Calculus Ch.
1 FUNCTIONS AND LIMITS
1.5 Continuity
이상준 교수
(경희대 수학과)
교재 : James Stewart,
Essential Calculus - Early Transcendentals (2 Ed)
1
Outline
(1) Definition
(2) More definitions and properties
(3) Composite function
(4) Intermediate value theorem
2
< Part 1 >
Definition
3
Continuity
The limit of a good 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.
4
Continuity
Definition 1 implicitly requires three things if 𝑓 is continuous at 𝑎:
1. 𝑓(𝑎) is defined (that is, 𝑎 is in the domain of 𝑓 )
2. lim 𝑓(𝑥) exists
!→#
3. lim 𝑓(𝑥) = 𝑓(𝑎)
!→#
5
Meaning
The definition says that 𝑓 is continuous at 𝑎
if 𝑓(𝑥) approaches 𝑓(𝑎) as 𝑥 approaches 𝑎.
Thus, a continuous function 𝑓 has the property
that a small change in 𝑥 produces only a small change in 𝑓(𝑥).
In fact, the change in 𝑓(𝑥) can be kept as small as we please
by keeping the change in 𝑥 sufficiently small.
6
Discontinuity
If 𝑓 is defined near 𝑎 (in other words, on an open interval containing 𝑎,
except perhaps at 𝑎), we say that 𝑓 is discontinuous at 𝒂
if 𝑓 is not continuous at 𝑎.
Geometrically, you can think of a function that is continuous
at every number in an interval
as a function whose graph has no break in it.
The graph can be drawn
without removing your pen from the paper.
7
Example 2
Where are each of the following functions discontinuous?
! ! $!$%
(a) 𝑓 𝑥 =
!$%
&
if 𝑥 ≠ 0
(b) 𝑓 𝑥 = ,! !
1 if 𝑥 = 0
! ! $!$%
(c) 𝑓 𝑥 = 1 !$% if 𝑥 ≠ 2
1 if 𝑥 = 2
(d) 𝑓(𝑥) = 𝑥
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Example 2 – Solution
(a) Notice that 𝑓(2) is not defined, so 𝑓 is discontinuous at 2.
(b) Here 𝑓 0 = 1 is defined but
does not exist.
So 𝑓 is discontinuous at 0.
9
Example 2 – Solution
(c) Here 𝑓 2 = 2 is defined and
! ! $!$%
lim 𝑓 𝑥 = lim !$%
!→% !→%
= lim 𝑥 + 1 = 3 exists.
!→%
But lim 𝑓 𝑥 ≠ 𝑓(2).
!→%
So 𝑓 is not continuous at 2.
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Example 2 – Solution
(d) The greatest integer function has 𝑓 𝑥 = 𝑥 discontinuities
at all integers because lim 𝑥 does not exist if 𝑛 is an integer.
!→'
11
Graph
In each case the graph can’t be drawn
without lifting the pen from the paper
because a hole or break or jump occurs in the graph.
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Definition
The kind of discontinuity illustrated in parts (a) and (c) is called removable
because we could remove the discontinuity
by redefining 𝑓 at just the single number 2.
The discontinuity in part (b) is called
an infinite discontinuity.
The discontinuities in part (d) are called
jump discontinuities because the function
“jumps” from one value to another.
13
Note
14
< Part 2 >
More definitions
and properties
15
One-sided continuity
16
Continuity on an interval
17
Theorem
18
Corollary
It follows from Theorem 4 and Definition 3 that if 𝑓 and 𝑔 are continuous
on an interval, then so are the functions 𝑓 + 𝑔, 𝑓– 𝑔, 𝑐𝑓, 𝑓𝑔 and 𝑓/𝑔 (if 𝑔 is
never 0).
Corollary
19
Continuity
20
Example 6
On what intervals is each function continuous?
(a) 𝑓 𝑥 = 𝑥 &(( − 2𝑥 )* + 75
! ! +%!+&*
(b) 𝑔 𝑥 =
! ! $&
!+& !+&
(c) ℎ 𝑥 = 𝑥 + !$& − !! +&
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Example 6 – Solution
(a) 𝑓 is a polynomial, so it is continuous on −∞, ∞ by Theorem 5(a).
(b) 𝑔 is a rational function, so, by Theorem 5(b), it is continuous
on its domain, which is
Thus, 𝑔 is continuous on −∞, −1 , −1,1 and 1, ∞ .
22
Example 6 – Solution
(c) We can write ℎ(𝑥) = 𝐹(𝑥) + 𝐺(𝑥) – 𝐻(𝑥), where
𝐹 is continuous on [0, ∞) by Theorem 6.
𝐺 is a rational function, so it is continuous everywhere
except when 𝑥 – 1 = 0, that is, 𝑥 = 1.
𝐻 is also a rational function, but its denominator
is never 0, so 𝐻 is continuous everywhere.
Thus, by parts 1 and 2 of Theorem 4,
ℎ is continuous on the intervals [0, 1) and (1, ∞).
23
Note
24
< Part 3 >
Composite function
25
Composite function
Another way of combining continuous functions 𝑓 and 𝑔 to get a new
continuous function is to form the composite function 𝑓 ° 𝑔.
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Continuity
Intuitively, Theorem 7 is reasonable because if 𝑥 is close to 𝑎,
then 𝑔(𝑥) is close to 𝑏, and since 𝑓 is continuous at 𝑏, if 𝑔(𝑥) is close to 𝑏,
then 𝑓(𝑔(𝑥)) is close to 𝑓(𝑏).
“A continuous function of a continuous function
is a continuous function.”
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Example 7
Where are the following functions continuous?
(a) ℎ 𝑥 = sin(𝑥 % )
&
(b) 𝐹 𝑥 =
! ! +*$,
We have ℎ(𝑥) = 𝑓(𝑔(𝑥)), where
28
Example 7 – Solution
(a) We have ℎ(𝑥) = 𝑓(𝑔(𝑥)), where
Now 𝑔 is continuous on ℝ since it is a polynomial,
and 𝑓 is also continuous every-where by Theorem 6.
Thus ℎ = 𝑓 ∘ 𝑔 is continuous on ℝ by Theorem 8.
29
Example 7 – Solution
(b) Notice that 𝐹 can be broken up as the composition of
four continuous functions:
where
We know that each of these functions is
continuous on its domain,
so by Theorem 8, 𝐹 is continuous on its domain,
which is 𝑥 ∈ ℝ L 𝑥 % + 7 ≠ 4 = {𝑥|𝑥 ≠ ±3}
30
Note
31
< Part 4 >
Intermediate Value Theorem
32
Intermediate Value Theorem
An important property of continuous functions
33
Continuity
Note that the value 𝑁 can be taken on once or more than once.
If we think of a continuous function as a function
whose graph has no hole or break,
then it is easy to believe that the Intermediate Value Theorem is true.
34
Example 8
Show that there is a root of the equation
between 1 and 2.
Solution:
Let 𝑓 𝑥 = 4𝑥 ) − 6𝑥 ) + 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.
35
Example 8 – Solution
Therefore, we take 𝑎 = 1, 𝑏 = 2, and 𝑁 = 0 in Theorem 9.
𝑓 1 = 4 − 6 + 3 = 2 = −1 < 0
𝑓 2 = 32 − 24 + 6 − 2 = 12 > 0
Thus 𝑓 (1) < 0 < 𝑓 (2); that is,
𝑁 = 0 is a number between 𝑓 1 and 𝑓(2).
Now 𝑓 is continuous since it 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 ) − 6𝑥 % + 3𝑥 − 2 = 0
has at least one root 𝑐 in the interval (1, 2).
36
Example 8 – Solution
In fact, we can locate a root more precisely
by using the Intermediate Value Theorem again.
Since 𝑓 1.2 = −0.128 < 0 and 𝑓 1.3 = 0.548 > 0,
a root must lie between 1.2 and 1.3.
A calculator gives, by trial and error,
𝑓 1.22 = −0.007008 < 0 and 𝑓 1.23 = 0.056068 > 0,
so a root lies in the interval (1.22, 1.23).
37
Note
38
