One-sided limits and
Infinite Limits
1
One-Sided Limits (continuation)
Recall
Intuitive Definition
The
limit of
f
(
x
)
as
x
approaches
a
from the left is
L
[limit of
f
(
x
)
as
x
approaches
a
from the right is
L
]
if the values of
f
(
x
)
get closer and closer to
L
as the values of
x
get closer and
closer to
a
, but are less [greater] than
a
.
Notation:
lim
x
→
a
−
f
(
x
) =
L
lim
x
→
a
+
f
(
x
) =
L
1
lim
x
→
a
f
(
x
) =
L
>
0
⇒
lim
x
→
a
np
f
(
x
) =
√
nL
2
lim
x
→
a
f
(
x
) =
L
<
0
⇒
lim
x
→
a
np
f
(
x
)
dne
Question:
lim
x
→
a
f
(
x
) =
0
⇒
lim
x
→
a
np
Consider:
f
(
x
) =
9
−
x
2
lim
x
→−
3
f
(
x
) =
0
and
lim
x
→
3
f
(
x
) =
0
(
−
∞
,
−
3
)
(
−
3, 3
)
(
3,
+
∞
)
(
3
+
x
)(
3
−
x
)
−
+
−
Notation
Suppose
lim
x
→
a
f
(
x
) =
0
.
If
f
(
x
)
approaches
0
through positive [negative] values,
we write
Consider:
f
(
x
) =
9
−
x
2
lim
x
→−
3
f
(
x
) =
0
and
lim
x
→
3
f
(
x
) =
0
(
−
∞
,
−
3
)
(
−
3, 3
)
(
3,
+
∞
)
(
3
+
x
)(
3
−
x
)
−
+
−
As
x
→ −
3
−
,
f
(
x
)
→
0
−
As
x
→ −
3
+
,
f
(
x
)
→
0
+
As
x
→
3
−
,
f
(
x
)
→
0
+
Consider:
f
(
x
) =
2
−
x
.
As
x
→
2
−
,
f
(
x
)
→
0
+
As
x
→
2
+
,
f
(
x
)
→
0
−
Consider:
g
(
x
) = (
2
−
x
)
2
.
Note
Let
n
be an even positive integer.
1
If as
x
→
a
,
f
(
x
)
→
0
+
⇒
lim
x
→
a
np
f
(
x
) =
0
2
If as
x
→
a
,
f
(
x
)
→
0
−
⇒
lim
x
→
a
np
lim
x
→−
1
−
√
2x
2
+
x
−
1
=
lim
x
→−
1
−
p
(
2x
−
1
)(
x
+
1
)
p
(
−
3
)(
0
−
)
√
0
+
=
0
lim
x
→−
1
+
√
2x
2
+
x
−
1
=
lim
x
→−
1
+
p
(
2x
−
1
)(
x
+
1
)
p
(
−
3
)(
0
+
)
√
0
−
dne
lim
x
→−
1
p
Example
f
(
x
) =
x
2
−
x
−
2
x
+
1
,
x
<
0
√
4
−
x,
0
≤
x
≤
4
x
2
−
5x
+
4,
x
>
4
lim
x
→
0
f
(
x
) =
?
dne
lim
x
→
0
−
f
(
x
) =
x
lim
→
0
−
x
2
−
x
−
2
x
+
1
=
−
2
lim
x
→
0
+
f
(
x
) =
x
lim
→
0
+
√
f
(
x
) =
x
2
−
x
−
2
x
+
1
,
x
<
0
√
4
−
x,
0
≤
x
≤
4
x
2
−
5x
+
4,
x
>
4
lim
x
→
4
f
(
x
) =
?0
lim
x
→
4
−
f
(
x
) =
x
lim
→
4
−
√
4
−
x
=
0
√
0
+
lim
x
→
4
+
f
(
x
) =
x
lim
→
4
+
(
x
Example
f
(
x
) =
x
2
−
x
−
2
x
+
1
,
x
<
0
√
4
−
x,
0
≤
x
≤
4
x
2
−
5x
+
4,
x
>
4
lim
x
→−
1
f
(
x
) =
?
−
3
lim
x
→−
1
f
(
x
) =
x
lim
→−
1
x
2
−
x
−
2
x
+
1
=
x
lim
→−
1
(
x
−
2
)(
x
+
1
)
Evaluate
lim
t
→−
4
t
+
4
|
t
+
4
|
.
Solution:
|
t
+
4
|
=
t
+
4,
t
+
4
≥
0t
≥ −
4
−
(
t
+
4
)
,
t
+
4
<
0t
<
−
4
lim
t
→−
4
−
t
+
4
|
t
+
4
|
=
t
→−
lim
4
−
t
+
4
−
(
t
+
4
)
=
t
→−
lim
4
−
(
−
1
) =
−
1
lim
t
→−
4
+
t
+
4
|
t
+
4
|
=
t
→−
lim
4
+
t
+
4
(
t
+
4
)
=
t
→−
lim
4
+
(
1
) =
1
∴
lim
t
→−
4
t
+
4
The Greatest Integer Function
Recall
[[
x
]] =
n
for
n
≤
x
<
n
+
1,
n
∈
Z
[[
x
]] =
..
.
−
3,
−
3
≤
x
<
−
2
−
2,
−
2
≤
x
<
−
1
−
1,
−
1
≤
x
<
0
0,
0
≤
x
<
1
1,
1
≤
x
<
2
..
.
1
lim
x
→−
2
−
[[
x
]] =
x
→−
lim
2
−
−
3
=
−
3
2
lim
x
→−
2
+
[[
x
]] =
x
→−
lim
2
+
−
2
=
−
2
3
lim
x
→
1 2[[
x
]] =
lim
x
→
1 2f
(
x
) = [[
2x
−
1
]] =
n
for
n
≤
2x
−
1
<
n
+
1,
n
∈
Z
n
+
1
≤
2x
<
n
+
2,
n
∈
Z
n
+
1
2
≤
x
<
n
+
2
2
,
n
∈
Z
[[
2x
−
1
]] =
..
.
−
1,
0
≤
x
<
1
2
0,
1
2
≤
x
<
1
1,
1
≤
x
<
3
2
..
.
1
lim
x
→
1 2−
[[
2x
−
1
]] =
lim
x
→
1 2−
−
1
=
−
1
2
lim
x
→
1 2+
[[
2x
−
1
]] =
lim
x
→
1 2+
0
=
0
3
lim
x
→
1 4[[
2x
−
1
]] =
lim
x
→
1 4[[
x
]] =
.
.
.
−
2,
−
2
≤
x
<
−
1
−
1,
−
1
≤
x
<
0
0,
0
≤
x
<
1
1,
1
≤
x
<
2
.
.
.
[[2
x
−
1]] =
.
.
.
−
1,
0
≤
x
<
1
2
0,
1
2
≤
x
<
1
1,
1
≤
x
<
3
2
2,
3
2
≤
x
<
2
.
.
.
Observe:
[[
f
(
x
)]]
“splits” at the points which make
f
(
x
)
an integer.
Suppose
f
(
x
)
→
n
as
x
approaches
a
in some direction, where
n
∈
Z
.
If
f
(
x
)
→
n
+
, then as
x
approaches
a
in the given direction,
[[
f
(
x
)]]
takes on a
constant value of
n
.
If
f
(
x
)
→
n
−
, then as
x
approaches
a
in the given direction,
[[
f
(
x
)]]
takes on a
constant value of
n
−
1
.
If
lim
x
→
a
f
(
x
)
∈
/
Z
, then
[[
f
(
x
)]]
is constant near
a
(that is, from both directions)
1
If
lim
x
→
a
+
/−
f
(
x
)
∈
/
Z
, replace
[[
f
(
x
)]]
by
x
→
lim
a
+
/−
f
(
x
)
.
2If
lim
x
→
a
+
/−
f
(
x
) =
n
∈
Z
,
Examples
1
Find
lim
x
→
5
+
−
x
5
.
Note that as
x
→
5
+
,
−
x
5
→ −
1
−
. Thus,
lim
x
→
5
+
−
x
5
=
lim
x
→
5
+
(
−
1
−
1
) =
−
2
.
2Find
lim
x
→
5
−
−
x
5
.
Note that as
x
→
5
−
,
−
x
5
→ −
1
+
. Thus,
x
lim
→
5
−
−
x
5
=
lim
x
→
5
−
−
1
=
−
1
.
3The previous computations show that
lim
x
→
5
−
x
5
does not exist.
4Find
lim
x
→−
2
+
−
5
x
.
Note that as
x
→ −
2
+
,
−
x
5
→
2
5
∈
/
Z
. Thus,
lim
x
→−
2
+
−
x
5
=
lim
x
→−
2
+
2
5
=
lim
x
→−
2
+
0
=
0
.
5lim
x
→−
2
−
−
x
5
=
0
and
lim
x
→−
2
−
x
5
1
lim
x
→−
1 3+
[[
3x
−
1
]] =
lim
x
→−
1 3+
(
−
2
) =
−
2
(
[[(
−
2
)
+
]]
)
2
lim
x
→
2
−
[[
3x
−
1
]] =
x
lim
→
2
−
(
4
) =
4
(
[[(
5
)
−
]]
)
3lim
x
→
3 2[[
3x
−
1
]] =
lim
x
→
3 2(
3
) =
3
(
7
2
)
4
lim
x
→
1
−
[[
5
−
x
]] =
x
lim
→
1
−
(
4
) =
4
5lim
x
→
3 2+
[[
3
−
2x
]] =
lim
x
→
3 2lim
x
→
2
−
[[
3x
−
1
]]
−
x
2
x
+ [[
−
x
]]
=
x
lim
→
2
−
4
−
x
2
x
−
2
=
lim
x
→
2
−
−
(
x
+
2
)
Consider
f
(
x
) =
1
x
2
. What happens as
x
→
0
?
x
f
(
x
)
1
1
0.5
4
0.1
100
0.001
1000000
0.00001
10000000000
x
f
(
x
)
−
1
1
−
0.5
4
−
0.1
100
−
0.001
1000000
−
0.00001
10000000000
−4
−3
−2
−1
1
2
3
−1
1
2
3
4
5
6
limit of
f
(
x
)
as
x
approaches
a
is positive infinity
[limit of
f
(
x
)
as
x
approaches
a
is negative infinity]
if the value of
f
(
x
)
increases [decreases] without bound whenever the values of
x
get closer and closer to
a
.
Notation:
lim
x
→
a
f
(
x
) = +
∞
[
x
lim
→
a
f
(
x
) =
−
∞
]
Example
Let
f
(
x
) =
1
x
2
.
x
f
(
x
)
1
1
0.5
4
0.1
100
0.001
1000000
0.00001
10000000000
↓
↓
0
+
+
∞
x
f
(
x
)
−
1
1
−
0.5
4
−
0.1
100
−
0.001
1000000
−
0.00001
10000000000
↓
↓
0
−
+
∞
We now write:
lim
x
→
0
1
Note that
∞
is NOT a number. So,
lim
x
→
a
f
(
x
) = +
∞
or
−
∞
does not mean that the
limit exists.
Vertical Asymptotes
The line with equation
x
=
a
is a
vertical asymptote
of the graph of
f
if at least
one of the following is true:
lim
x
→
a
−
f
(
x
) = +
∞
x
=
a
lim
x
→
a
−
f
(
x
) =
−
∞
x
=
a
lim
x
→
a
+
f
(
x
) = +
∞
x
=
a
lim
x
→
a
+
f
(
x
) =
−
∞
x
−
1
lim
x
→
1
−
3x
x
−
1
=
−
∞
3
0
−
lim
x
→
1
+
3x
x
−
1
= +
∞
3
0
+
−6
−4
−2
2
4
6
8
−2
2
4
6
In general:
Theorem
Suppose
lim
x
→
a
f
(
x
) =
c
and
x
lim
→
a
g
(
x
) =
0
.
1
If
c
>
0
1
and
g
(
x
)
→
0
+
as
x
→
a
, then
lim
x
→
a
f
(
x
)
g
(
x
)
= +
∞
2
and
g
(
x
)
→
0
−
as
x
→
a
, then
lim
x
→
a
f
(
x
)
g
(
x
)
=
−
∞
2
If
c
<
0
1
and
g
(
x
)
→
0
+
as
x
→
a
, then
lim
x
→
a
f
(
x
)
g
(
x
)
=
−
∞
2
and
g
(
x
)
→
0
−
as
x
→
a
, then
lim
x
→
a
f
(
x
)
lim
x
→−
2
−
5x
4
−
x
2
=
lim
x
→−
2
−
5x
(
2
+
x
)(
2
−
x
)
= +
∞
−
10
(
0
−
)(
4
)
lim
x
→−
2
+
5x
4
−
x
2
=
lim
x
→−
2
+
5x
(
2
+
x
)(
2
−
x
)
=
−
∞
−
10
(
0
+
)(
4
)
Theorem
1
If
lim
x
→
a
f
(
x
)
exists and
x
lim
→
a
g
(
x
) =
±
∞
, then
lim
x
→
a
(
f
(
x
) +
g
(
x
)) =
±
∞
.
2
If
lim
x
→
a
f
(
x
)
exists and
x
lim
→
a
g
(
x
) =
±
∞
, then
lim
x
→
a
(
f
(
x
)
−
g
(
x
)) =
∓
∞
.
3
If
lim
x
→
a
f
(
x
) = +
∞
and
x
lim
→
a
g
(
x
) = +
∞
, then
x
lim
→
a
(
f
(
x
) +
g
(
x
)) = +
∞
.
4
If
lim
x
→
a
f
(
x
) = +
∞
and
x
lim
→
a
g
(
x
) =
−
∞
, then
x
lim
→
a
(
f
(
x
)
−
g
(
x
)) = +
∞
, and
lim
x
→
a
(
g
(
x
)
−
f
(
x
)) =
−
∞
.
lim
x
→
a
f
(
x
)
lim
x
→
a
g
(
x
)
x
lim
→
a
[
f
(
x
) +
g
(
x
)]
lim
x
→
a
[
f
(
x
)
−
g
(
x
)]
c
+
∞
+
∞
−
∞
c
−
∞
−
∞
+
∞
+
∞
+
∞
+
∞
?
Let
c
∈
R
\ {
0
}. Suppose
lim
x
→
a
f
(
x
) =
c
and
x
lim
→
a
g
(
x
) =
±
∞
.
1
If
c
>
0
, then
lim
x
→
a
f
(
x
)
g
(
x
) =
±
∞
.
2
If
c
<
0
, then
lim
x
→
a
f
(
x
)
g
(
x
) =
∓
∞
.
2
If
lim
x
→
a
f
(
x
) = +
∞
and
x
lim
→
a
g
(
x
) = +
∞
, then
x
lim
→
a
f
(
x
)
g
(
x
) = +
∞
.
3
If
lim
x
→
a
f
(
x
) = +
∞
and
x
lim
→
a
g
(
x
) =
−
∞
, then
x
lim
→
a
f
(
x
)
g
(
x
) =
−
∞
.
lim
x
→
a
f
(
x
)
x
lim
→
a
g
(
x
)
lim
x
→
a
[
f
(
x
)
g
(
x
)]
c
>
0
±
∞
±
∞
c
<
0
±
∞
∓
∞
+
∞
+
∞
+
∞
+
∞
−
∞
−
∞
Example
Evaluate
lim
x
→
1
−
x
2
−
2
x
−
1
.
lim
x
→
1
−
x
2
=
1
lim
x
→
1
−
2
x
−
1
=
−
∞
2
0
−
lim
x
→
1
−
x
2
−
2
x
−
1
Evaluate
lim
x
→−
3
−
1
x
2
−
9
+
2x
x
+
3
.
lim
x
→−
3
−
1
x
2
−
9
+
2x
x
+
3
=
lim
x
→−
3
−
1
(
x
+
3
)(
x
−
3
)
+
2x
x
+
3
1
(
0
−
)(
−
6
)
+
−
6
0
−
Recall
lim
x
→
a
f
(
x
)
lim
x
→
a
g
(
x
)
x
lim
→
a
[
f
(
x
) +
g
(
x
)]
lim
x
→
a
[
f
(
x
)
−
g
(
x
)]
+
∞
+
∞
+
∞
?
+
∞
−
∞
?
+
∞
lim
x
→
a
f
(
x
)
x
lim
→
a
g
(
x
)
lim
x
→
a
[
f
(
x
)
g
(
x
)]
Definition
1
Suppose
lim
x
→
a
f
(
x
) = +
∞
and
x
lim
→
a
g
(
x
) = +
∞
. Then
lim
x
→
a
(
f
(
x
)
−
g
(
x
))
is called an
indeterminate form of type
∞
−
∞
.
2
Suppose
lim
x
→
a
f
(
x
) =
0
and
x
lim
→
a
g
(
x
) =
±
∞
.Then
lim
x
→
a
f
(
x
)
g
(
x
)
Example
lim
x
→−
1
−
1
x
+
1
+
3
2x
2
+
x
−
1
=
lim
x
→−
1
−
1
x
+
1
+
3
(
2x
−
1
)(
x
+
1
)
1
0
−
+
3
(
−
3
)(
0
−
)
(
∞
−
∞
)
=
lim
x
→−
1
−
(
2x
−
1
) +
3
(
2x
−
1
)(
x
+
1
)
=
lim
x
→−
1
−
2x
+
2
(
2x
−
1
)(
x
+
1
)
0
0
=
lim
x
→−
1
−
2
2x
−
1
=
−
2
lim
x
→
2
+
3
x
−
2
−
2x
x
2
−
4
=
lim
x
→
2
+
3
x
−
2
−
2x
(
x
−
2
)(
x
+
2
)
3
0
+
−
4
(
0
+
)(
4
)
(
∞
−
∞
)
=
lim
x
→
2
+
3x
+
6
−
2x
(
x
−
2
)(
x
+
2
)
=
lim
x
→
2
+
x
+
6
(
x
−
2
)(
x
+
2
)
8
(
0
+
)(
4
)
Example
lim
t
→
4
+
1
4
−
t
t
t
−
1
−
8
t
+
2
1
0
−
4
3
−
8
6
(
−
∞
·
0
)
=
lim
t
→
4
+
1
4
−
t
t
(
t
+
2
)
−
8
(
t
−
1
)
(
t
−
1
)(
t
+
2
)
=
lim
t
→
4
+
t
2
−
6t
+
8
(
4
−
t
)(
t
−
1
)(
t
+
2
)
0
0
=
lim
t
→
4
+
(
t
−
2
)(
t
−
4
)
(
4
−
t
)(
t
−
1
)(
t
+
2
)
=
lim
t
→
4
+
−
(
t
−
2
)
(
t
−
1
)(
t
+
2
)
=
−
2
Evaluate the following limits:
1lim
x
→
1 2−
x
2
−
1
1
−
2x
2
lim
x
→
2
+
x
2
−
2x
[[
x
]]
−
x
3
lim
x
→−
3
−
p