Limit of a Function and
One-sided limits
Mathematics 53
Institute of Mathematics (UP Diliman)
For today
1
Limit of a Function: An intuitive approach
2
Evaluating Limits
3
One-sided Limits
For today
1
Limit of a Function: An intuitive approach
2
Evaluating Limits
3
One-sided Limits
Introduction
Given a function
f
(x)
and
a
∈
R
,
what is the value of
f
at
x
near
a
,
but not equal to
a
?
Introduction
Given a function
f
(x)
and
a
∈
R
,
what is the value of
f
at
x
near
a
,
but not equal to
a
?
Introduction
Given a function
f
(x)
and
a
∈
R
,
what is the value of
f
at
x
near
a
,
but not equal to
a
?
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(x)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
Consider
f
(x) =
3
x
−
1
.
What can we say about values of
f
(
x
)
for values of
x
near
1
but not equal to
1
?
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
Based on the table, as
x
gets closer and closer to
1
,
f
(x)
gets closer and closer
to
2
.
Illustration 1
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
−
1
1
2
3
−
1
1
2
3
4
Illustration 1
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
−
1
1
2
3
−
1
1
2
3
4
Illustration 1
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
−
1
1
2
3
−
1
1
2
3
4
Illustration 1
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
−
1
1
2
3
−
1
1
2
3
4
Illustration 1
x
f
(x)
0
−
1
0.5
0.5
0.9
1.7
0.99
1.97
0.99999
1.99997
−
1
1
2
3
−
1
1
2
3
4
Illustration 1
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
f
(x)
gets closer and closer to
2
.
Illustration 1
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
f
(x)
gets closer and closer to
2
.
Illustration 1
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
f
(x)
gets closer and closer to
2
.
Illustration 1
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
f
(x)
gets closer and closer to
2
.
Illustration 1
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
f
(x)
gets closer and closer to
2
.
Illustration 1
x
f
(x)
2
5
1.5
3.5
1.1
2.3
1.001
2.003
1.00001
2.00003
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
f
(x)
gets closer and closer to
2
.
Illustration 2
Consider:
g(x) =
3
x
2
−
4
x
+
1
x
−
1
=
(
3
x
−
1
)(
x
−
1
)
x
−
1
=
3
x
−
1,
x
6
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
g(x)
gets closer and closer to
2
.
Illustration 2
Consider:
g(x) =
3
x
2
−
4
x
+
1
x
−
1
=
(
3
x
−
1
)(x
−
1
)
x
−
1
=
3
x
−
1,
x
6
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
g(x)
gets closer and closer to
2
.
Illustration 2
Consider:
g(x) =
3
x
2
−
4
x
+
1
x
−
1
=
(
3
x
−
1
)(x
−
1
)
x
−
1
=
3
x
−
1,
x
6
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
g(x)
gets closer and closer to
2
.
Illustration 2
Consider:
g(x) =
3
x
2
−
4
x
+
1
x
−
1
=
(
3
x
−
1
)(x
−
1
)
x
−
1
=
3
x
−
1,
x
6
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
g(x)
gets closer and closer to
2
.
Illustration 2
Consider:
g(x) =
3
x
2
−
4
x
+
1
x
−
1
=
(
3
x
−
1
)(x
−
1
)
x
−
1
=
3
x
−
1,
x
6
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
g(x)
gets closer and closer to
2
.
Illustration 2
Consider:
g(x) =
3
x
2
−
4
x
+
1
x
−
1
=
(
3
x
−
1
)(x
−
1
)
x
−
1
=
3
x
−
1,
x
6
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
g(x)
gets closer and closer to
2
.
Illustration 2
Consider:
g(x) =
3
x
2
−
4
x
+
1
x
−
1
=
(
3
x
−
1
)(x
−
1
)
x
−
1
=
3
x
−
1,
x
6
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
g(x)
gets closer and closer to
2
.
Illustration 2
Consider:
g(x) =
3
x
2
−
4
x
+
1
x
−
1
=
(
3
x
−
1
)(x
−
1
)
x
−
1
=
3
x
−
1,
x
6
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
g(x)
gets closer and closer to
2
.
Illustration 2
Consider:
g(x) =
3
x
2
−
4
x
+
1
x
−
1
=
(
3
x
−
1
)(x
−
1
)
x
−
1
=
3
x
−
1,
x
6
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
g(x)
gets closer and closer to
2
.
Illustration 3
Consider:
h(x) =
3
x
−
1,
x
6
=
1
0,
x
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
h(x)
gets closer and closer to
2
.
Illustration 3
Consider:
h(x) =
3
x
−
1,
x
6
=
1
0,
x
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
h(x)
gets closer and closer to
2
.
Illustration 3
Consider:
h(x) =
3
x
−
1,
x
6
=
1
0,
x
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
h(x)
gets closer and closer to
2
.
Illustration 3
Consider:
h(x) =
3
x
−
1,
x
6
=
1
0,
x
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
h(x)
gets closer and closer to
2
.
Illustration 3
Consider:
h(x) =
3
x
−
1,
x
6
=
1
0,
x
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
h(x)
gets closer and closer to
2
.
Illustration 3
Consider:
h(x) =
3
x
−
1,
x
6
=
1
0,
x
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
h(x)
gets closer and closer to
2
.
Illustration 3
Consider:
h(x) =
3
x
−
1,
x
6
=
1
0,
x
=
1
−
1
1
2
3
−
1
1
2
3
4
As
x
gets closer and closer to
1
,
h(x)
gets closer and closer to
2
.
Limit
Intuitive Notion of a Limit
a
∈
R
,
L
∈
R
f
(x)
: function defined on some open interval containing
a
, except possibly at
a
The limit of
f
(x)
as
x
approaches
a
is
L
if the values of
f
(x)
get closer and closer to
L
as
x
assumes values getting closer
and closer to
a
but not reaching
a
.
Notation:
lim
x
→
a
f
(x) =
L
Limit
Intuitive Notion of a Limit
a
∈
R
,
L
∈
R
f
(x)
: function defined on some open interval containing
a
, except possibly at
a
The limit of
f
(x)
as
x
approaches
a
is
L
if the values of
f
(x)
get closer and closer to
L
as
x
assumes values getting closer
and closer to
a
but not reaching
a
.
Notation:
lim
x
→
a
f
(x) =
L
Limit
Intuitive Notion of a Limit
a
∈
R
,
L
∈
R
f
(x)
: function defined on some open interval containing
a
, except possibly at
a
The limit of
f
(x)
as
x
approaches
a
is
L
if the values of
f
(x)
get closer and closer to
L
as
x
assumes values getting closer
and closer to
a
but not reaching
a
.
Notation:
lim
x
→
a
f
(x) =
L
Limit
Intuitive Notion of a Limit
a
∈
R
,
L
∈
R
f
(x)
: function defined on some open interval containing
a
, except possibly at
a
The limit of
f
(x)
as
x
approaches
a
is
L
if the values of
f
(x)
get closer and closer to
L
as
x
assumes values getting closer
and closer to
a
but not reaching
a
.
Notation:
lim
x
→
a
f
(x) =
L
Limit
Intuitive Notion of a Limit
a
∈
R
,
L
∈
R
f
(x)
: function defined on some open interval containing
a
, except possibly at
a
The limit of
f
(x)
as
x
approaches
a
is
L
if the values of
f
(x)
get closer and closer to
L
as
x
assumes values getting closer
and closer to
a
but not reaching
a
.
Notation:
lim
x
→
a
f
(x) =
L
Limit
Intuitive Notion of a Limit
a
∈
R
,
L
∈
R
f
(x)
: function defined on some open interval containing
a
, except possibly at
a
The limit of
f
(x)
as
x
approaches
a
is
L
if the values of
f
(x)
get closer and closer to
L
as
x
assumes values getting closer
and closer to
a
but not reaching
a
.
Notation:
lim
x
→
a
f
(x) =
L
Limit
Intuitive Notion of a Limit
a
∈
R
,
L
∈
R
f
(x)
: function defined on some open interval containing
a
, except possibly at
a
The limit of
f
(x)
as
x
approaches
a
is
L
if the values of
f
(x)
get closer and closer to
L
as
x
assumes values getting closer
and closer to
a
but not reaching
a
.
Notation:
lim
x
→
a
f
(x) =
L
Examples
f
(x) =
3
x
−
1
−
1
1
2
3
−
1
1
2
3
4
lim
x
→
1
(
3
x
−
1
) =
2
Note: In this case,
lim
x
→
1
f
(x) =
f
(
1
)
.
Examples
f
(x) =
3
x
−
1
−
1
1
2
3
−
1
1
2
3
4
x
lim
→
1
(
3
x
−
1
)
=
2
Note: In this case,
lim
x
→
1
f
(x) =
f
(
1
)
.
Examples
f
(x) =
3
x
−
1
−
1
1
2
3
−
1
1
2
3
4
x
lim
→
1
(
3
x
−
1
) =
2
Note: In this case,
lim
x
→
1
f
(x) =
f
(
1
)
.
Examples
f
(x) =
3
x
−
1
−
1
1
2
3
−
1
1
2
3
4
x
lim
→
1
(
3
x
−
1
) =
2
Note: In this case,
lim
x
→
1
f
(x)
=
f
(
1
)
.
Examples
f
(x) =
3
x
−
1
−
1
1
2
3
−
1
1
2
3
4
x
lim
→
1
(
3
x
−
1
) =
2
Note: In this case,
lim
x
→
1
f
(x) =
f
(
1
)
.
Examples
g(x) =
3
x
2
−
4
x
+
1
x
−
1
−
1
1
2
3
−
1
1
2
3
4
lim
x
→
1
3
x
2
−
4
x
+
1
x
−
1
=
2
Note: Though
g(
1
)
is undefined,
lim
x
→
1
g(x)
exists.
Examples
g(x) =
3
x
2
−
4
x
+
1
x
−
1
−
1
1
2
3
−
1
1
2
3
4
lim
x
→
1
3
x
2
−
4
x
+
1
x
−
1
=
2
Note: Though
g(
1
)
is undefined,
lim
x
→
1
g(x)
exists.
Examples
g(x) =
3
x
2
−
4
x
+
1
x
−
1
−
1
1
2
3
−
1
1
2
3
4
lim
x
→
1
3
x
2
−
4
x
+
1
x
−
1
=
2
Note: Though
g(
1
)
is undefined,
lim
x
→
1
g(x)
exists.
Examples
g(x) =
3
x
2
−
4
x
+
1
x
−
1
−
1
1
2
3
−
1
1
2
3
4
lim
x
→
1
3
x
2
−
4
x
+
1
x
−
1
=
2
Note: Though
g(
1
)
is undefined,
lim
x
→
1
g(x)
exists.
Examples
h(x) =
3
x
−
1,
x
6
=
1
0,
x
=
1
−
1
1
2
3
−
1
1
2
3
4
lim
x
→
1
h(x) =
2
Note:
h(
1
)
6
=
lim
x
→
1
h(x)
.
Examples
h(x) =
3
x
−
1,
x
6
=
1
0,
x
=
1
−
1
1
2
3
−
1
1
2
3
4
lim
x
→
1
h(x)
=
2
Note:
h(
1
)
6
=
lim
x
→
1
h(x)
.
Examples
h(x) =
3
x
−
1,
x
6
=
1
0,
x
=
1
−
1
1
2
3
−
1
1
2
3
4
lim
x
→
1
h(x) =
2
Note:
h(
1
)
6
=
lim
x
→
1
h(x)
.
Examples
h(x) =
3
x
−
1,
x
6
=
1
0,
x
=
1
−
1
1
2
3
−
1
1
2
3
4
lim
x
→
1
h(x) =
2
Note:
h(
1
)
6
=
lim
x
→
1
h(x)
.
Some Remarks
Remark
In finding
lim
x
→
a
f
(x)
:
We only need to consider values of
x
very close to
a
but not exactly at
a
.
Thus,
lim
x
→
a
f
(x)
is
NOT NECESSARILY
the same as
f
(a)
.
We let
x
approach
a
from
BOTH SIDES
of
a
.
Some Remarks
Remark
In finding
lim
x
→
a
f
(x)
:
We only need to consider values of
x
very close to
a
but not exactly at
a
.
Thus,
lim
x
→
a
f
(x)
is
NOT NECESSARILY
the same as
f
(a)
.
We let
x
approach
a
from
BOTH SIDES
of
a
.
Some Remarks
Remark
In finding
lim
x
→
a
f
(x)
:
We only need to consider values of
x
very close to
a
but not exactly at
a
.
Thus,
lim
x
→
a
f
(x)
is
NOT NECESSARILY
the same as
f
(a)
.
We let
x
approach
a
from
BOTH SIDES
of
a
.
Some Remarks
Remark
In finding
lim
x
→
a
f
(x)
:
We only need to consider values of
x
very close to
a
but not exactly at
a
.
Thus,
lim
x
→
a
f
(x)
is
NOT NECESSARILY
the same as
f
(a)
.
We let
x
approach
a
from
BOTH SIDES
of
a
.
Some Remarks
If
f
(x)
does not approach any
particular real number as
x
approaches
a
, then we say
lim
x
→
a
f
(x)
does not exist (dne).
e.g.
H(x) =
1,
x
≥
0
0,
x
<
0
−3
−2
−1
1
2
3
1
2
3
0
lim
x
→
0
H(x) =
0
? No.
lim
x
→
0
H(x) =
1
? No.
lim
x
→
0
H(x)
dne
Some Remarks
If
f
(x)
does not approach any
particular real number as
x
approaches
a
, then we say
lim
x
→
a
f
(x)
does not exist (dne).
e.g.
H(x) =
1,
x
≥
0
0,
x
<
0
−3
−2
−1
1
2
3
1
2
3
0
lim
x
→
0
H(x) =
0
? No.
lim
x
→
0
H(x) =
1
? No.
lim
x
→
0
H(x)
dne
Some Remarks
If
f
(x)
does not approach any
particular real number as
x
approaches
a
, then we say
lim
x
→
a
f
(x)
does not exist (dne).
e.g.
H(x) =
1,
x
≥
0
0,
x
<
0
−3
−2
−1
1
2
3
1
2
3
0
lim
x
→
0
H(x) =
0
? No.
lim
x
→
0
H(x) =
1
? No.
lim
x
→
0
H(x)
dne
Some Remarks
If
f
(x)
does not approach any
particular real number as
x
approaches
a
, then we say
lim
x
→
a
f
(x)
does not exist (dne).
e.g.
H(x) =
1,
x
≥
0
0,
x
<
0
−3
−2
−1
1
2
3
1
2
3
0
lim
x
→
0
H(x) =
0
?
No.
lim
x
→
0
H(x) =
1
? No.
lim
x
→
0
H(x)
dne
Some Remarks
If
f
(x)
does not approach any
particular real number as
x
approaches
a
, then we say
lim
x
→
a
f
(x)
does not exist (dne).
e.g.
H(x) =
1,
x
≥
0
0,
x
<
0
−3
−2
−1
1
2
3
1
2
3
0
lim
x
→
0
H(x) =
0
? No.
lim
x
→
0
H(x) =
1
? No.
lim
x
→
0
H(x)
dne
Some Remarks
If
f
(x)
does not approach any
particular real number as
x
approaches
a
, then we say
lim
x
→
a
f
(x)
does not exist (dne).
e.g.
H(x) =
1,
x
≥
0
0,
x
<
0
−3
−2
−1
1
2
3
1
2
3
0
lim
x
→
0
H(x) =
0
? No.
lim
x
→
0
H
(
x
) =
1
?
No.
lim
x
→
0
H(x)
dne
Some Remarks
If
f
(x)
does not approach any
particular real number as
x
approaches
a
, then we say
lim
x
→
a
f
(x)
does not exist (dne).
e.g.
H(x) =
1,
x
≥
0
0,
x
<
0
−3
−2
−1
1
2
3
1
2
3
0
lim
x
→
0
H(x) =
0
? No.
lim
x
→
0
H
(
x
) =
1
? No.
lim
x
→
0
H(x)
dne
Some Remarks
If
f
(x)
does not approach any
particular real number as
x
approaches
a
, then we say
lim
x
→
a
f
(x)
does not exist (dne).
e.g.
H(x) =
1,
x
≥
0
0,
x
<
0
−3
−2
−1
1
2
3
1
2
3
0
lim
x
→
0
H(x) =
0
? No.
lim
x
→
0
H
(
x
) =
1
? No.
lim
x
→
0
H(x)
dne
For today
1
Limit of a Function: An intuitive approach
2
Evaluating Limits
3
One-sided Limits
Limit Theorems
Theorem
If
lim
x
→
a
f
(x)
exists, then it is unique.
If
c
∈
R
, then
lim
x
→
a
c
=
c
.
lim
x
→
a
x
=
a
Limit Theorems
Theorem
If
lim
x
→
a
f
(x)
exists, then it is unique.
If
c
∈
R
, then
lim
x
→
a
c
=
c
.
lim
x
→
a
x
=
a
Limit Theorems
Theorem
If
lim
x
→
a
f
(x)
exists, then it is unique.
If
c
∈
R
, then
lim
x
→
a
c
=
c
.
lim
x
→
a
x
=
a
Limit Theorems
Theorem
If
lim
x
→
a
f
(x)
exists, then it is unique.
If
c
∈
R
, then
lim
x
→
a
c
=
c
.
lim
x
→
a
x
=
a
Limit Theorems
Theorem
If
lim
x
→
a
f
(x)
exists, then it is unique.
If
c
∈
R
, then
lim
x
→
a
c
=
c
.
lim
x
→
a
x
=
a
Limit Theorems
Theorem
If
lim
x
→
a
f
(x)
exists, then it is unique.
If
c
∈
R
, then
lim
x
→
a
c
=
c
.
lim
x
→
a
x
=
a
Limit Theorems
Theorem
Suppose
lim
x
→
a
f
(x) =
L
1
and
x
lim
→
a
g(x) =
L
2
. Let
c
∈
R,
n
∈
N.
lim
x
→
a
[
f
(x)
±
g(x)] =
x
lim
→
a
f
(x)
±
lim
x
→
a
g(x) =
L
1
±
L
2
lim
x
→
a
[
f
(
x
)
g
(
x
)] =
lim
x
→
a
f
(
x
)
x
lim
→
a
g
(
x
)
=
L
1
L
2
lim
x
→
a
[c f
(x)] =
c
x
lim
→
a
f
(x) =
cL
1
lim
x
→
a
f
(x)
g(x)
=
lim
x
→
a
f
(x)
lim
x
→
a
g(x)
=
L
1
L
2
, provided
L
2
6
=
0
lim
x
→
a
(
f
(x))
n
=
lim
x
→
a
f
(x)
n
= (L
1
)
n
Limit Theorems
Theorem
Suppose
lim
x
→
a
f
(x) =
L
1
and
x
lim
→
a
g(x) =
L
2
. Let
c
∈
R,
n
∈
N.
lim
x
→
a
[
f
(x)
±
g(x)]
=
lim
x
→
a
f
(x)
±
lim
x
→
a
g(x) =
L
1
±
L
2
lim
x
→
a
[
f
(
x
)
g
(
x
)] =
lim
x
→
a
f
(
x
)
x
lim
→
a
g
(
x
)
=
L
1
L
2
lim
x
→
a
[c f
(x)] =
c
x
lim
→
a
f
(x) =
cL
1
lim
x
→
a
f
(x)
g(x)
=
lim
x
→
a
f
(x)
lim
x
→
a
g(x)
=
L
1
L
2
, provided
L
2
6
=
0
lim
x
→
a
(
f
(x))
n
=
lim
x
→
a
f
(x)
n
= (L
1
)
n
Limit Theorems
Theorem
Suppose
lim
x
→
a
f
(x) =
L
1
and
x
lim
→
a
g(x) =
L
2
. Let
c
∈
R,
n
∈
N.
lim
x
→
a
[
f
(x)
±
g(x)] =
x
lim
→
a
f
(x)
±
lim
x
→
a
g(x) =
L
1
±
L
2
lim
x
→
a
[
f
(
x
)
g
(
x
)] =
lim
x
→
a
f
(
x
)
x
lim
→
a
g
(
x
)
=
L
1
L
2
lim
x
→
a
[c f
(x)] =
c
x
lim
→
a
f
(x) =
cL
1
lim
x
→
a
f
(x)
g(x)
=
lim
x
→
a
f
(x)
lim
x
→
a
g(x)
=
L
1
L
2
, provided
L
2
6
=
0
lim
x
→
a
(
f
(x))
n
=
lim
x
→
a
f
(x)
n
= (L
1
)
n
Limit Theorems
Theorem
Suppose
lim
x
→
a
f
(x) =
L
1
and
x
lim
→
a
g(x) =
L
2
. Let
c
∈
R,
n
∈
N.
lim
x
→
a
[
f
(x)
±
g(x)] =
x
lim
→
a
f
(x)
±
lim
x
→
a
g(x)
=
L
1
±
L
2
lim
x
→
a
[
f
(
x
)
g
(
x
)] =
lim
x
→
a
f
(
x
)
x
lim
→
a
g
(
x
)
=
L
1
L
2
lim
x
→
a
[c f
(x)] =
c
x
lim
→
a
f
(x) =
cL
1
lim
x
→
a
f
(x)
g(x)
=
lim
x
→
a
f
(x)
lim
x
→
a
g(x)
=
L
1
L
2
, provided
L
2
6
=
0
lim
x
→
a
(
f
(x))
n
=
lim
x
→
a
f
(x)
n
= (L
1
)
n
Limit Theorems
Theorem
Suppose
lim
x
→
a
f
(x) =
L
1
and
x
lim
→
a
g(x) =
L
2
. Let
c
∈
R,
n
∈
N.
lim
x
→
a
[
f
(x)
±
g(x)] =
x
lim
→
a
f
(x)
±
lim
x
→
a
g(x) =
L
1
±
L
2
lim
x
→
a
[
f
(
x
)
g
(
x
)] =
lim
x
→
a
f
(
x
)
x
lim
→
a
g
(
x
)
=
L
1
L
2
lim
x
→
a
[c f
(x)] =
c
x
lim
→
a
f
(x) =
cL
1
lim
x
→
a
f
(x)
g(x)
=
lim
x
→
a
f
(x)
lim
x
→
a
g(x)
=
L
1
L
2
, provided
L
2
6
=
0
lim
x
→
a
(
f
(x))
n
=
lim
x
→
a
f
(x)
n
= (L
1
)
n
Limit Theorems
Theorem
Suppose
lim
x
→
a
f
(x) =
L
1
and
x
lim
→
a
g(x) =
L
2
. Let
c
∈
R,
n
∈
N.
lim
x
→
a
[
f
(x)
±
g(x)] =
x
lim
→
a
f
(x)
±
lim
x
→
a
g(x) =
L
1
±
L
2
lim
x
→
a
[
f
(x)g(x)]
=
lim
x
→
a
f
(
x
)
x
lim
→
a
g
(
x
)
=
L
1
L
2
lim
x
→
a
[c f
(x)] =
c
x
lim
→
a
f
(x) =
cL
1
lim
x
→
a
f
(x)
g(x)
=
lim
x
→
a
f
(x)
lim
x
→
a
g(x)
=
L
1
L
2
, provided
L
2
6
=
0
lim
x
→
a
(
f
(x))
n
=
lim
x
→
a
f
(x)
n
= (L
1
)
n
Limit Theorems
Theorem
Suppose
lim
x
→
a
f
(x) =
L
1
and
x
lim
→
a
g(x) =
L
2
. Let
c
∈
R,
n
∈
N.
lim
x
→
a
[
f
(x)
±
g(x)] =
x
lim
→
a
f
(x)
±
lim
x
→
a
g(x) =
L
1
±
L
2
lim
x
→
a
[
f
(x)g(x)] =
lim
x
→
a
f
(x)
x
lim
→
a
g(x)
=
L
1
L
2
lim
x
→
a
[c f
(x)] =
c
x
lim
→
a
f
(x) =
cL
1
lim
x
→
a
f
(x)
g(x)
=
lim
x
→
a
f
(x)
lim
x
→
a
g(x)
=
L
1
L
2
, provided
L
2
6
=
0
lim
x
→
a
(
f
(x))
n
=
lim
x
→
a
f
(x)
n
= (L
1
)
n
Limit Theorems
Theorem
Suppose
lim
x
→
a
f
(x) =
L
1
and
x
lim
→
a
g(x) =
L
2
. Let
c
∈
R,
n
∈
N.
lim
x
→
a
[
f
(x)
±
g(x)] =
x
lim
→
a
f
(x)
±
lim
x
→
a
g(x) =
L
1
±
L
2
lim
x
→
a
[
f
(x)g(x)] =
lim
x
→
a
f
(x)
x
lim
→
a
g(x)
=
L
1
L
2
lim
x
→
a
[c f
(x)] =
c
x
lim
→
a
f
(x) =
cL
1
lim
x
→
a
f
(x)
g(x)
=
lim
x
→
a
f
(x)
lim
x
→
a
g(x)
=
L
1
L
2
, provided
L
2
6
=
0
lim
x
→
a
(
f
(x))
n
=
lim
x
→
a
f
(x)
n
= (L
1
)
n
Limit Theorems
Theorem
Suppose
lim
x
→
a
f
(x) =
L
1
and
x
lim
→
a
g(x) =
L
2
. Let
c
∈
R,
n
∈
N.
lim
x
→
a
[
f
(x)
±
g(x)] =
x
lim
→
a
f
(x)
±
lim
x
→
a
g(x) =
L
1
±
L
2
lim
x
→
a
[
f
(x)g(x)] =
lim
x
→
a
f
(x)
x
lim
→
a
g(x)
=
L
1
L
2
lim
x
→
a
[c f
(x)] =
c
lim
x
→
a
f
(x) =
cL
1
lim
x
→
a
f
(x)
g(x)
=
lim
x
→
a
f
(x)
lim
x
→
a
g(x)
=
L
1
L
2
, provided
L
2
6
=
0
lim
x
→
a
(
f
(x))
n
=
lim
x
→
a
f
(x)
n
= (L
1
)
n
Limit Theorems
Theorem
Suppose
lim
x
→
a
f
(x) =
L
1
and
x
lim
→
a
g(x) =
L
2
. Let
c
∈
R,
n
∈
N.
lim
x
→
a
[
f
(x)
±
g(x)] =
x
lim
→
a
f
(x)
±
lim
x
→
a
g(x) =
L
1
±
L
2
lim
x
→
a
[
f
(x)g(x)] =
lim
x
→
a
f
(x)
x
lim
→
a
g(x)
=
L
1
L
2
lim
x
→
a
[c f
(x)] =
c
x
lim
→
a
f
(x) =
cL
1
lim
x
→
a
f
(x)
g(x)
=
lim
x
→
a
f
(x)
lim
x
→
a
g(x)
=
L
1
L
2
, provided
L
2
6
=
0
lim
x
→
a
(
f
(x))
n
=
lim
x
→
a
f
(x)
n
= (L
1
)
n
Limit Theorems
Theorem
Suppose
lim
x
→
a
f
(x) =
L
1
and
x
lim
→
a
g(x) =
L
2
. Let
c
∈
R,
n
∈
N.
lim
x
→
a
[
f
(x)
±
g(x)] =
x
lim
→
a
f
(x)
±
lim
x
→
a
g(x) =
L
1
±
L
2
lim
x
→
a
[
f
(x)g(x)] =
lim
x
→
a
f
(x)
x
lim
→
a
g(x)
=
L
1
L
2
lim
x
→
a
[c f
(x)] =
c
x
lim
→
a
f
(x) =
cL
1
lim
x
→
a
f
(x)
g(x)
=
lim
x
→
a
f
(x)
lim
x
→
a
g(x)
=
L
1
L
2
, provided
L
2
6
=
0
lim
x
→
a
(
f
(x))
n
=
lim
x
→
a
f
(x)
n
= (L
1
)
n
Limit Theorems
Theorem
Suppose
lim
x
→
a
f
(x) =
L
1
and
x
lim
→
a
g(x) =
L
2
. Let
c
∈
R,
n
∈
N.
lim
x
→
a
[
f
(x)
±
g(x)] =
x
lim
→
a
f
(x)
±
lim
x
→
a
g(x) =
L
1
±
L
2
lim
x
→
a
[
f
(x)g(x)] =
lim
x
→
a
f
(x)
x
lim
→
a
g(x)
=
L
1
L
2
lim
x
→
a
[c f
(x)] =
c
x
lim
→
a
<