1. f(x) =
x
+1
2
x
+
x logx
5
3
overtakes , so f(x) =
=
=
x
5
x
5
x
+1
2
x
2
x
5
x
3
Therefore, since n is equal to 3, O(
) = O(
)
x
n
x
3
3 is the smallest integer n.
2.
ln
(
x
)
x
2
Using the limit theorem: O [
] = O [
] * O [
]
(
x
)
f
(
x
)
f
1
*
2
(
x
)
f
1
f
(
x
)
2
So if
and
, then
= O(
) and
x
f
1
=
2
ln
(
x
)
f
2
=
f
1
x
2
O
(
x
)
f
2
=
Next, O(
ln(x)) = O(
) * O(x) = O(
)
x
2
x
2
x
3
Therefore
is big-O of
ln
(
x
)
x
2
x
3
3. By definition of O(n), we must prove that
for some
n
(
ln
n
0
n
1
n
n
2
+1)
n
3
n
2
2
.
n
3

6 A) No, there is no smallest real number a for which
is big-O of
.
6
x
2
x
a
x
B) Yes, there is a smallest integer such
is big-O of
. If we pick a=6, then |
,
6
x
2
x
a
x
| ≤ |
x
6 |
a
x
2
x
meaning that
is 0(
).
6
x
2
x
a
x
7.
O
means there are positive constants