moseley (cmm3869) – ReviewExam01 – Gilbert – (56380)
1
This
printout
should
have
16
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
0.0 points
When
f, g, F
and
G
are functions such that
lim
x
→
1
f
(
x
) = 0
,
lim
x
→
1
g
(
x
) =
∞
,
lim
x
→
1
F
(
x
) = 2
,
lim
x
→
1
G
(
x
) =
∞
,
which, if any, of
A.
lim
x
→
1
g
(
x
)
f
(
x
)
,
B.
lim
x
→
1
f
(
x
)
F
(
x
)
,
C.
lim
x
→
1
G
(
x
)
−
g
(
x
)
,
are indeterminate forms?
1.
B and C only
2.
A and C only
3.
none of them
4.
A and B only
5.
B only
6.
C only
correct
7.
all of them
8.
A only
Explanation:
A. By properties of limits,
lim
x
→
1
g
(
x
)
f
(
x
)
=
∞
0
=
∞
,
so this limit is not an indeterminate form.
B. By properties of limits,
lim
x
→
1
f
(
x
)
F
(
x
)
= 0
2
= 0
,
so this limit is not an indeterminate form.
C. Since
lim
x
→
1
G
(
x
)
−
g
(
x
) =
∞ − ∞
,
this limit is an indeterminate form.
002
0.0 points
Determine
lim
x
→
1
e
5
x
−
e
5
ln(3
x
−
2)
.
1.
limit =
5
3
e
5
correct
2.
limit =
e
5
3.
limit =
5
4
e
5
4.
limit does not exist
5.
limit =
5
3
6.
limit =
e
5
3
Explanation:
Since
e
5
x
−
e
5
ln(3
x
−
14)
=
f
(
x
)
g
(
x
)
where
f
and
g
are differentiable on (0
,
∞
) and
lim
x
→
1
f
(
x
) = 0
,
lim
x
→
1
g
(
x
) = 0
,
L’Hospital’s Rule can be applied. Now
f
′
(
x
) = 5
e
5
x
,
g
′
(
x
) =
3
3
x
−
2
,
so
lim
x
→
1
f
′
(
x
) = 5
e
5
,
lim
x
→
1
g
′
(
x
) = 3
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
moseley (cmm3869) – ReviewExam01 – Gilbert – (56380)
2
Consequently,
lim
x
→
1
e
5
x
−
e
5
ln(3
x
−
2)
=
5
3
e
5
.
003
0.0 points
Find the value of
lim
x
→
0
1
−
cos
x
5 sin
2
3
x
.
1.
limit does not exist
2.
limit =
1
90
correct
3.
limit =
1
45
4.
limit =
1
30
5.
limit =
2
45
Explanation:
Set
f
(
x
) = 1
−
cos
x,
g
(
x
) = 5 sin
2
3
x .
Then
f, g
are differentiable functions such
that
lim
x
→
0
f
(
x
) =
lim
x
→
0
g
(
x
) = 0
.
Thus L’Hospital’s Rule can be applied:
lim
x
→
0
f
(
x
)
g
(
x
)
=
lim
x
→
0
f
′
(
x
)
g
′
(
x
)
=
lim
x
→
0
sin
x
30 sin 3
x
cos 3
x
.
To compute this last limit we can either apply
L’Hospital’s Rule again or use the fact that
lim
x
→
0
sin
x
x
= 1
,
lim
x
→
0
sin 3
x
x
= 3
.
Consequently,
lim
x
→
0
1
−
cos
x
5 sin
2
3
x
=
1
90
.
004
0.0 points
Determine if the improper integral
I
=
integraldisplay
1
0
4 ln 3
x dx
converges, and if it does, find its value.
1.
I
= 3(ln 4
−
1)
2.
I
= 4 ln 3
−
3
3.
I
= 4(ln 3 + 1)
4.
I
does not converge
5.
I
= 4(ln 3
−
1)
correct
6.
I
= 3 ln 4 + 4
7.
I
= 3(ln 4 + 1)
Explanation:
Since ln 3
x
→ −∞
as
x
→
0 through pos
itive values of
x
, the graph of 4 ln 3
x
has a
vertical asymptote at
x
= 0.
It is this that
makes
I
an improper integral. So we set
I
=
lim
t
→
0+
integraldisplay
1
t
4 ln 3
x dx
and check if the limit exists.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 Sadler
 Multivariable Calculus, Taylor Series, Limits, Mathematical Series, Limit, Mathematical analysis, lim g

Click to edit the document details