For today
1
Continuity of Functions
Continuity at a point
Definition
A function
f
(
x
)
is said to be
continuous
at
x
=
a
if the following conditions are all
satisfied:
1
f
is defined at
x
=
a
2lim
x
→
a
f
(
x
)
exists
3f
(
a
) =
lim
x
→
a
f
(
x
)
1
2
−
1
1
2
3
4
0
f
(
x
) =
3
x
−
1
Graphical Example
1
2
3
−
1
1
2
3
4
0
g
(
x
) =
3
x
2
−
4
x
+
1
x
−
1
g
(
x
) =
3
x
2
−
4
x
+
1
−
3
−
2
−
1
1
2
3
−
3
−
2
−
1
1
2
3
0
h
(
x
) =
1
x
h
(
x
) =
1
Graphical Example
−
3
−
2
−
1
1
2
3
−
1
1
0
The Heaviside Function
H
(
x
)
is discontinuous at
x
=
0
since
lim
Let
f
(
x
) =
x
3
+
x
2
−
2
.
f
(
1
) =
0
Also,
lim
x
→1
f
(
x
) =
f
(
1
) =
0
.
Therefore,
f
is continuous at
x
=
1
.
In fact, for any
a
∈
R
,
lim
Example
Let
f
(
x
) =
x
2
−
x
−
2
x
−
2
.
1
f
is discontinuous at
x
=
2
. (why?)
2Let
a
∈
R
\ {
2
}
f
(a)
is defined since
a
∈
dom
f
.
From a previous theorem,
lim
x→a
f(x) =
f
(a)
Example
Let
g
(
x
) =
x
2
−
x
−
2
x
−
2
,
x
6
=
2
0,
x
=
2
g
(
2
) =
0
lim
x
→2
g
(
x
) =
x
lim
→2
x
2
−
x
−
2
x
−
2
=
lim
x
→2
(
x
+
1
) =
3
lim
x
→2
g
(
x
) =
3
6
=
0
=
g
(
2
)
Let
h
(
x
) =
x
2
−
x
−
2
x
−
2
,
x
6
=
2
3,
x
=
2
h
(
2
) =
3
lim
x
→2
h
(
x
) =
lim
x
→2
x
2
−
x
−
2
x
−
2
=
x
lim
→2
(
x
+
1
) =
3
lim
x
→2
h
(
x
) =
3
=
h
(
2
)
Removable and Essential Discontinuity
Definition
1
If
lim
x
→
a
f
(
x
)
exists but either
f
(
a
)
is undefined or
f
(
a
)
6
=
lim
x
→
a
f
(
x
)
, then we say
that
f
has a removable discontinuity at
x
=
a
.
2
If
lim
x
→
a
f
(
x
)
does not exist, then we say that
f
has an essential discontinuity
at
x
=
a
.
Moreover,
1if
lim
x→a
−f
(x)
and
x→a
lim
+f
(x)
both exist but are not equal, then
f
is said to have a
jump essential discontinuity at
x
=
a
.
2
if
lim
x→a
−f
(x) = +
∞
or
−
∞
or
x→a
lim
+f
(x) = +
∞
or
−
∞
, then
f
is said to have an
1
2
−
1
1
2
3
4
0
f
(
x
) =
3
x
−
1
Example
1
2
3
−
1
1
2
3
4
0
g
(
x
) =
3
x
2
−
4
x
+
1
x
−
1
g
(
x
)
has a removable discontinuity at
x
=
1
since
g
(
1
)
is
undefined but
lim
−
3
−
2
−
1
1
2
3
−
3
−
2
−
1
1
2
3
0
f
(
x
) =
1
x
h
(
x
) =
1
Example
−
3
−
2
−
1
1
2
3
−
1
1
0
Removable discontinuity: discontinuity can be removed by redefining the value of
f
at
a
so that
f
(
a
) =
lim
x
→
a
f
(
x
)
Recall:
g
(
x
) =
3
x
2
−
4
x
+
1
x
−
1
has a removable discontinuity at
x
=
1
.
1
2
3
−
1
1
2
3
4
0
g
(
x
) =
3
x
2
−
4
x
+
1
Example
Remark
Removable discontinuity: discontinuity can be removed by redefining the value of
f
at
a
so that
f
(
a
) =
lim
x
→
a
f
(
x
)
Redefining
g
at
1
such that
g
(
1
) =
2
, the function is now continuous!
1
2
3
−
1
1
2
3
4
0
G
(
x
) =
3
x
2−
4
x
+
1
x
−
1
,
x
6
=
1
Discuss the continuity of the following function at
x
=
1
and
x
=
2
:
f
(
x
) =
4
x
−
3,
x
<
1
x
−
2
2
x
2
−
5
x
+
2
,
x
≥
1
At
x
=
1
1
f
(
1
) =
1
−
2
2
(
1
)
2
−
5
(
1
) +
2
=
1
2
lim
x→
1
−f
(x) =
x→
lim
1
−(
4
x
−
3
) =
4
(
1
)
−
3
=
1
3
lim
x→
1
+f
(x) =
x→
lim
1
+x
−
2
2
x
2
−
5
x
+
2
=
1
Thus,
lim
Example
Discuss the continuity of the following function at
x
=
1
and
x
=
2
:
f
(
x
) =
4
x
−
3,
x
<
1
x
−
2
2
x
2
−
5
x
+
2
,
x
≥
1
At
x
=
2
1
f
(
2
)
: undefined
2
lim
x→
2
f
(x) =
x→
lim
2
x
−
2
(x
−
2
)(
2
x
−
1
)
=
lim
x→
2
1
2
x
−
1
=
1
3
Thus,
lim
−
1
1
2
3
Example
Redefine
f
at point with removable discontinuity:
f
(
x
) =
4
x
−
3,
x
<
1
x
−
2
2
x
2
−
5
x
+
2
,
x
≥
1
Since
f
has a removable discontinuity at
x
=
2
, we redefine
f
as follows:
F
(
x
) =
4
x
−
3,
x
<
1
x
−
2
2
x
2
−
5
x
+
2
,
x
≥
1,
x
6
=
2
1
Redefine
f
at point with removable discontinuity:
f
(
x
) =
4
x
−
3,
x
<
1
x
−
2
2
x
2
−
5
x
+
2
,
x
≥
1
or simply:
F
(
x
) =
f
(
x
)
,
x
6
=
2
1
Theorem
Let
f
and
g
be continuous at
x
=
a
and let
c
∈
R
. Then the following are also
continuous at
x
=
a
:
1
f
±
g
2f g
3
f
g
, provided
g
(
a
)
6
=
0
Examples
Let
f
(
x
) =
√
2
−
x
.
f
(
2
) =
√
2
−
2
=
0
lim
x
→2
−
√
2
−
x
=
0
(
√
0
+
)
lim
x
→2
+
√
2
−
x
dne
(
√
0
−
)
Thus,
f
is continuous from the left at
x
=
2
but not from the right at
x
=
2
.
−
4
−
3
−
2
−
1
1
2
3
4
−
2
−
1
1
2
3
Let
g
(
x
) = [[
x
]]
and let
n
∈
Z
. Then
g
is continuous from the right at
x
=
n
but
not from the left at
x
=
n
.
−
4
−
3
−
2
−
1
1
2
3
4
−
2
−
1
1
2
3
Continuity on an Interval
Definition
A function
f
is said to be continuous
1
everywhere if
f
is continuous at every real number
2on
(
a
,
b
)
if
f
is continuous at every point in
(
a
,
b
)
3
on
[
a
,
b
)
if
f
is continuous on
(
a
,
b
)
and from the right of
a
4on
(
a
,
∞
)
if
f
is continuous at every
x
>
a
5
on
[
a
,
∞
)
if
f
is continuous on
(
a
,
∞
)
and from the right of
a
Similar definitions for:
1
Polynomial functions are continuous everywhere.
2
Rational functions are continuous on their respective domains.
3f
(
x
) =
|x|
is continuous everywhere.
4
f
(
x
) =
√
x
is continuous on
[
0,
∞
)
.
Theorem
If
lim
x
→
a
g
(
x
) =
b
and
f
is continuous at
b
, then
x
lim
→
a
f
(
g
(
x
)) =
f
(
b
)
.
That is, if
f
is continuous at
lim
x
→
a
g
(
x
)
, then
lim
x
→
a
f
(
g
(
x
)) =
f
lim
x
→
a
g
(
x
)
Evaluate:
lim
x
→2
x
−
2
x
2
−
4
Solution:
lim
x
→2
x
−
2
x
2
−
4
=
x
lim
→2
1
x
+
2
=
1
4
The absolute value function is continuous everywhere.
By the previous theorem,
lim
x
→2
x
−
2
x
2
−
4
=
x
lim
→2
x
−
2
x
2
−
2
Example
Evaluate:
lim
x
→
1 3[[
3
x
2
]]
Solution:
lim
x
→
1 33
x
2
=
1
3
, which is not an integer.
The greatest integer function is continuous at
1
3
.
By the previous theorem,
lim
x
→
1 3[[
3
x
2
]] =
"
"
lim
x
→
1 33
x
2
Example
Determine the largest interval for which
h
(
x
) =
√
x
2
−
1
is continuous.
The square root function is continuous at every
x
>
0
.
The function
p
(
x
) =
x
2
−
1
is continuous everywhere.
h
(
x
) =
p
p
(
x
)
is continuous at all
x
such that
p
(
x
)
>
0
or
x
2
−
1
>
0
or
x
∈
(
−
∞
,
−
1
)
∪
(
1,
∞
)
(why?)
Example
Discuss the continuity of
f
(
x
) =
x
−
2
2
x
+
6
,
x
≤
1
2
x
2
+
x
−
6
|
2
x
−
3
|
,
x
>
1
Note:
|
2
x
−
3
|
=
(
2
x
−
3,
2
x
−
3
≥
0
⇔
x
≥
3
2
Discuss the continuity of
f
(
x
) =
x
−
2
2
x
+
6
,
x
≤
1
2
x
2
+
x
−
6
|
2
x
−
3
|
,
x
>
1
f
is undefined:
x
=
−
3
and
x
=
3
2
Endpoints of intervals:
x
=
1
(and
x
=
3
2
)
Example
Discuss the continuity of
f
(
x
) =
x
−
2
2
x
+
6
,
x
≤
1
2
x
2
+
x
−
6
|
2
x
−
3
|
,
x
>
1
At
x
=
−
3
1
f
(
−
3
)
is undefined
2
lim
x→−
3
−f
(x) =
x→−
lim
3
−x
−
2
2
x
+
6
(
−5 0−)
=
+
∞
3
lim
x→−
3
+f
(x) =
x→−
lim
3
+x
−
2
2
x
+
6
(
−5 0+)
f
(
x
) =
x
−
2
2
x
+
6
,
x
≤
1
2
x
2
+
x
−
6
|
2
x
−
3
|
,
x
>
1
At
x
=
3
2
1
f
3
2
is undefined
2lim
x→
3 2−
f
(x) =
lim
x→
3 2−
2
x
2
+
x
−
6
|
2
x
−
3
|
=
x→
lim
3 2−
(
2
x
−
3
)(x
+
2
)
−
(
2
x
−
3
)
=
−
7
2
3
lim
x→
3 2+
f
(x) =
lim
x→
3 2+
2
x
2
+
x
−
6
|
2
x
−
3
|
=
x→
lim
32
+
(
2
x
−
3
)(x
+
2
)
2
x
−
3
=
7
2
Example
Discuss the continuity of
f
(
x
) =
x
−
2
2
x
+
6
,
x
≤
1
2
x
2
+
x
−
6
|
2
x
−
3
|
,
x
>
1
At
x
=
1
1
f
(
1
) =
1
−
2
2
(
1
) +
6
=
−
1
8
2
lim
x→
1
−f
(x) =
x→
lim
1
−x
−
2
2
x
+
6
=
−
1
8
3
lim
x→
1
+f
(x) =
x→
lim
1
+2
x
2
+
x
−
6
|
2
x
−
3
|
=
2
x
2
+
x
−
6
−
(
2
x
−
3
)
=
−
3
−
7
−
6
−
5
−
4
−
3
−
2
−
1
1
2
3
4
Example
Discuss the continuity of
g
(
x
) =
1
x
2
−
x
+
1
x
,
x
<
0
[[
x
−
1
]]
,
0
≤
x
≤
2
√
x
−
2,
x
>
2
Note:
[[
x
−
1
]] =
−
1, 0
≤
x
<
1
0,
1
≤
x
<
2
g
(
x
) =
1
x
2
−
x
+
1
x
,
x
<
0
[[
x
−
1
]]
,
0
≤
x
≤
2
√
x
−
2,
x
>
2
At
x
=
0
1
g
(
0
) = [[
0
−
1
]] =
−
1
2
lim
x→
0
−g(x) =
x→
lim
0
−1
x
2
−
x
+
1
x
=
lim
x→
0
−1
+
x
−
1
x(x
−
1
)
=
x→
lim
0
−1
x
−
1
=
−
1
3
lim
x→
0
+g(x) =
x→
lim
0
+[[x
−
1
]] =
x→
lim
0
+(
−
1
) =
−
1
Example
Discuss the continuity of
g
(
x
) =
1
x
2
−
x
+
1
x
,
x
<
0
[[
x
−
1
]]
,
0
≤
x
≤
2
√
x
−
2,
x
>
2
At
x
=
2
1
g
(
2
) = [[
2
−
1
]] =
1
2
lim
x→
2
−g(x) =
x→
lim
2
−[[x
−
1
]] =
x→
lim
2
−0
=
0
3
lim
x→
2
+g(x) =
x→
lim
2
+√
g
(
x
) =
1
x
2
−
x
+
1
x
,
x
<
0
[[
x
−
1
]]
,
0
≤
x
≤
2
√
x
−
2,
x
>
2
At
x
=
1
1
g
(
1
) = [[
1
−
1
]] =
0
2
lim
x→
1
−g(x) =
x→
lim
1
−[[x
−
1
]] =
x→
lim
1
−−
1
=
−
1
3
lim
x→
1
+g(x) =
x→
lim
1
+[[x
−
1
]] =
x→
lim
1
+0
=
0
Example
−
4
−
3
−
2
−
1
1
2
3
4
5
6
Given:
f
(
x
) =
2
+
x
if
x
<
−
1
[[
x
−
1
]]
if
−
1
≤
x
<
1
x
2
−
4
x
4
−
x
if
x
≥
1
1
Find all the possible points of discontinuity of
f
.
