, b] such that

AM
f (x1 ) = inf f (x) and f (x2 ) = sup f (x).
x∈[a,b] x∈[a,b]

PH
Proof. We shall prove the existence of x1 (the proof for x2 are left as an exercise).
Denote α = inf x∈[a,b] f (x). Suppose that there is no such x1 . Then

T.
f (x) > α ∀x ∈ [a, b].

g. f (t) + α
Let t ∈ [a, b]. Then f (t) > α, thus f (t) > > α. Since f is continuous
2
on
at t, there is δt > 0 such that
Du

 
1 f (t) + α
|f (x) − f (t)| < f (t) − ∀x ∈ (t − δt , t + δt ) ∩ [a, b].
2 2
Denote It = (t − δt , t + δt ). This implies
f (t) + α
f (x) > ∀x ∈ It ∩ [a, b]. (14)
2
Duong T. PHAM 109 / 123
Continuous Functions
The collection H = {It | t ∈ [a, b]} is an open covering of [a, b]. Since [a, b] is
compact, Heine–Borel Theorem suggests that there are finitely many
t1 , . . . , tn ∈ [a, b] such that the intervals It1 , . . . , Itn cover [a, b]. Define

AM
f (ti ) + α
α1 = min .
i=1,...,n 2

PH
Sn
Since [a, b] ⊂ i=1 Iti , inequality (14) implies that

T.
f (x) > α1 ∀x ∈ [a, b].

Noting that α1 > α, we derive

g.
on
∀x ∈ [a, b].
Du

f (x) > α1 > α

This contradicts the definition of α (α = inf x∈[a,b] f (x)). Hence, there must be
a x1 ∈ [a, b] such that
f (x1 ) = inf f (x).
x∈[a,b]

Duong T. PHAM 110 / 123

The Intermediate Value Theorem

Theorem 11.
Let f : [a, b] → R be continuous on [a, b]. Let µ be any number between
f (a) and f (b), where f (a) 6= f (b). Then, there exists a c ∈ (a, b) such

AM
that f (c) = µ .

PH
y = f (x)

T.
f (b)
g.
on
µ
Du

f (a)
a c b

Duong T. PHAM 111 / 123

The Intermediate Value Theorem
Proof. Assume that f (a) < µ < f (b). The set
S = {x | a ≤ x ≤ b and f (x) ≤ µ}

AM
is bounded and nonempty. Let c = sup S. We will show that f (c) = µ.
If f (c) > µ, then c > a. Since f is continuous at c, there is a δ > 0 such

PH
that
f (c) − µ
|f (x) − f (c)| < whenever x ∈ (c − δ, c + δ).
2

T.
This implies that f (x) > µ whenever c − δ < x ≤ c. Thus, c − δ/2 is an
g.
upper bound of S. This contradicts the definition of c.
on
If f (c) < µ , then c < b. Since f is continuous at c, there is a δ > 0 such
that
Du

µ − f (c)
|f (x) − f (c)| < whenever x ∈ (c − δ, c + δ).
2
This implies that f (x) < µ whenever c ≤ x ≤ c + δ. This means that
c + δ/2 ∈ S and thus c is not an upper bound for S. This is also a
contradiction. Therefore, f (c) = µ.

Duong T. PHAM 112 / 123

The Intermediate Value Theorem

AM
Ex: Prove that the equation x 3 − x + 2 = 0 has a root between −2 and 0.
Ans:

PH
The function f (x) = x 3 − x + 2 is continuous on [−2, 0]. Moreover,
f (−2) = −4 and f (0) = 2.

T.
Number 0 satisfies −2 < 0 < 2. By using Intermediate Value Theorem,
g.
there is a c ∈ (−2, 0) such that f (c) = 0 .
on
In other words, the equation: x 3 − x + 2 = 0 has a solution c ∈ (−2, 0).
Du

Duong T. PHAM 113 / 123

The Intermediate Value Theorem

y = x3 − x + 2

AM
PH
−2
0
c

T.
g.
on
Du

−4

Duong T. PHAM 114 / 123

Exercises

AM
2.5:

PH
1–6; 7, 9, 10–12, 15–20, 31–32;

T.
35, 37, 38, 40, 42
47–50; 55, 62, 63 g.
on
Du

Duong T. PHAM 115 / 123

Horizontal Asymptote

Def: The line y = L is called a horizontal asymptote of the curve y =

f (x) if either

AM
lim f (x) = L or lim f (x) = L
x→∞ x→−∞

PH
y
x2 − 1
Ex: Consider y = 2 .

T.
x +1 1
2
x −1
Since lim 2
x→∞ x + 1
= g.
on
x2 − 1 x
lim 2 = 1, the
Du

x→−∞ x + 1
line y = 1 is a horizontal −1

asymptote of the curve.

Duong T. PHAM 116 / 123

Exercises

AM
2.6:

PH
1, 3, 4, 11, 13, 14

T.
15–30; 37–44
53, 55, 57, 61, 68, 69, 71.
g.
on
Du

Duong T. PHAM 117 / 123

Curves defined by parametric equations

Imagine that a particle moves along

the curve C shown in the figure

AM
Impossible to describe C by an
equation of the form y = f (x) as C

PH
fails the Vertical Line Test
the x- and y -coordinates of the

T.
particle are functions of time and so
g. we can write x = f (t) and y = g (t)
on
Suppose that x and y are both given as functions of a third variable t
Du

x = f (t), y = g (t) −→ parametric equations

When t varies, the point (x, y ) = (f (t), g (t)) varies and trace out a curve, which
we call a parametric curve

Duong T. PHAM 118 / 123

Parametric curves
Ex: Sketch and identify the curve defined by the parametric equations

x = t 2 − 2t, y =t +1

AM
Ans.
y
t=4

PH
t x y
t=3
-2 8 -1
-1 3 0 t=2

T.
0 0 1 t=1
1 -1 2 g.
2 0 3 t=0
on
8
3 3 4 0 t = -1 x
Du

4 8 5 −1
t=-2
Is the curve a parabola?

y = t + 1 =⇒ t = y − 1 =⇒ x = (y − 1)2 − 2(y − 1)
=⇒ x = y 2 − 4y + 3 (15)

Duong T. PHAM 119 / 123

Parametric curves

In general, the curve with parametric equations

x = f (t), y = g (t), a≤t≤b

AM
has initial point (f (a), g (a)) and terminal point (f (b), g (b)).

PH
y t t==π/4
0
xx ==cos
cos0π/4
=1

T.
t = π/2
Ex: What curve is represented by the yy ==sin
sin0π/4
=0
t = π/4
following parametric equations?
g.
on
x = cos t, y = sin t, t ∈ [0, 2π]. t =π t=0
x
Du

t = 2π
Ans.
x 2 + y 2 = cos2 t + sin2 t = 1
−→ It is a CIRCLE
t = 3π/2

Duong T. PHAM 120 / 123

Parametric curves

Ex: Consider the circle centered at (a, b) and of radius r .

Ans.
y
Equation: (x − a)2 + (y − b)2 = r 2

AM
t = π/2
b+r
Parametric equation:

PH
t = 0,starting point (
t =π r
b x = a + r cos t
t = 2π t ∈ [0, 2π] (16)
y = b + r sin t,

T.
t = 3π/2
g. t = 0 −→ x = a + r cos 0 = a + r ,
a a+r x
y = b + r sin 0 = b → (a + r , b)
on
t = π/2 −→ x = a + r cos π/2 = a, b = b + r sin π/2 = b + r
Du

→ (a, b + r )

Equation (16) represents the circle centered at (a, b) with radius r , following the
counterclockwise direction and the starting point (a + r , b).

Duong T. PHAM 121 / 123

Graphing devices
Ex: Use Wolfram Alpha to graph the curve x = y 4 − 3y 2 .
Hint: Denote y = t. Then x = t 4 − 3t 2 . Type
plot(x = t^4-3t^2, y = t, t=[-2.5,2.5])

AM
Ex: Graph the curve: x = t + 2 sin 2t, y = t + 2 cos 5t
Hint: Type

PH
plot(x=t+2*sin(2*t), y = t + 2*cos(5*t), t=[-10,10])
Ex: Graph the curve:

T.
x = 16 sin3 t, y = 13 cos t − 5 cos(2t) − 2 cos(3t) − cos(4t)
Hint: Type g.
on
plot(x=16(sin(t))^3, y=13cos t - 5cos(2t) - 2cos(3t)-cos(4t),
t=[-5,5])
Du

Ex: Graph the curve: x = 1.5 cos t − cos 30t, y = 1.5 sin t − sin 30t
Hint: Type
plot( x=1.5*cos(t)-cos(30t), y=1.5*sin(t) - sin(30*t), t=[-10,10])
Ex: Graph the curve: x = sin(t + cos 100t), y = cos(t + sin 100t)
Hint: Type
Duong T. PHAM 122 / 123
Exercises

AM
PH
Section 10.1 (Page 626)

T.
g.
on
Du

Duong T. PHAM 123 / 123