Calculus 1 (Math 105)

2.3 Continuity

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2.3 Continuity

2.3.1 Continuity at a Point and on Open Interval

2.3.1 Right and left continuity and Continuity on Closd Interval

2.3.2 Continuous Extension and Removable Discontinuities

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 2.3 Continuity

2.3.1 Continuity at a Point and on an Open Interval

• In mathematics, the term continuous has much the same meaning as it has in
everyday usage.
• To say that a function 𝑓 is continuous at 𝑥 = 𝑐 means that there is no
interruption in the graph of 𝑓 at 𝑐.
• That is, its graph is unbroken at 𝑐 and there are no holes, jumps, or gaps.
• The Figure below identifies three values of 𝑥 at which the graph of 𝑓 is not
continuous.
• At all other points in the interval (𝑎, 𝑏), the graph of 𝑓 is uninterrupted and
continuous.

In the Figure above, it appears that continuity at 𝑥 = 𝑐 can be destroyed

by any one of the following conditions.

• The function is not defined at 𝑥 = 𝑐.

• The limit of 𝑓(𝑥) does not exist at 𝑥 = 𝑐.

• The limit of 𝑓(𝑥) exists at 𝑥 = 𝑐, but it is not equal to 𝑓(𝑐).

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If none of the above three conditions is true, the function 𝑓 is called
continuous at 𝑐, as indicated in the following important definition.

Definition 1:

Continuity at a Point:
Assume that 𝑓(𝑥) is defined on an open interval containing 𝑥 = 𝑐. Then 𝑓
is continuous at 𝑥 = 𝑐 if
lim 𝑓(𝑥) = 𝑓(𝑐)
𝑥→𝑐

If the limit does not exist, or if it exists but is not equal to 𝑓(𝑐), we say
that 𝑓 is discontinuous at 𝑥 = 𝑐.

Continuity on an Open Interval:

A function is continuous on an open interval (𝑎, 𝑏) if it is continuous at each
point in the interval.

A function that is continuous on the entire real line (−∞, ∞) is everywhere

continuous.

Example 1

Show that the following functions are continuous:

(a) 𝑓(𝑥) = 𝑘 (𝑘 any constant)
(b) 𝑔(𝑥) = 𝑥 𝑛 (𝑛 a whole number)

Solution

(a) We have
lim𝑥→𝑐 𝑓(𝑥) = lim𝑥→𝑐 𝑘 = 𝑘,
and 𝑓(𝑐) = 𝑘.

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The limit exists and is equal to the function value for all 𝑐, so 𝑓(𝑥) is
continuous (Figure 3).
(b) lim𝑥→𝑐 𝑔(𝑥) = lim𝑥→𝑐 𝑥 𝑛 = 𝑐 𝑛 for all 𝑐.
Also 𝑔(𝑐) = 𝑐 𝑛 , so again, the limit exists and is equal to the function
value. Therefore, 𝑔(𝑥) is continuous. (Figure 4 illustrates the case 𝑛 = 1 ).

2.3.2 Right and left continuity and Continuity on a Closd Interval

Theorem 1:

Right and left continuity

We say that 𝑓 is right continuous at 𝑐 if lim+ 𝑓(𝑥) = 𝑓(𝑐).
𝑥→𝑐

We say that 𝑓 is left continuous at 𝑐 if lim− 𝑓(𝑥) = 𝑓(𝑐).

𝑥→𝑐

Continuity on a Closd Interval:

A function 𝑓 is continuous on a closed interval [𝑎, 𝑏] if it is continuous on
(𝑎, 𝑏) and, in addition,

lim 𝑓(𝑥) = 𝑓(𝑎) and lim− 𝑓(𝑥) = 𝑓(𝑏).

𝑥→𝑎+ 𝑥→𝑏

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Theorem 2:

Function 𝑓 is continuous at 𝑐 if and only if it is both right and left

continuous at 𝑐.

Example 2

Discuss the continuity of 𝑓(𝑥) = √1 − 𝑥 2 .

Solution

The domain of 𝑓 is the closed interval [−1,1].

At all points 𝑐 ∈ (−1,1),

lim √1 − 𝑥 2 = √1 − 𝑐 2 = 𝑓(c)
𝑥→𝑐

Moreover, because

lim + √1 − 𝑥 2 = 0 = 𝑓(−1) Continuous from the right

𝑥→−1

and

lim √1 − 𝑥 2 = 0 = 𝑓(1) Continuous from the left

𝑥→1−

you can conclude that 𝑓 is continuous on the closed interval [−1,1], as

shown in the Figure below.

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Check Point 1

Discuss the continuity of 𝑓(𝑥) = √9 − 𝑥 2 .

Solution

There Are Lots of Continuous Functions

The following functions are continuous wherever they are defined:

(a) all polynomials;
(b) all rational functions;
𝑛
(c) all rational powers 𝑥 𝑚/𝑛 = √𝑥 𝑚 ;
(d) the sin, cos, tan, sec, csc and cot functions.

(e) the absolute value function |𝑥|.

Example 3

Show that the function 𝑓(𝑥) = √𝑥 is a continuous function.

Solution

The Domain is [0, ∞).The function is continuous at the left endpoint 0

because it is right continuous there.

lim 𝑓(𝑥) = lim+ √𝑥 = 0 = 𝑓(0).

𝑥→0+ 𝑥→0

Also, 𝑓 is continuous at every number 𝑐 > 0 since:

lim 𝑓(𝑥) = lim √𝑥 = √𝑐 = 𝑓(𝑐), 𝑐 > 0.

𝑥→𝑐 𝑥→𝑐

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Theorem 3:

If 𝑓 and 𝑔 are continuous at 𝑐, then

(i) 𝑓 ± 𝑔 is continuous at 𝑐;
(iv) 𝑓 ⋅ 𝑔 is continuous at 𝑐;
(v) 𝑓/𝑔 is continuous at 𝑐 provided 𝑔(𝑐) ≠ 0.

Example 4

Discuss the continuity of the function 𝑓(𝑥) = 𝑥 + 𝑒 𝑥 .

Solution

By the Theorem above, it follows that 𝑓(𝑥 ) is continuous at every point in

ℝ because the functions 𝑦 = 𝑥 and 𝑦 = 𝑒 𝑥 are continuous at every real
number and the sum of continuous functions is continuous.

Theorem 4:

Suppose that lim 𝑔(𝑥) = 𝐿 and 𝑓 is continuous at 𝐿. Then,

𝑥→c

lim 𝑓(𝑔(𝑥)) = 𝑓 (lim 𝑔(𝑥)) = 𝑓(𝐿).

𝑥→𝑐 𝑥→𝑐

In particular, if 𝑔 is continuous at 𝑐 (so 𝐿 = 𝑔(𝑐) ), then the composition

𝑓 ∘ 𝑔 is continuous at 𝑐 :
lim 𝑓(𝑔(𝑥)) = 𝑓(𝑔(𝑐 )) = 𝑓(𝐿).
𝑥→𝑐

Example 5

1
If 𝑓(𝑥 ) = 𝑥 and 𝑔(𝑥 ) = 𝑥 2 + 4. Find lim 𝑓(𝑔(𝑥 )).
𝑥→1

Solution

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∵ lim𝑔(𝑥 ) = lim (𝑥 2 + 4) = 12 + 4 = 5
𝑥→1 𝑥→1

and because 𝑓 is continuous at 5, therefore,

1
lim 𝑓(𝑔(𝑥 )) = 𝑓 (lim 𝑔(𝑥)) = 𝑓(5) = 5.
𝑥→1 𝑥→1

Note that
1
𝑓(𝑔(𝑥 )) = 𝑓(𝑥 2 + 4) =
𝑥2 + 4
and we can use the direct substitution to find the limit above,
1 1
lim 𝑓(𝑔(𝑥 )) = lim 𝑥 2 +4 = 5 .
𝑥→1 𝑥→1

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2.3.3 Continuous Extension and Removable Discontinuities
• Consider an open interval 𝐼 that contains a real number 𝑐.
• If a function 𝑓 is defined on 𝐼 (except possibly at 𝑐 ), and 𝑓 is not
continuous at 𝑐, then 𝑓 is said to have a discontinuity at 𝑐.
• Discontinuities fall into two categories: removable and
nonremovable.
• A discontinuity at 𝑐 is called removable if 𝑓 can be made
continuous by appropriately defining (or redefining) 𝑓(𝑐).

For instance, the functions shown in Figure (a) and (c) have removable
discontinuities at 𝑐, and the function shown in Figure (b) has a
nonremovable discontinuity at 𝑐.

Continuous Extension
If 𝑓(𝑐) is not defined, but lim𝑥→𝑐 𝑓(𝑥) = 𝐿 exists, we can define a new function
𝐹(𝑥) by

𝑓(𝑥), if 𝑥 is in the domain of 𝑓

𝐹(𝑥) = {
𝐿 if 𝑥 = 𝑐.
𝐹(𝑥) is continuous at 𝑥 = 𝑐. It is called the continuous extension of 𝑓(𝑥) to
𝑥 = 𝑐. For rational functions 𝑓(𝑥), continuous extensions are usually found
by cancelling common factors.

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Example 6

𝑥 2 −1
Show that 𝑓(𝑥) = has a continuous extension to 𝑥 = 1 and find that
𝑥−1
extension.

Solution

1−1 0
Although 𝑓(1) = 1−1 = 0 is not defined, if 𝑥 ≠ 1 we have

𝑥 2 − 1 (𝑥 + 1)(𝑥 − 1)
𝑓(𝑥) = = =𝑥+1
𝑥−1 𝑥−1
The function

𝐹(𝑥) = 𝑥 + 1

is equal to 𝑓(𝑥) for 𝑥 ≠ 1 but it is also continuous at 𝑥 = 1, having there

the value 2.
The graph of 𝑓(𝑥) is shown in the figure below.
The continuous extension of 𝑓(𝑥) to 𝑥 = 1 is 𝐹(𝑥). It has the same graph
as except with no hole at (1,2).

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Check Point 2

𝑥 2 −𝑥
Show that 𝑓(𝑥) = 𝑥 2 −1 has a continuous extension to x = 1 and find that
extension.

Solution

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2.3.4 Intermediate value theorem
Theorem 5:

If 𝑓 is continuous on the closed interval [𝑎, 𝑏] and 𝑘 is any number

between 𝑓(𝑎) and 𝑓(𝑏), then there is at least one number 𝑐 in [𝑎, 𝑏] such
that

𝑓(𝑐) = 𝑘.

NOTE The Intermediate Value Theorem tells you that at least one 𝑐 exists,
but it does not give a method for finding 𝑐. Such theorems are called existence
theorems.

COROLLARY:
If 𝑓(𝑎) and 𝑓(𝑏) have opposite signs, then there is a number 𝑐 (zero or
root) in [𝑎, 𝑏] such that
𝑓(𝑐) = 0.

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Example 7

Use the Intermediate Value Theorem to show that the polynomial function
𝑓(𝑥) = 𝑥 3 + 2𝑥 − 1 has a zero in the interval [0,1].

Solution

Note that 𝑓 is continuous on the closed interval [0,1]. Because

𝑓(0) = 03 + 2(0) − 1 = −1 and 𝑓(1) = 13 + 2(1) − 1 = 2

it follows that 𝑓(0) < 0 and 𝑓(1) > 0.

We can therefore apply the Intermediate Value Theorem to conclude that
there must be some 𝑐 in [0,1] such that
𝑓(𝑐) = 0 (𝑓 has a zero in the closed interval [0,1].)
as shown in the Figure below.

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COROLLARY:
If a function 𝑓 is continuous and has no zeroes on an interval, 𝐼, then
either 𝑓(𝑥) > 0 or and 𝑓(𝑥) < 0 for every 𝑥 in 𝐼.

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 Appendix

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Example 8

Prove that there is a solution to the equation 𝑓(𝑐) = 1 in [−1,4]. Then find
the values of 𝑐.

Solution

The polynomial function 𝑓(𝑥) = 𝑥 2 − 𝑥 − 5 is continuous on the interval

[−1,4] and 𝑓(−1) = −3, 𝑓(4) = 7.
∵−3 ≤ 1 ≤ 7,
the Theorem above guarantees that there is a solution to the equation
𝑓(𝑐) = 1, that is, 𝑐 2 − 𝑐 − 5 = 1 in [−1,4].

We can solve the equation 𝑐 2 − 𝑐 − 5 = 1 to find 𝑐.

𝑐 2 − 𝑐 − 6 = 0 or (𝑐 − 3)(𝑐 + 2) = 0.

Although the latter equation has two solutions, 𝑐 = 3 or 𝑐 = −2, only the
value 𝑐 = 3 is between −1 and 4.

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