18.100B/C: Fall 2008
Solutions to Homework 9
1.
Let
f
(
x
) = sin
x
. Then
f
(0) = 0,
f
(0) = cos 0 = 1 and
f
(0) =

sin 0 = 0. Using Taylor’s
Theorem for the degree3 expansion for
f
(
x
) around the point
x
= 0, we have for any
x
∈
R
that
f
(
x
) =
x
1!
+
f
(3)
(
t
)
3!
x
3
for some
t
between 0 and
x
. We have
f
(3)
(
x
) =

cos
x
, so we can rewrite this equation as
f
(
x
)

x
=

cos
t
6
x
3
.
Taking absolute values and using the fact that

cos
t
 ≤
1 for all
t
∈
R
gives

sin(
x
)

x
 ≤

x

3
6
,
as needed.
For the second part of the question, we always consider
x
→ ∞
and so also
x >
0. From above
we have

sin(
x

c
)

x

c
 ≤
1
6
x

3
c
and so

x
a
sin(
x

c
)

x
a

c
 ≤
1
6
x
a

3
c
.
If
a < c
then
x
a

c
→
0 and so

x
a
sin(
x

c
)

has the same limit as

x
a
sin(
x

c
)

x
a

c

. This latter
expression is bounded between 0 and
1
6
x
a

3
c
→
0, so in fact in this case
x
a
sin
x

c
→
0.
If
a
=
c
then we have

x
a
sin(
x

c
)

1
 ≤
1
6
x

2
a
→
0
and so
x
a
sin
x

c
→
1.
Finally, if
a > c
then
x
a
sin
x

c
=
x
a

c
(
x
c
sin
x

c
).
Since
x
a

c
→ ∞
and
x
c
sin
x

c
→
1 we
have
x
a
sin
x

c
→ ∞
.
2.
For the first part of the question, we proceed by induction on the degree of the polynomial.
We know that
lim
x
→∞
e
x
=
∞
and so
lim
x
→∞
c
e
x
= 0
for any
c
∈
R
. Suppose that
lim
x
→∞
Q
(
x
)
e
x
= 0
1
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for all polynomials
Q
of degree
n
. Let
P
be a polynomial of degree
n
+ 1. Then as
x
→ ∞
,
P
(
x
)
e
x
is
of the form
±∞
∞
and so we can apply l’Hospital’s rule to get
lim
x
→∞
P
(
x
)
e
x
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 Fall '10
 Prof.KatrinWehrheim
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