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pokharel (yp624) – HW01 – Radin – (57410)
1
This printout should have 20 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
HW01 is a review oF M408K material. It
overlaps HW02 and HW03.
001
10.0 points
Determine iF
lim
x
→∞
b
ln(3 + 4
x
)

ln(8 + 5
x
)
B
exists, and iF it does fnd its value.
1.
limit = ln
5
4
2.
limit = ln
3
8
3.
limit = ln
8
3
4.
limit = ln
4
5
correct
5.
limit does not exist
Explanation:
By properties oF logs,
ln(3 + 4
x
)

ln(8 + 5
x
)
= ln
p
3 + 4
x
8 + 5
x
P
= ln
p
3
/x
+ 4
8
/x
+ 5
P
.
But
lim
x
→∞
3
/x
+ 4
8
/x
+ 5
=
4
5
.
Consequently, the limit exists and
limit = ln
4
5
.
002
10.0 points
Determine iF
lim
x
→∞
sin
−
1
p
1 +
x
3 + 2
x
P
exists, and iF it does, fnd its value.
1.
limit = 0
2.
limit =
π
6
correct
3.
limit =
π
4
4.
limit does not exist
5.
limit =
π
3
6.
limit =
π
2
Explanation:
Since
lim
x
→∞
1 +
x
3 + 2
x
=
1
2
,
we see that
lim
x
→∞
sin
−
1
p
1 +
x
3 + 2
x
P
exists, and that the
limit = sin
−
1
1
2
=
πover
6
.
003
10.0 points
Determine iF
lim
x
→∞
tan
−
1
p
3 + 2
x
2
1 + 4
x
P
exists, and iF it does, fnd its value.
1.
limit =
π
4
2.
limit =
π
2
correct
3.
limit =
π
3
4.
limit =
π
6
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View Full Documentpokharel (yp624) – HW01 – Radin – (57410)
2
5.
limit does not exist
6.
limit = 0
Explanation:
Since
3 + 2
x
2
1 + 4
x
→ ∞
as
x
→ ∞
, we see that
lim
x
→∞
tan
−
1
p
3 + 2
x
2
1 + 4
x
P
exists, and that the
limit = tan
−
1
∞
=
π
2
.
004
10.0 points
Express the function
f
(
x
) = 5 sin
2
x

cos
2
x
in terms of cos 2
x
.
1.
f
(
x
) = 3 + 2 cos2
x
2.
f
(
x
) =

3

2 cos 2
x
3.
f
(
x
) = 2 + 3 cos2
x
4.
f
(
x
) =

2

3 cos 2
x
5.
f
(
x
) = 2

3 cos2
x
correct
6.
f
(
x
) = 3

2 cos2
x
Explanation:
Since
sin
2
x
=
1
2
(1

cos 2
x
)
,
cos
2
x
=
1
2
(1 + cos 2
x
)
,
we can rewrite
f
(
x
) as
f
(
x
) =
5
2
(1

cos 2
x
)

1
2
(1 + cos 2
x
)
.
Consequently,
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 Fall '08
 RAdin

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