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Granillo, Yvette – Homework 15 – Due: Dec 9 2005, 3:00 am – Inst: Edward Odell
1
This
printout
should
have
22
questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 1) 10 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
)
G
(
x
)
,
B.
lim
x
→
1
g
(
x
)
f
(
x
)
,
C.
lim
x
→
1
f
(
x
)
F
(
x
)
g
(
x
)
,
are indeterminate Forms?
1.
B and C only
2.
all oF them
3.
B only
4.
A and B only
5.
C only
6.
A only
correct
7.
A and C only
8.
none oF them
Explanation:
A. Since
lim
x
→
1
g
(
x
)
G
(
x
)
=
∞
∞
,
this limit is an indeterminate Form.
B. By properties oF limits,
lim
x
→
1
g
(
x
)
f
(
x
)
=
∞
0
=
∞
,
so this limit is not an indeterminate Form.
C. By properties oF limits,
lim
x
→
1
f
(
x
)
F
(
x
)
g
(
x
)
=
0
·
2
∞
= 0
,
so this limit is not an indeterminate Form.
keywords: Stewart5e,
002
(part 1 oF 1) 10 points
When
f, g, F
and
G
are Functions such that
lim
x
→
2
f
(
x
) = 0
,
lim
x
→
2
g
(
x
) = 1
,
lim
x
→
2
F
(
x
) = 2
,
lim
x
→
2
G
(
x
) =
∞
,
which oF the Following is an indeterminate
Form?
1.
lim
x
→
2
F
(
x
)
G
(
x
)
2.
lim
x
→
2
F
(
x
)
f
(
x
)
3.
lim
x
→
2
g
(
x
)
G
(
x
)
correct
4.
lim
x
→
2
f
(
x
)
g
(
x
)
5.
lim
x
→
2
g
(
x
)
F
(
x
)
Explanation:
Since
lim
x
→
2
g
(
x
)
G
(
x
)
= 1
∞
,
lim
x
→
2
F
(
x
)
G
(
x
)
= 2
∞
,
lim
x
→
2
f
(
x
)
g
(
x
)
= 0
1
,
lim
x
→
2
g
(
x
)
F
(
x
)
= 1
2
,
lim
x
→
2
F
(
x
)
f
(
x
)
= 2
0
,
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View Full DocumentGranillo, Yvette – Homework 15 – Due: Dec 9 2005, 3:00 am – Inst: Edward Odell
2
we see that only
lim
x
→
2
g
(
x
)
G
(
x
)
is an indeterminate form.
keywords: Stewart5e,
003
(part 1 of 1) 10 points
Determine the value of
lim
x
→∞
x
√
x
2
+ 1
.
1.
limit = 4
2.
limit =
1
4
3.
limit = 1
correct
4.
limit =
1
2
5.
limit = 0
6.
limit = 2
7.
limit =
∞
Explanation:
Since
lim
x
→∞
x
√
x
2
+ 1
→
∞
∞
,
the limit is of indeterminate form. We might
Frst try to use L’Hospital’s Rule
lim
x
→∞
f
(
x
)
g
(
x
)
=
lim
x
→∞
f
0
(
x
)
g
0
(
x
)
with
f
(
x
) =
x,
g
(
x
) =
p
x
2
+ 1
to evaluate the limit. But
f
0
(
x
) = 1
,
g
0
(
x
) =
x
√
x
2
+ 1
,
so
lim
x
→∞
f
0
(
x
)
g
0
(
x
)
=
lim
x
→∞
√
x
2
+ 1
x
=
∞
∞
,
which is again of indeterminate form.
Let’s
try using L’Hospital’s Rule again but now
with
f
(
x
) =
p
x
2
+ 1
,
g
(
x
) =
x,
and
f
0
(
x
) =
x
√
x
2
+ 1
,
g
0
(
x
) = 1
.
In this case,
lim
x
→∞
√
x
2
+ 1
x
=
lim
x
→∞
x
√
x
2
+ 1
,
which is the limit we started with. So, this is
an example where L’Hospital’s Rule applies,
but doesn’t work!
We have to go back to algebraic methods:
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