(26)
The problem now is that we have two deltas, and the definition of the limit requires that we give
a single delta. The way to solve this is to recall that if we have found a delta which works, any
smaller delta will also work. Hence by setting
δ
= min
{
δ
f
,
δ
g
}
we can guarantee that
δ
δ
f
and
δ
δ
g
so that both equations in (
26
) hold. Hence if 0
<

x

c

<
δ
then
[
f
(
x
) +
g
(
x
)]

[
L
+
M
]

f
(
x
)

L

+

g
(
x
)

M

<
✏
2
+
✏
2
=
✏
.
This is what we wanted to show, and so the proof is complete.
Exercise 2.25.
Complete the proofs for Theorem
2.24
.
We can immediately use this to prove some very powerful and useful results.
Corollary 2.26
If
f
(
x
) =
p
(
x
)
q
(
x
)
is any rational functions (so that
p
(
x
)
and
q
(
x
)
are polynomials), and
c
2
R
is
such that
q
(
c
)
6
= 0
then
lim
x
!
c
p
(
x
)
q
(
x
)
=
p
(
c
)
q
(
c
)
.
Proof.
The key is to first show this for polynomials and apply the limit laws. It is easy to see (the
student should check!) that
lim
x
!
c
x
=
c
and so by induction, (the student can show that)
lim
x
!
c
x
n
=
c
n
.
Let
p
(
x
) =
a
n
x
n
+
a
n

1
x
n
1
+
· · ·
+
c
1
x
+
c
0
be an arbitrary polynomial. Our discussion tells us that
we know how to deal with every occurrence of the function
x
7!
x
, and so using the Theorem
2.24
c 2015 Tyler Holden
41
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2
Limits
2.5
Limits at Infinity
we have
lim
x
!
c
p
(
x
) = lim
x
!
c
[
a
n
x
n
+
· · ·
+
a
1
x
+
a
0
]
=
a
n
h
lim
x
!
c
x
i
n
+
a
n

1
h
lim
x
!
c
x
i
n

1
+
· · ·
+
a
1
h
lim
x
!
c
x
i
+
a
0
=
a
n
c
n
+
a
n

1
c
n

1
+
· · ·
+
a
1
c
+
a
0
=
p
(
c
)
.
Thus the result holds for any polynomial. Now if
p
(
x
) and
q
(
x
) are two polynomials and
q
(
c
)
6
= 0,
then the limit laws for quotients implies
lim
x
!
c
p
(
x
)
q
(
x
)
=
lim
x
!
c
p
(
x
)
lim
x
!
c
q
(
x
)
=
p
(
c
)
q
(
c
)
,
as required.
2.5
Limits at Infinity
There are two notions of infinity that we will be interested in tackling, both with definitions similar
to the
✏

δ
definition we saw before.
Definition 2.27.
•
We say that lim
x
!1
f
(
x
) =
L
if for every
✏
>
0 there exists an
M
2
R
such that whenever
x > M
then

f
(
x
)

L

<
✏
.
•
We say that lim
x
!
c
f
(
x
) =
1
if for every
M
2
R
there exists a
δ
>
0 such that if
0
<

x

c

<
δ
then
f
(
x
)
> M
.
In both cases, to consider
1
we replace
x > M, f
(
x
)
> M
with
x < M, f
(
x
)
< M
.
Exercise 2.28.
Determine an appropriate statement for lim
x
!1
f
(
x
) =
1
.
Example 2.29.
Show that lim
x
!1
x
x
+ 1
= 1.
42
c 2015 Tyler Holden
2.5
Limits at Infinity
2
Limits
Solution.
Let
✏
>
0 be given and take
M
= max

1
,
✏

1

1
. If
x > M
then since
x
≥
M
≥ 
1
we know that

x
+ 1

=
x
+ 1. Furthermore, since
x >
✏

1

1 we know that
x
+ 1
>
1
✏
, thus
x
x
+ 1

1 =

1
x
+ 1
=
1

x
+ 1

<
✏
as required.
⌅
Example 2.30.
Show that lim
x
!
0
1
x
2
=
1
.
Solution.
Let
M
2
R
be given and let
δ
=
1
p

M

. If

x

<
δ
=
1
p

M

then 0
< x
2
<
1

M

and so
f
(
x
) =
1
x
2
>
1

M

>
1
M
.
⌅
Just as it was possible to use the
✏

δ
definition to show that finite limits do not exist, we can
use our rigorous definitions to show that limits at infinity do not exist. Indeed, we say that
lim
x
!1
f
(
x
) does not exist
if for every
L
2
R
there exists an
✏
>
0 such that for every
M
2
R
there exists an
x
such that
x > M
and

f
(
x
)

L

≥
✏
.