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Lecture 9

# MATH 140 Lecture Notes - Lecture 9: Intermediate Value Theorem, Classification Of Discontinuities, Oliver Heaviside

Department
Mathematics & Statistics (Sci)
Course Code
MATH 140
Professor
Ewa Duma
Lecture
9

Page:
of 3
MATH140 - Lecture 9 Notes
Continuity:
The limit of a function as x approaches a can be often found by calculating the value of
the function at “a”. Functions with this property are usually called “continuous at a”.
Deﬁnition: A function f is continuous at a number “a” if:
lim (x > a) f(x) = f(a)
The deﬁnition requires three things if f is continuous at a:
f(a) is deﬁned (that is, a is in the domain of f)
lim (x > a) f(x) exists
lim (x > a) f(x) = f(a)
This implies that a continuous function f has the property that a small change in x
produces only a small change in f(x).
If f is deﬁned near a (f is deﬁned on an open interval containing a, except perhaps at a),
we say that f is discontinuous at a (or f has a discontinuity at a) if f is not continuous at a
Example:
SECTION 2.5 CONTINUITY
||||
119
CONTINUITY
We noticed in Section 2.3 that the limit of a function as approaches can often be found
simply by calculating the value of the function at . Functions with this property are called
continuous at a. We will see that the mathematical deﬁnition of continuity corresponds
closely with the meaning of the word continuity in everyday language. (A continuous
process is one that takes place gradually, without interruption or abrupt change.)
DEFINITION A function is continuous at a number aif
Notice that Deﬁnition l implicitly requires three things if is continuous at a:
1. is deﬁned (that is, ais in the domain of )
2. exists
3.
The deﬁnition says that is continuous at if approaches as xapproaches a.
Thus a continuous function has the property that a small change in xproduces only a
small change in . In fact, the change in can be kept as small as we please by keep-
ing the change in sufciently small.
If is deﬁned near (in other words, is deﬁned on an open interval containing ,
except perhaps at ), we say that is discontinuous at a(or has a discontinuity at ) if
is not continuous at .
Physical phenomena are usually continuous. For instance, the displacement or velocity
of a vehicle varies continuously with time, as does a person’s height. But discontinuities
do occur in such situations as electric currents. [See Example 6 in Section 2.2, where the
Heaviside function is discontinuous at because does not exist.]
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.
EXAMPLE 1 Figure 2 shows the graph of a function f. At which numbers is fdiscontinu-
ous? Why?
SOLUTION It looks as if there is a discontinuity when a!1 because the graph has a break
there. The ofﬁcial reason that fis discontinuous at 1 is that is not deﬁned.
The graph also has a break when , but the reason for the discontinuity is differ-
ent. Here, is deﬁned, but does not exist (because the left and right limits
are different). So fis discontinuous at 3.
What about ? Here, is deﬁned and exists (because the left and
right limits are the same). But
So is discontinuous at 5. M
Now let’s see how to detect discontinuities when a function is deﬁned by a formula.
f
lim
x l 5 f!x""f!5"
limxl5 f!x"f!5"a!5
limxl3 f!x"f!3"
a!3
f!1"
limtl0 H!t"0
af
affa
afaf
x
f!x"f!x"
f
f!a"f!x"af
lim
x l a f!x"!f!a"
lim
x l a f!x"
ff !a"
f
lim
x l a f!x"!f!a"
f
1
a
ax
2.5
NAs illustrated in Figure 1, if is continuous,
then the points on the graph of
approach the point on the graph. So
there is no gap in the curve.
!a, f!a""
f!x, f!x""
f
f(a)
x
0
y
a
y=ƒ
ƒ
approaches
f(a).
As x approaches a,
F IGUR E 1
F IGUR E 2
y
0x
1 2 3 4 5
Based on the graph, we can see that when x = 1 and when
x = 5. When x = 1, we see that it is discontinuous because f(1) is not deﬁned. When x =
5, this is a jump discontinuity because f(5) is different from the left/right hand limits.
The following graphs are more examples of jump/inﬁnite/removable discontinuity.
EXAMPLE 2 Where are each of the following functions discontinuous?
(a) (b)
(c) (d)
SOLUTION
(a) Notice that is not deﬁned, so fis discontinuous at 2. Later we’ll see why is
continuous at all other numbers.
(b) Here is deﬁned but
does not exist. (See Example 8 in Section 2.2.) So fis discontinuous at 0.
(c) Here is deﬁned and
exists. But
so is not continuous at 2.
(d) The greatest integer function has discontinuities at all of the integers
because does not exist if is an integer. (See Example 10 and Exercise 49 in
Section 2.3.) M
Figure 3 shows the graphs of the functions in Example 2. 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. The kind of discontinuity illustrated in parts (a) and (c) is called removable
because we could remove the discontinuity by redeﬁning at just the single number 2.
[The function is continuous.] The discontinuity in part (b) is called an inﬁ-
nite discontinuity. The discontinuities in part (d) are called jump discontinuities because
the function “jumps” from one value to another.
1 2 3
1
x
y
0
(d) ƒ=[x]
1 2
1
x
y
0
(c) ƒ= if x2
1if x=2
-x-2
x-2
(b) ƒ= if x0
1if
1
x=0
1
x
y
0
1 2 x
y
0
1
(a) ƒ=-x-2
x-2
FIGURE 3 Graphs of the functions in Example 2
t!x"!x!1
f
nlimxln #x\$
f!x"!#x\$
f
lim
xl2 f!x""f!2"
lim
xl2 f!x"!lim
xl2 x2"x"2
x"2!lim
xl2 !x"2"!x!1"
x"2!lim
xl2 !x!1"!3
f!2"!1
lim
xl0 f!x"!lim
xl0 1
x2
f!0"!1
ff !2"
f!x"!#x\$f!x"!
%
x2"x"2
x"2if x"2
1 if x!2
f!x"!
%
1
x2if x"0
1 if x!0
f!x"!x2"x"2
x"2
V
120
||||
CHAPTER 2 LIMITS AND DERIVATIVES
Deﬁnition: A function f is continuous from the right at a number a if lim (x > a+) f(x) = f(a)
And f is continuous from the left at a if: lim (x > a-) f(x) = f(a)
Deﬁnition: A function f is continuous on an interval if it is continuous at every number in
the interval. If f is deﬁned only on one side of an endpoint of the interval, we understand
continuous at the endpoint to mean continuous from the right or continuous from the
left.
Example: Show that the function f(x) = 1 - sqrt(1-x^2) is continuous on the interval [-1,1].
Using limit laws, we can see that
SECTION 2.5 CONTINUITY
|| ||
121
DEFINITION A function is continuous from the right at a number aif
and is continuous from the left at aif
EXAMPLE 3 At each integer , the function [see Figure 3(d)] is continuous
from the right but discontinuous from the left because
but M
DEFINITION A function is continuous on an interval if it is continuous at
every number in the interval. (If fis deﬁned only on one side of an endpoint of the
interval, we understand continuous at the endpoint to mean continuous from the
right or continuous from the left.)
EXAMPLE 4 Show that the function is continuous on the
interval
SOLUTION If , then using the Limit Laws, we have
(by Laws 2 and 7)
(by 11)
(by 2, 7, and 9)
Thus, by Deﬁnition l, is continuous at if . Similar calculations show that
and
so is continuous from the right at !1 and continuous from the left at 1. Therefore,
according to Deﬁnition 3, is continuous on .
The graph of is sketched in Figure 4. It is the lower half of the circle
M
Instead of always using Deﬁnitions 1, 2, and 3 to verify the continuity of a function as
we did in Example 4, it is often convenient to use the next theorem, which shows how to
build up complicated continuous functions from simple ones.
x2"!y!1"2!1
f
#!1, 1\$f
f
lim
xl1! f!x"!1!f!1"lim
x l !1" f!x"!1!f!!1"
!1#a#1
af
!f!a"
!1!s1!a2
!1!
s
lim
xla !1!x2"
!1!lim
xla s1!x2
lim
xla f!x"!lim
xla
(
1!s1!x2
)
!1#a#1
#!1, 1\$.
f!x"!1!s1!x2
f
3
lim
x l n! f!x"!lim
x l n!%x&!n!1"f!n"
lim
x l n" f!x"!lim
x l n" %x&!n!f!n"
f!x"!%x&n
lim
x l a! f!x"!f!a"
f
lim
x l a"
f!x"!f!a"
f
2
1-1
1
x
y
0
ƒ=1-œ„„„„„
1-
F I G U R E 4
Therefore, f is continuous from the right at -1 and continuous from the left at 1.
Theorem: If f and g are continuous at a and c is a constant then the following functions
are also continuous at a:
F + g
F - g
Cf
Fg
F/g if g(a) 0
Theorem based on direct substitution property:
Any polynomial is continuous everywhere; it is continuous on XER
Any rational function is continuous wherever it is deﬁned; it is continuous on its
domain
The Intermediate Value Theorem:
Suppose that f is continuous on the closed interval [a, b] and let N be any number
between f(a) and f(b), where f(a) f(b). Then there exists a number c in (a,b) such that
f(c) = N.
which is precisely the statement that the function is continuous at ; that
is, is continuous at . M
EXAMPLE 9 Where are the following functions continuous?
(a) (b)
SOLUTION
(a) We have , where
Now is continuous on since it is a polynomial, and is also continuous everywhere.
Thus is continuous on by Theorem 9.
(b) We know from Theorem 7 that is continuous and
is continuous (because both and are continuous). Therefore, by
Theorem 9, is continuous wherever it is deﬁned. Now is
deﬁned when . So it is undeﬁned when , and this happens
when . Thus Fhas discontinuities when xis an odd multiple of and
is continuous on the intervals between these values (see Figure 7). M
An important property of continuous functions is expressed by the following theorem,
whose proof is found in more advanced books on calculus.
THE INTERMEDIATE VALUE THEOREM Suppose that is continuous on the
closed interval and let be any number between and , where
. Then there exists a number in such that .
The Intermediate Value Theorem states that a continuous function takes on every inter-
mediate value between the function values and . It is illustrated by Figure 8. Note
that the value can be taken on once [as in part (a)] or more than once [as in part (b)].
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. In geometric terms it
says that if any horizontal line is given between and as in Fig-
ure 9, then the graph of can’t jump over the line. It must intersect somewhere.
It is important that the function in Theorem 10 be continuous. The Intermediate Value
Theorem is not true in general for discontinuous functions (see Exercise 44).
One use of the Intermediate Value Theorem is in locating roots of equations as in the
following example.
f
y!Nf
y!f!b"y!f!a"y!N
(b)
0x
y
f(b)
N
f(a)
a c£ b
y=ƒ
c™
(a)
0x
y
f(b)
N
f(a)
a c b
y=ƒ
FIGURE 8
N
f!b"f!a"
f!c"!N!a, b"cf !a""f!b"
f!b"f!a"N#a, b\$
f
10
!
x!"
!
, "3
!
, . . .
cos x!#11 \$cos x%0
ln!1\$cos x"F!x"!f!t!x""
y!cos xy !1
t!x"!1\$cos xf !x"!ln x
!h!f"tf!t
f!x"!sin xandt!x"!x2
h!x"!f!t!x""
F!x"!ln!1\$cos x"h!x"!sin!x2"
V
af "tah!x"!f!t!x""
126
||||
CHAPTER 2 LIMITS AND DERIVATIVES
FIGURE 7
y=ln(1+cos x)
2
_6
_10 10
b
0x
y
f(a)
N
f(b)
a
y=ƒ
y=N
FIGURE 9