Module 1: Limits and Continuity

After discussing the lesson you should

 know the definitions of limit and continuous,
 know basic properties of limits, of continuity, and the connection between
continuity and open sets,
 be able to use these properties to instantly recognize continuous functions and
open or closed sets,
 be able to determine whether lim 𝑓(𝑥) exists, and if so, identify the limit, and
𝑥→𝑎
 be able to solve problems on limits and continuity using different theorems

I. Definitions of limits and continuity theorems

a. The Limit Process

Let L be some real number. We say that the limits of 𝑓(𝑥) as x tends to c
is L and write
𝐥𝐢𝐦 𝒇(𝒙) = 𝑳
𝒙→∞

provided that, as x approaches c, 𝒇(𝒙) approaches L or provided that 𝒇(𝒙) is

closed to L for all 𝑥 ≠ 𝑐, which are close to c.

Figure 1. Limit of a Function

Do we always have to have a limit?

How to find the limit?

1. Direct Substitution
2. Cancellation
3. Graph
 b. One – sided limit, infinite limit and limits at infinity

1. Left – hand limit

The lim 𝑓 (𝑥 ) = 𝐿 : as x approaches c from the left, 𝑓(𝑥 ) approaches L.
𝑛→𝑐−

2. Right – hand
The lim 𝑓 (𝑥 ) = 𝐿 : as x approaches c from the right, 𝑓 (𝑥 ) approaches L.
𝑛→𝑐−

Example 1. lim (4𝑥 + 5) = 13

𝑥→2

𝑥 3 −2𝑥+4 5
Example 2. lim =2
𝑥→3 𝑥 2 +1

𝑥 2 −3𝑥+2
Example 3. lim =1
𝑥→2 𝑥−2

𝑥 2 −9
Example 4. lim =6
𝑥→3 𝑥−3

Example 5.

c. Some Limit Theorems

ANSWERS TO EXAMPLE 1 TO 4
1. (1) 8 (2) -2
1
2. (1) 5 (2) 0
3. (1) 5 (2) 0
4. (1) 6 (2) 1

The following theorem would help you expand your ability to evaluate limits
because it shows how to analyze the limit of a composite functions.

Example: Limit of a Composite Functions

Example: Limits of Trigonometric Functions

Problem Set No. 1. Limits and Continuity

𝒙𝟐 −𝟏
1. 𝐥𝐢𝐦
𝒙→ −𝟏 𝒙+𝟏
𝒔𝒊𝒏 𝒙 ( 𝟏−𝒄𝒐𝒔 𝒙)
8. 𝐥𝐢𝐦
𝒙→ 𝟎 𝒙𝟐
𝒙𝟑 −𝟖
2. 𝐥𝐢𝐦
𝒙→ 𝟐 𝒙−𝟐
𝒔𝒊𝒏𝟐 𝒙
9. 𝐥𝐢𝐦
𝒙→ 𝟎 𝒙
𝒙𝟐 +𝒙−𝟔
3. 𝐥𝐢𝐦
𝒙→ −𝟑 𝒙𝟐 − 𝟗
(𝟏−𝒄𝒐𝒔 𝒙)𝟐
10. 𝐥𝐢𝐦
𝒙→ 𝟎 𝒙
√𝒙+𝟓 −𝟑
4. 𝐥𝐢𝐦 𝒙−𝟒
𝒙→ 𝟒

𝒕𝒂𝒏 𝒙
5. 𝐥𝐢𝐦
𝒙→ 𝟎 𝒙

𝒔𝒊𝒏 𝟒𝒙
6. 𝐥𝐢𝐦
𝒙→ 𝟎 𝒙

𝒔𝒊𝒏 𝒙
7. 𝐥𝐢𝐦
𝒙→ 𝟎 𝟓𝒙
II. Continuity of a function and the Intermediate Value Theorem

Many functions have the property that their graphs can be traced with a pencil
without lifting the pencil from the page. Normally, such functions are called continuous.
Other functions have points at which a break in the graph occurs, but satisfy this property
over intervals contained in their domains. They are continuous on these intervals and are
said to have a discontinuity at a point where a break occurs.

We begin our investigation of continuity by exploring what it means for a function

to have continuity at a point. Intuitively, a function is continuous at a particular point if
there is no break in its graph at that point.
Continuity at a Point

Before we look at a formal definition of what it means for a function to be

continuous at a point, let’s consider various functions that fail to meet our intuitive notion
of what it means to be continuous at a point. We then create a list of conditions that
prevent such failures.

Our first function of interest is shown in Figure. We see that the graph of f(x) has
a hole at a. In fact, f(a) is undefined. At the very least, for f(x) to be continuous at a, we
need the following condition:

Figure 3. The function f(x) is not continuous at

Figure 2. The function f(x) is not continuous at a a because limx→a f(x) does not exist.
because f(a) is undefined
As we see in Figure, these two
The figure shows that although f(a) is conditions by themselves do not
defined, the function has a gap at a. In this guarantee continuity at a point. The
example, the gap exists because limx→a f(x) function in this figure satisfies both of
does not exist. our first two conditions, but is still not
continuous at a.
 When is the limit continuous?

A function f(x) is continuous at

a point a if and only if the following
three conditions are satisfied:
1. f(a) is defined
2. limx→a f(x) exists
3. limx→a f(x) = f(a)

Figure 3, The function f(x) is not continuous at a

because limx→a f(x) ≠ f(a)

Problem-Solving Strategy: Determining Continuity at a Point

1. Check to see if f(a) is defined. If f(a) is undefined, we need go no further. The

function is not continuous at a. If f(a) is defined, continue to step 2.

2. Compute lim 𝑓(𝑥). In some cases, we may need to do this by first computing
𝑥→𝑎
lim 𝑓(𝑥) and lim 𝑓(𝑥). If lim 𝑓(𝑥) does not exist (that is, it is not a real number),
𝑥→𝑎− 𝑥→𝑎+ 𝑥→𝑎
then the function is not continuous at a and the problem is solved. If lim 𝑓(𝑥) exists,
𝑥→𝑎
then continue to step 3.

3. Compare f(a) and lim 𝑓(𝑥). If lim 𝑓(𝑥) ≠ f(a), then the function is not continuous
𝑥→𝑎 𝑥→𝑎
at a. If lim 𝑓(𝑥) = f(a), then the function is continuous at a.
𝑥→𝑎

Determining Continuity at a Point (Condition1)

𝑥2− 4
Example 1. Using the definition, determine whether the function 𝑓 (𝑥) = is
𝑥−2
continuous at x=2. Justify the conclusion.

0
Solution: Substituting x = 2 to the function, then 𝑓(2) = 0 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑, therefore,
𝑥2 − 4
𝑓 (𝑥 ) = 𝑥−2 is discontinuous at 2 because 𝑓 (2) is undefined. The graph of f(x) is shown
in Figure.
 Fig. 4. The function f(x) is discontinuous at 2 because f(2) is undefined.

Determining Continuity at a Point (Condition1)

Example 2. Using the definition, determine whether the function

is continuous at x = 3. Justify your answer.

Solution
Let’s begin by trying to calculate f(3).

f(3) = −(32) + 4 = −(9)+4=−5.

Thus, f(3) is defined. Next, we calculate lim 𝑓(𝑥). To do this, we must

𝑥→3
compute lim 𝑓(𝑥) and lim 𝑓(𝑥):
𝑥→3− 𝑥→3+

lim 𝑓(𝑥) =−(32) + 4 = −5 and lim 𝑓(𝑥) = −(32) + 4 = −5

𝑥→3− 𝑥→3+
and
lim 𝑓(𝑥) = 4(3) – 8 = 4.
𝑥→3+

Therefore, lim 𝑓(𝑥) does not exist. Thus, f(x) is not continuous at 3. The graph of f(x) is
𝑥→3
shown in Figure.
 Figure 5. The function f(x) is not continuous at 3 because, lim 𝑓(𝑥) does not exist.
𝑥→3

Problem Set No. 2 Continuity of a function

1. Using the definition, determine whether the function

is continuous at x=1. If the function is not continuous at 1, indicate the condition for
continuity at a point that fails to hold.

2. Determine whether or not the function is continuous at the indicated point.

Types of Discontinuity

a. Removable Discontinuity

b. Jump Discontinuity
c. Infinite Discontinuity

The Intermediate Value Theorem

If f(x) is continuous over a closed, bounded interval [a, b] and z is any real
number between f(a) and f(b), then there is a number c in [a, b] satisfying
f(c) = z.
Examples:
a) Show that 𝒇(𝒙) = 𝒙𝟒 + 𝒙 − 𝟑 has a root between - 2 and 0, and another one
between 0 and 2.

Solution:

𝟏
b) 𝒇(𝒙) = 𝒙
Key Concepts
 For a function to be continuous at a point, it must be defined at that point, its limit
must exist at the point, and the value of the function at that point must equal the
value of the limit at that point.
 Discontinuities may be classified as removable, jump, or infinite.
 A function is continuous over an open interval if it is continuous at every point in the
interval. It is continuous over a closed interval if it is continuous at every point in its
interior and is continuous at its endpoints.
 The composite function theorem states: If f(x) is continuous at L and 𝐥𝐢𝐦 𝒈(𝒙) = 𝑳 ,
𝒙→𝒂
then 𝐥𝐢𝐦 𝒇(𝒈(𝒙)) = 𝒇(𝐥𝐢𝐦 𝒈(𝒙)) = 𝑳 .
𝒙→𝒂 𝒙→𝒂
 The Intermediate Value Theorem guarantees that if a function is continuous over a
closed interval, then the function takes on every value between the values at its
endpoints.

Problem Set No. 3. Intermediate Value Theorem

Show that there is a root of the given equation on the given

interval

1. 𝑥 3 − 3𝑥 + 1 = 0 [0, 1]

2. 𝑥 2 = √𝑥 + 1 [1, 2]

3. Show that 𝑓 (𝑥) = 𝑥 3 − 𝑥 2 − 3𝑥 + 1 has a zero over the

interval [0, 1]
The Squeeze Theorem
In the graph below, the lower and upper functions have the same limit value at x=a.
The middle function has the same limit value because it is trapped between the two
outer functions.

Theorem: The Squeeze Theorem (Sandwich Theorem)

Reference
1. creativecommons.org.
2. The Calculus 7th Edition by Leitholds
3. The Calculus 9th Edition by Larson & Edwards