hussain (tsh476) – test1review – rusin – (55565)
1
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before answering.
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with.
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homework assignment. Together, they cover
all the topics I consider to be fair game for the
exam on Tuesday. The test questions will be
similar (but fewer, and NOT multiplechoice).
Note that this homework is NOT required and
NOT graded – it is for your practice only. Ob
viously you don’t have to do all fifty problems!
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001
0.0points
Use L’Hospital’s Rule to determine which of
the inequalitites
A.
100
x
> e
−
x
,
B.
e
2
x
> xe
x
+ 100,
C.
e
x
< x
2
+ 100,
holds for all large
x
.
1.
none of them
2.
all of them
3.
C only
4.
A and B only
correct
5.
B and C only
6.
A and C only
7.
B only
8.
A only
Explanation:
The notion of
limit at infinity
tells us that
if
lim
x
→ ∞
f
(
x
)
g
(
x
)
=
∞
,
then
f
(
x
)
g
(
x
)
>
1
for all large
x
. But then
f
(
x
)
> g
(
x
)
holds for all large
x
so long as
g
(
x
)
>
0 for
large
x
, allowing us to multiply through the
inequality by
g
(
x
).
Similarly, if
lim
x
→ ∞
f
(
x
)
g
(
x
)
= 0
,
then
f
(
x
)
g
(
x
)
<
1
for all large
x
, so if
g
(
x
)
>
0 for all large
x
,
then
f
(
x
)
< g
(
x
)
for all large
x
.
On the other hand, if
lim
x
→ ∞
f
(
x
) =
∞
=
lim
x
→ ∞
g
(
x
)
,
then we can use L’Hospital’s Rule to deter
mine
lim
x
→ ∞
f
(
x
)
g
(
x
)
.
Similarly, if
lim
x
→ ∞
f
(
x
) = 0 =
lim
x
→ ∞
g
(
x
)
,
then we can use L’Hospital’s Rule to deter
mine
lim
x
→ ∞
f
(
x
)
g
(
x
)
.
In this way, we can use L’Hospital’s Rule to
compare the rates of growth or decay of
f
(
x
)
and
g
(
x
) when
x
→ ∞
.
For the three given inequalities, therefore,
we have to choose appropriate
f
and
g
and
make sure that
g
(
x
)
>
0 for all large
x
.
A.
TRUE: set
f
(
x
) =
e
−
x
,
g
(
x
) =
100
x
.
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hussain (tsh476) – test1review – rusin – (55565)
2
Then
lim
x
→ ∞
f
(
x
) = 0 =
lim
x
→ ∞
g
(
x
)
,
and by applying L’Hospital’s Rule we see that
lim
x
→ ∞
f
(
x
)
g
(
x
)
= 0
.
Thus the inequality
100
x
> e
−
x
holds for all large
x
.
B.
TRUE: set
f
(
x
) =
e
2
x
,
g
(
x
) =
xe
x
+ 100
.
Then
lim
x
→ ∞
f
(
x
) =
∞
=
lim
x
→ ∞
g
(
x
)
,
and by applying L’Hospital’s Rule twice we
see that
lim
x
→ ∞
f
(
x
)
g
(
x
)
=
∞
.
Thus the inequality
e
2
x
> xe
x
+ 100
holds for all large
x
.
C.
FALSE: set
f
(
x
) =
e
x
,
g
(
x
) =
x
2
+ 100
.
Then
lim
x
→ ∞
f
(
x
) =
∞
=
lim
x
→ ∞
g
(
x
)
,
and by applying L’Hospital’s Rule twice we
see that
lim
x
→ ∞
f
(
x
)
g
(
x
)
=
∞
.
Thus the inequality
e
x
> x
2
+ 100
,
not
e
x
< x
2
+ 100
,
holds for all large
x
.
keywords:
002
0.0points
Find the value of
lim
x
→
0
+
5
x
−
ln
x
2
x
.
1.
limit =
∞
correct
2.
limit = 5
3.
none of the other answers
4.
limit =
−∞
5.
limit =
5
2
6.
limit = 2
7.
limit = 0
Explanation:
Let’s first check if the given limit is an
indeterminate form.
Now ln
x
is defined for
x >
0 and the graph of ln
x
has a vertical
asymptote at
x
= 0; in addition,
lim
x
→
0
+
ln
x
=
−∞
.
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 Spring '09
 Chu
 Mathematical Series, lim g

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