Chapter 2
Limits of
Functions and
Continuity
Section 2.5:
Continuity
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Continuity of a Function
Objectives
• Understand the concept of continuity (K1).
• Analyze functions using continuity (S1)
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Definition of a continuous function
• The limit of a function as 𝑥 approaches 𝑎 can often be found
simply by calculating the value of the function at 𝑎. Functions
with this property are called continuous at 𝑎.
• A continuous process is one that takes place gradually,
without interruption or abrupt change.
Definition 1. A function 𝒇 is continuous at 𝒂 if
lim 𝑓(𝑥) = 𝑓 𝑎 .
𝑥→𝑎
We say that f is discontinuous at a (or f has a discontinuity at a)
if f is not continuous at a
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Continuity
Notice that a function f is continuous at a number a if
• 𝑓(𝑎) is defined.
• lim 𝑓(𝑥) exists as a real number.
𝑥→𝑎
• lim 𝑓(𝑥) = 𝑓 𝑎
𝑥→𝑎
If one of the previous conditions is not satisfied then 𝑓 is
discontinuous at 𝑎.
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Continuity: Types of discontinuity
Suppose that a function 𝑓 has a discontinuity at a number a.
Removable discontinuity: if lim 𝑓(𝑥) exists.
𝑥→𝑎
Jump discontinuity: if both lim+ 𝑓(𝑥) and lim− 𝑓(𝑥) exist but
𝑥→𝑎
𝑥→𝑎
are different.
Infinite discontinuity: if lim+ 𝑓(𝑥) or lim− 𝑓(𝑥) is ∞ or −∞.
𝑥→𝑎
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Example 1
Determine the points of discontinuity of the function 𝑓 and precise
their types.
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Solution
The points of discontinuity are :
a) The point 𝑥 = −2 which is a removable discontinuity
b) The point 𝑥 = −1 which is an infinite discontinuity
c) The point 𝑥 = 0 which is a jump discontinuity
d) The point 𝑥 = 1 which is a jump discontinuity
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Example 2
In each of the following cases, determine whether the function is
continuous at the given point. If it is not continuous, then state the
type of discontinuity.
b) 𝑓 𝑥
𝑥 2 −𝑥−2
=
, 𝑎=2
𝑥−2
1
if 𝑥 ≠ 0
= ൝𝑥 2
, 𝑎=0
c) 𝑓 𝑥
𝑥 2 −𝑥−2
= ൝ 𝑥−2
a) 𝑓 𝑥
1
if 𝑥 = 0
1
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if 𝑥 ≠ 2 , 𝑎 = 2
if 𝑥 = 2
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Solution
a) Notice that 𝑓 (2) is not defined, so 𝑓 is discontinuous at 2.
b) Here 𝑓 (0) = 1 is defined but
1
𝑥→0 𝑥
lim 𝑓(𝑥) = lim ( 2 )
𝑥→0
does not exist. So 𝑓 is discontinuous at 0.
c) Here 𝑓 (2) = 1 is defined and
𝑥 2 −𝑥−2
lim 𝑓(𝑥) = lim
𝑥→2
𝑥→2 𝑥−2
(𝑥−2)(𝑥+1)
= lim
𝑥−2
𝑥→2
= lim (𝑥 + 1) = 3.
𝑥→2
But
lim 𝑓(𝑥) ≠ 𝑓 (2)
𝑥→2
So 𝑓 is not continuous at 2.
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Solution
Figure 3 shows the graphs of the functions in
Example 2.
Figure 3. Graphs of the functions
in Example 2
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Continuity
Definition 2. A function 𝒇 is continuous from the right at 𝒂 if
lim+ 𝑓(𝑥) = 𝑓 𝑎
𝑥→𝑎
and 𝒇 is continuous from the left at 𝒂 if
lim− 𝑓(𝑥) = 𝑓 𝑎
𝑥→𝑎
Definition 3. A function is continuous on an interval:
𝑎, 𝑏 if it is continuous at every point of the interval.
[𝑎, 𝑏) if it is continuous on (𝑎, 𝑏) and right continuous at 𝑎.
(𝑎, 𝑏] if it is continuous on (𝑎, 𝑏) and left continuous at 𝑏.
[𝑎, 𝑏] if it is continuous on (a,b), right continuous at 𝑎 and left
continuous at 𝑏.
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Example 3
Determine the intervals of continuity of the function f whose graph is
given.
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Solution
We saw in Example 2 that the points of discontinuity of this
functions are given by : 𝑥 = −2, −1 , 0, 1.
Consequently, the intervals of the continuity are:
[−3, −2), (−2, −1), (−1,0], (0,1) and (1,3].
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Continuity
Theorem 1.
a) If f and 𝑔 are continuous at a and c is a constant, then the
functions 𝑓 + 𝑔, 𝑓 − 𝑔, 𝑐𝑔 and 𝑓𝑔 are continuous at 𝑎.
𝑓
If moreover 𝑔 𝑎 ≠ 0 then is continuous at 𝑎.
𝑔
b) If f and 𝑔 are continuous on an interval and c is a constant,
then the functions 𝑓 + 𝑔, 𝑓 − 𝑔 , 𝑐𝑓 and 𝑓𝑔 are continuous
on the interval. If moreover 𝑔 𝑥 ≠ 0 for all 𝑥 in the
𝑓
interval, then is continuous on the interval.
𝑔
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Continuity
It follows from Theorem 1 and Definition 3 that if f and 𝑔 are
continuous on an interval, then so are the functions 𝑓 + 𝑔,
𝑓
𝑓 − 𝑔, 𝑐𝑓, 𝑓𝑔 and (if 𝑔 is never 0) .
𝑔
Theorem 2.
a) A polynomial function is continuous everywhere; that is
continuous on −∞, ∞ .
b) A rational function is continuous wherever is defined; that is
continuous on its domain.
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Example 4
Determine the intervals of continuity of the function 𝑓 𝑥
𝑥 3 +2𝑥−1
=
.
5−3𝑥
Solution
The function 𝑓 is a rational function.
Hence, by application of Theorem 2, it is continuous on its
domain of definition.
5
Or, 𝑓 is defined for all real numbers except 𝑥 = .
3
Therefore,
5
5
Domain of continuity is (− ∞, ) ∪ ( , ∞).
3
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Continuity – More basic functions!
It turns out that most of the familiar functions are
continuous at every number in their domains.
From the appearance of the graphs of the sine and cosine
functions, we would certainly guess that they are continuous.
We know from the definitions of
sin  and cos  that the coordinates
of the point P in Figure 5 are
(cos , sin  ). As  → 0, we see
that P approaches the point (1, 0)
and so cos  → 1 and sin  → 0.
Figure 5
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Continuity – More basic functions!
Thus,
lim cos = 1 and lim sin  = 0.
→0
→0
Since cos 0 = 1 and sin 0 = 0, the above equations assert
that the cosine and sine functions are continuous at 0.
The addition formulas for cosine and sine can then be used to
deduce that these functions are continuous everywhere.
It follows from Theorem 1 that
sin 𝑥
tan 𝑥 =
cos 𝑥
is continuous everywhere except for x such that cos x = 0.
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Continuity – More basic functions!
𝜋
This happens when 𝑥 is an odd integer multiple of .
2
So, 𝑦 = tan 𝑥 has infinite discontinuities when
𝜋
3𝜋
5𝜋
𝑥 = ± , ± , ± , and so on (see Figure 6).
2
2
2
𝑦 = tan 𝑥
Figure 6
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Continuity
Theorem 3. The following type of functions are continuous at
every point of their domain:
• polynomials
• rational functions
• root functions
• trigonometric functions
• inverse trigonometric functions
• exponential functions
• logarithmic functions
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Example 5
Determine the domain of continuity of the following function
𝑓 𝑥
ln 𝑥−tan−1 𝑥
=
.
𝑥 2 −1
Solution
The function 𝑓 is a fraction of two functions.
In the numerator, the functions 𝑥 ↦ ln 𝑥 and 𝑥 ↦ tan−1 𝑥 are
continuous on their domains of definition which are, respectively,
0, ∞ and −∞, ∞ . So the numerator is defined on the
intersection which is 0, ∞ .
For the denominator, the function 𝑥 ↦ 𝑥 2 − 1 is continuous
everywhere but it must be not equal to 0. So, 𝑥 ≠ ±1.
By application of Theorem 1, the domain of continuity is
0, 1 ∪ 1, ∞ .
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Continuity of a composite function
Theorem 4. (Continuity of a composite function)
a) If lim 𝑔 𝑥 = 𝑏 and 𝑓 is continuous at 𝑏, then
𝑥→𝑎
lim 𝑓 𝑔 𝑥
= 𝑓 𝑏 . In other words,
𝑥→𝑎
lim 𝑓 𝑔 𝑥
𝑥→𝑎
= 𝑓 lim 𝑔 𝑥
𝑥→𝑎
.
b) If 𝑔 is continuous at a and f is continuous at 𝑔(𝑎), then the
composite function 𝑓 𝜊𝑔 given by (𝑓𝜊𝑔)(𝑥) = 𝑓(𝑔(𝑥)) is
continuous at a.
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Example 6
Where are these functions being continuous?
a) ℎ 𝑥 = sin 𝑥 2
𝑏)𝐹(𝑥) = ln(1 + cos 𝑥)
Solution
a) We have ℎ 𝑥 = 𝑓 𝑔 𝑥 with 𝑓(𝑥) = sin(𝑥) and 𝑔 𝑥 = 𝑥 2 .
Since 𝑓 and 𝑔 are continuous on −∞, ∞ , so by Theorem 4, we
conclude that ℎ is continuous on −∞, ∞ .
b) We need to have 1 + cos 𝑥 > 0 (Remember −1 ≤ cos 𝑥 ≤ 1) .
Or this happens for all real numbers except when cos 𝑥 = −1!
This gives: domain of continuity of 𝐹 = ℝ ∖ { 2𝑘 + 1 𝜋; 𝑘 ∈ ℤ}.
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Continuity - The Intermediate Value
Theorem (IVT)
Theorem 5. (The Intermediate Value theorem)
Suppose that 𝑓 is continuous on a closed interval [𝑎, 𝑏] and let N be
any number between 𝑓(𝑎) and𝑓(𝑏), where 𝑓 𝑎 ≠ 𝑓 𝑏 . Then there
exists a number c in (𝑎, 𝑏) such that 𝑓 𝑐 = 𝑁.
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Continuity - The Intermediate Value
Theorem (IVT)
The Intermediate Value Theorem states that a continuous
function takes on every intermediate value between the function
values f (a) and f (b). It is illustrated by Figure 8.
Note that the value N can be taken on once [as in part (a)] or
more than once [as in part (b)].
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Continuity - The Intermediate Value
Theorem (IVT)
Corollary 6. (The Intermediate Value theorem)
Suppose that 𝑓 is continuous on a closed interval [𝑎, 𝑏] and
𝑓 𝑎 × 𝑓 𝑏 < 0, then there exists a number c in (𝑎, 𝑏) such that
𝑓 𝑐 = 0.
Corollary 6 is a consequence of Theorem 5 for 𝑁 = 0!
Example 7
Show that the equation
4𝑥 3 − 6𝑥 2 + 3𝑥 − 2 = 0
has a root between 1 and 2.
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Solution
Let 𝑓 𝑥 = 4𝑥 3 − 6𝑥 2 + 3𝑥 − 2.
We are looking for a solution of the given equation, that is, a
number c between 1 and 2 such that 𝑓 (𝑐) = 0. Therefore, we take
a = 1, b = 2, and N = 0 in Theorem 10 (or equivalently Corollary
6). We have
𝑓 1 = −1 < 0
𝑓 2 = 12 > 0
Thus, 𝑓(1) < 0 < 𝑓(2); that is, N = 0 is a number between
𝑓 1 and 𝑓 2 . The function f is a polynomial, so the Intermediate
Value Theorem says there is a number c between 1 and 2 such that
𝑓(𝑐) = 0.
In other words, the equation 4𝑥 3 − 6𝑥 2 + 3𝑥 − 2 = 0 has at least
one solution c in the interval (1,2).
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Practice
Exercise 47 page 125 (Textbook)
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END OF LECTURE
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