Math 111- Math Analysis 1 Chapter 1

Author Olga DG Unay LIMITS Module 1

LIMITS

Overview:

This chapter introduces concepts on functions and limits. The purpose of this module is to
impart a thorough knowledge and understanding of these concepts.

Learning Outcomes

After working with the module, you will be able to:

1. Differentiate relations and functions.

2. Understand limits and its theorems.

1. 1 FUNCTIONS AND RELATIONS

Functions and Relations:

A relation is a set of inputs and outputs, often written as ordered pairs (input, output).
We can also represent a relation as a mapping diagram or graph.

A relation is any set of ordered pairs. The set of all the first components of the ordered
pairs is called the domain of the relation and the set of all the second components is
called the range.

A relation in which each number of the domain corresponds to exactly one member of
the range is called a FUNCTION.

The following table shows how relations and functions are the same and how they are different.
RELATION FUNCTION

A RELATION is a rule that relates values A FUNCTION is a relation where each

from a set of values called the DOMAIN to a element in the domain is related to only
second set of values called the RANGE ONE value in the range by some rule.
A RELATION is a set of ordered pairs (x,y). A FUNCTION is a set of ordered pairs (x,y)
such that NO TWO ordered pairs have the
same x-value but different y-values.
The elements of the domain can be The elements of the domain can be
imagined as input to a machine that applies imagined as input to a machine that applies
a rule to these inputs to generate one or a rule so that each input corresponds to
more outputs. only one output.

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 Math 111- Math Analysis 1 Chapter 1

Author Olga DG Unay LIMITS Module 1

LET’S INVESTIGATE!

A group of UEP STEM students in their PE Class sold ticket for the UEPAA Games 2020. The
entrance ticket is sold Php50.00. How many tickets should be sold in order to finance the total cost
for the food and other expenses in the amount of Php7,500.00. How much profit they will get if they
sold 250 tickets?

No. of tickets 20 50 100 150 200 250

sold

Amount of 1000 2500 5000 7,500 10,000 12,500

sales

The given data can be written in an ordered pair where the first components are the values
of the independent quantity while the second components are the values of the dependent quantity.

Thus, the first component is the number of tickets sold while the second component is the
amount of sales.

{(20, 1000), (50, 2500), (100, 5000), (150, 7,500), (20, 10000), (250, 12500)}

Now this set of ordered pairs is an example of a relation.

A relation is any set of ordered pairs. The set of all the first components of the ordered pairs
is called the domain of the relation and the set of all the second components is called the range.

1 Suppose we have to find the domain and range:

1. {(1,3), (2,4), (5,7), (6,8)}
2. {(-2,4), (-1,1), (-2,0), (0,5), (2,-2)}

Solution:
The domain in number 1 is {1,2,5,6}
The domain in number 2 is ________
The range in number 1 is { 3,4,7,8}
The range in number 2 is __________

A function is a relation in which NO TWO ORDERED PAIRS HAVE THE SAME FIRST
COMPONENT and DIFFERENT SECOND COMPONENTS.

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For example, the ordered pairs (2,3) and (2,1) is NOT a function!

2 Find the domain and range of the function given its set of ordered pairs
{(1,-1), (2,-2), (3,-3), (4,-4), (5,-5)}

Solution:
Domain: {1,2,3,4,5}
Range: {-1,-2,-3,-4,-5}

3 Given the equation -x2 + y2 = 16, determine if the relation define y as a function of x

Solution:

Solving for y in terms of x in -x2 + y2 = 16 results to

y2 = 16 +x2
y = ±√x2 + 16

This implies that the relation produces two values of y for each x, a positive value and a
negative value.
Thus, the equation does not define a function. Why?

Note that if an equation is solved for y and two or more values of y are obtained for a given
x, then the equation does not define a function.

4 Given the figure, determine if the relation defines y as a function of x.

Input x output y

-2 0
-1 1
0 4
1

Solution:

In the figure, each x is paired with exactly one y. thus, the correspondence defines y as a
function of x.

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 Math 111- Math Analysis 1 Chapter 1

Author Olga DG Unay LIMITS Module 1

VERTICAL LINE TEST

Given a graph, we can use the vertical line test to determine whether it represents a function.
Note: A graph in the xy plane is a function if and only if no vertical line intersects the graph in
more than one point.

1 Determine if the relation defines y as a function of x.

Solution:
Since no vertical line intersects the graph in more than one point, the graph defines y as a
function of x

FUNCTION NOTATION

Functions are usually expressed in equation form. The equation form of a function is written
in a special notation. Think of a function as a machine which is programmed with a rule that defines
the relationship between the input and the output. Consequently, the machine accepts the member
of the domain which is the input and releases the member of the range which is the output.

In the figure below, the letter “f” is used to name the function. The input is represented by “x”
and the output by “f(x)”. The special notation f(x), read as “f of x” or “f at x”, represents the value of
the function at x.

Functions in the form of equations use the f(x) notation. For example, f(x) = 4x + 25
describes the function f that takes an input x which is multiplied by 4 then add by 25.

1 Evaluate f(5) in the function f(x) = 4x + 25

Solution: we take the input 4

f(x) = 4x + 25
f(5) = 4(5) + 25
f(5) = 20 + 25
f(5) = 45

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Author Olga DG Unay LIMITS Module 1

2 Evaluate a. f(0) b. f(-2) c. (a + 1) defined by f(x) = 3x2 – x + 6

Solution:
a.
f(0) = 3x2 – x + 6
f(0) = 3(0)2 - 0 + 6
f(5) = 20 + 25
f(5) = 45
b.
f(-2) = 3x2 -x + 6
f(-2) = 3(-2)2 – (-2) + 6
f(-2) = 3(4) +2 + 6
f(-2) = 12 + 8
f(-2) = 20

c. f(a + 1) = 3x2 -x + 6
f(a + 1) = 3(a +1)2 – (a +1) + 6
f(a + 1) = 3 (a2 +2a+1) -a – 1 +6
f(a + 1) = 3a2 + 6a + 3 – a +5
f(a + 1) = 3a2 +5a + 8

GRAPH OF A FUNCTION

The graph of a function f on the xy-plane consists of points (x,y) such that x is in the domain of f and
y=f(x). To graph a function through a table of values, follow these steps.

1. Select at least five values of x in the domain. The choice of x-values is arbitrary.
2. find the corresponding values of the function for each selected x.
3. plot the obtained ordered pairs or points on the coordinate plane, then join them with a smooth
curve.

Illustrative example:

1 Sketch the graph of f(x) = 2x + 4. Then, find the domain.

Solution:
Make a table of values then, sketch the graph
x f(x)= 2x + 4 points
-2 2(-2) + 4 = 0 (-2,4)
-1 2(-1) + 4 = 2 (-1,0)
0 2(0) + 4 = 4 (0,4)
1 2(1) + 4 = 6 (1,6)
2 2(2) + 4 = 8 (2,8)

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Author Olga DG Unay LIMITS Module 1

Graph of f(x) = 2x + 4

The domain is the set of all real numbers.

2 Sketch the graph of f(x) = x2 – 2x. Then, find the domain.

Solution:
Make a table of values then, sketch the graph
x f(x)= x2 – 2x points
-2 (-2)2 – 2(-2) = 8 (-2,8)
-1 (-1)2 – 2(-1) = 3 (-1,3)
0 (0)2 – 2(0) = 0 (0,0)
1 (1)2 – 2(1) = -1 (1,-1)
2 (2)2 – 2(2) = 0 (2,0)
3 (3)2 – 2(3) = 3 (3,3)
4 (4)2 – 2(4) = 8 (4,8)

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 Math 111- Math Analysis 1 Chapter 1

Author Olga DG Unay LIMITS Module 1

Graph of f(x) = x2 – 2x.

Since x2 – 2x is a real number for each x, the domain of f is the set of all real numbers. Kindly sketch
the graph.

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Author Olga DG Unay LIMITS Module 1

REAL-LIFE APPLICATION OF FUNCTIONS

Functions are used to mathematically model real-life situations. These situations involve
one quantity depending on another quantity. Examples are the numbers of days worked in relation
to the salary earned; the area of the circle depends on its radius; and the interest of investment
affected by the rate.

1 a. A rectangle has a perimeter of 60 ft. Express its area A as a function of the length.

b. The volume V of a right cylinder is twice its radius, express the volume V as a function of radius
r.

Solution:
a. Given: Perimeter of a rectangle P= 2 length + 2 Width

solve w in terms of l :
P = 2l + 2w= 60
2(l + w) = 60
l + w = 30
w = 30 - l

Since the area of the rectangle is A= lw

A = lw
A = l(30-l)
A(l) = l(30-l)
A(l) = 30l – l2

b. Given: V= 𝜋𝑟2h. If the height of the cylinder is twice its radius, h = 2r

V = 𝜋r2h
V(r) = 𝜋r2(2r)
V(r) = 2𝜋r3

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 Math 111- Math Analysis 1 Chapter 1

Author Olga DG Unay LIMITS Module 1

1.1 Self Assessment Questions

SAQ

A. Which of the following relations are functions? Justify your answer.

1. {(3,5), (1,0), (9,5), (2,0)}

2. {(-1,-2), (3,5), (1,2), (-1,7)}

3. 2x + 3y = 8

4. -x + y2 = 2

5.

B. Evaluate each function at x = 4

1. f(x) = x2

2. g(x) = x2 – 3x

3. Express the distance D traveled in t hour by a car whose speed is 60km/hr.

4. Express the area A of an equilateral triangle as a function of its side x.

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 Math 111- Math Analysis 1 Chapter 1

Author Olga DG Unay LIMITS Module 1

1.2 LIMITS OF A FUNCTION

Familiarity with the limit concept is absolutely essential for a deeper understanding of the
calculus. In this section, we shall begin our discussion of the limit of a function and its theorems

Consider the function defined by the equation

f(x) = 3x + 1

and assign some values to x near, but not equal to a specific value, say 3. We construct a table of values
shown here:

X 2.500 2.670 2.900 3.005 3.230 3.300

f(x) 8.500 9.01 9.7 10.015 10.69 10.9

The table shows that when x is near 3, whether a little less or a little greater than 3, f(x) = 3x
+ 1 = 10. In other words, “3x + 1 approaches the number 7 as a limit when x approaches 2”. The
abbreviated symbolic form for this statement is

3x + 7 → 7 as x → 3

We may also say that “the limit of 3x + 1 as x approaches 2 is 10.” In symbol, we write this as

lim (3x + 1) = 7
x→3

Definition. Let f(x) be any function and let a and L be numbers. If we can make f(x) as
close to L as we please by choosing x sufficiently close to a then we say that the limit
of f(x) as x approaches a is L or symbolically.

lim f(x) = L
x → a

1.3 THEOREMS ON LIMITS

This section deals with several theorems by means of which we shall be able to evaluate the
limits of functions rapidly and efficiently. To evaluate or to find

lim f(x)
x→a

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 Math 111- Math Analysis 1 Chapter 1

Author Olga DG Unay LIMITS Module 1

means that we are to find the number L that f(x) is near, whenever x is near a but not equal
to a. Of course, when x = a, the value of the function is f(a). It may be that f(a) is also the limit, that is,
L=f(a). Thus to evaluate
lim (4 - x2)
x→1

means to find a number which 4 -x2 is near whenever x is near the number 1. By definition
1.1, we know that

lim (4 -x2) = 3
x→ 1

since by choosing x sufficiently close to 1, 4 – x2 come as close to 3 as we please.

We shall use the following theorems which we shall state state symbolically without proof.

L1. lim c = c c= any constant a= any real number

x→a

L2. lim x = a
x→a

L3. lim c f(x) = c lim f(x)

x →a x →a

L4. lim [f(x) + g(x)] = lim f(x) + lim g(x)

x →a x →a x →a

L5. lim [ f(x) . g(x)] = lim f(x) . lim g(x)

x →a x →a

𝑓(𝑥)
L6. lim = lim 𝑓(𝑥)
𝑥 → 𝑎 𝑔(𝑥) 𝑥→𝑎

lim 𝑔(𝑥)
𝑥→𝑎

L7. lim 𝑛ඥ𝑓(𝑥) = 𝑛ඥ𝑙𝑖𝑚 𝑓(𝑥) n = any positive integer and f(x) ≥0 if n is even
x →a x →a

L8. lim [f(x)]n = [lim f(x)]n

x→a x→a

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 Math 111- Math Analysis 1 Chapter 1

Author Olga DG Unay LIMITS Module 1

In stating the above theorems, we assume that f(x) and g(x) are defined for all values of x in
some interval containing a, except possibly at a itself. These theorems may be stated briefly in words.
For instance, L4 is sometimes stated as “the limit of a sum is the sum of the limits”. To illustrate the
use of these theorems, we have the following examples.

1 Evaluate the following limits:

lim (x2 +3x + 4)

x→2
Solutions:
Example

lim (x2 +3x + 4) = lim x2 + lim 3x + lim 4

x→2 x→2 x→2

= lim x 2 + 3limx + 4
x→ 2

= (2)2 + 3(2) +4
= 14

2 Evaluate the following limits:

lim (3x + 5)2

x→2
Solutions:
Example

lim (3x + 5)2 = lim (3x + 4) 2

x→2 x→2

= lim 3x + lim 4 2

x→2 x→2

= 3limx + lim 4 2

x→2 x→2

= 3(3) + 4 2

= 169

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Author Olga DG Unay LIMITS Module 1

Note that the limits of the functions in the above examples can be obtained by straight
substitution. For instance,

lim (4x + 5) √𝑥 + 5 = [(4)3 + 5] (3 + 5) = 17(8) = 136

x→3

1.2 Self Assessment Questions

SAQ

1. Evaluate each of the following:

1. lim (x2 – 4x + 3)
x→2

4𝑥 + 2
2. lim
𝑥+4
x→3

3
3. lim ( 2x + √𝑥 − 4)
x→8

4. lim 3x + 2
x→0 x – 2x + 4
2

5. lim (tan x + sin x)

x→ᴫ/4

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Author Olga DG Unay LIMITS Module 1

PROBLEM SET No. 1

Test your skill.

1. Let f(x) = x2 – 4 and g(x) = 2x + 1. Evaluate the following:

a. (f + g)(5)
1
b. (fg)(2)

c. (f ⃘ g)(3)

2. Let f(x) = 3x + 5, g(x) = x3 + 2x and h(x) = -5x. Form the following composite functions.

d. ( f ⃘ g)(x)

e. (g ⃘ h) (x)

3. Find the indicated function.

f. Given f(x) = x + 6 and f[g(x)]= 3x + 6, find g(x).

3𝑥 + 6
g. Given g(x) = 3x + 5 and f[g(x)]= 3𝑥 + 5 , find f(x).

3. Show your solution for each.

4
a. The volume V of a sphere with radius x is V = 3ᴫx3. Do the following:

i. Express the volume V of the balloon as a function of radius x.

ii. Use the function V(x) to determine the volume of the balloon when the radius is 5.

4. Graph the functions f(x) = 3x -2

5. Graph the piecewise function: fx = x + 2 if x ≤ -1

x if /x/ < 1
-x + 3 if x≥ 1

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 Math 111- Math Analysis 1 Chapter 1

Author Olga DG Unay LIMITS Module 1

PROBLEM SET No. 1

Name: ______________________________________________________________ Grade and Section: _________________________

Write your answers here.
Detached this Page

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 Math 111- Math Analysis 1 Chapter 1

Author Olga DG Unay LIMITS Module 1

PROBLEM SET No. 1

Name: ______________________________________________________________ Grade and Section: _________________________

Write your answers here.
Detached this Page

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 Math 111- Math Analysis 1 Chapter 1

Author Olga DG Unay LIMITS Module 1

SUMMARY

Functions and Relations

A relation is a set of inputs and outputs, often written as ordered pairs (input, output). We can also
represent a relation as a mapping diagram or graph.

A relation is any set of ordered pairs. The set of all the first components of the ordered pairs is called
the domain of the relation and the set of all the second components is called the range.

A relation in which each number of the domain corresponds to exactly one member of the range is
called a FUNCTION.

RELATION FUNCTION

A RELATION is a rule that relates values A FUNCTION is a relation where each

from a set of values called the DOMAIN to a element in the domain is related to only
second set of values called the RANGE ONE value in the range by some rule.
A RELATION is a set of ordered pairs (x,y). A FUNCTION is a set of ordered pairs (x,y)
such that NO TWO ordered pairs have the
same x-value but different y-values.
The elements of the domain can be The elements of the domain can be
imagined as input to a machine that applies imagined as input to a machine that applies
a rule to these inputs to generate one or a rule so that each input corresponds to
more outputs. only one output.

Vertical Line Test

Given a graph, we can use the vertical line test to determine whether it represents a function.
Note: A graph in the xy plane is a function if and only if no vertical line intersects the graph in
more than one point.

Function Notation
Functions are usually expressed in equation form. The equation form of a function is written
in a special notation. Think of a function as a machine which is programmed with a rule that defines
the relationship between the input and the output. Consequently, the machine accepts the member
of the domain which is the input and releases the member of the range which is the output.

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 Math 111- Math Analysis 1 Chapter 1

Author Olga DG Unay LIMITS Module 1

In the figure below, the letter “f” is used to name the function. The input is represented by “x”
and the output by “f(x)”. The special notation f(x), read as “f of x” or “f at x”, represents the value of
the function at x.

Functions in the form of equations use the f(x) notation. For example, f(x) = 4x + 25
describes the function f that takes an input x which is multiplied by 4 then add by 25.

Limits

Definition. Let f(x) be any function and let a and L be numbers. If we can make f(x) as
symbolically.
close to L as we please by choosing x sufficiently close to a then we say that the limit
of f(x) as x approaches a is=
lim f(x) LLor symbolically.
x → a
lim f(x) = L
x → a

Theorems on Limits

L1. lim c = c c= any constant a= any real number

x→a

L2. lim x = a
x→a

L3. lim c f(x) = c lim f(x)

x →a x →a

L4. lim [f(x) + g(x)] = lim f(x) + lim g(x)

x →a x →a x →a

L5. lim [ f(x) . g(x)] = lim f(x) . lim g(x)

x →a x →a

𝑓(𝑥)
L6. lim = lim 𝑓(𝑥)
𝑥 → 𝑎 𝑔(𝑥) 𝑥→𝑎

lim 𝑔(𝑥)
𝑥→𝑎

L7. lim 𝑛ඥ𝑓(𝑥) = 𝑛ඥ𝑙𝑖𝑚 𝑓(𝑥) n = any positive integer and f(x) ≥0 if n is even
x →a x →a

L8. lim [f(x)]n = [lim f(x)]n

x→a x→a

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 Math 111- Math Analysis 1 Chapter 1

Author Olga DG Unay LIMITS Module 1

FEEDBACK

As you go along this module, what lesson did you feel the most difficult for you? Try to list
them down and give time to consult your teacher for further discussion.

References:

Feliciano and Uy, Differential and Integral Calculus. Merriam & Webster Bookstore, Inc, 1983.

Clyde E. Love and Earl Rainville. Differential and Integral Calculus. Collier-McMillan International
Edition.

Schaums Outline in Calculus

Calculus by Leithold.

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