handa (nh5757) – Review Assignment – bormashenko – (54880)
1
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printout
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have
32
questions.
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001
10.0 points
If
f
oscillates faster and faster when
x
ap
proaches 0 as indicated by its graph
determine which, if any, of
L
1
:
lim
x
→
0+
f
(
x
)
,
L
2
:
lim
x
→
0

f
(
x
)
exist.
1.
neither
L
1
nor
L
2
exists
2.
both
L
1
and
L
2
exist
3.
L
1
exists, but
L
2
doesn’t
4.
L
1
doesn’t exist, but
L
2
does
correct
Explanation:
For
x >
0 the graph of
f
oscillates but the
oscillations do not get smaller and smaller as
x
approaches 0; so
L
1
does not exist. But for
x <
0, the graph oscillates and the oscillations
get smaller and smaller as
x
approaches 0; in
fact, the oscillation goes to 0 as
x
approaches
0, so
L
2
exists. Consequently,
L
1
does not exist, but
L
2
does
.
002
10.0 points
Consider the function
f
(
x
) =
3

x,
x <

1
x,

1
≤
x <
3
(
x

3)
2
,
x
≥
3
.
Find all the values of
a
for which the limit
lim
x
→
a
f
(
x
)
exists, expressing your answer in interval no
tation.
1.
(
∞
,

1]
∪
[3
,
∞
)
2.
(
∞
,

1)
∪
(

1
,
3)
∪
(3
,
∞
)
correct
3.
(
∞
,
3)
∪
(3
,
∞
)
4.
(
∞
,

1)
∪
(

1
,
∞
)
5.
(
∞
,
∞
)
Explanation:
The
graph
of
f
is
a
straight
line
on
(
∞
,

1), so
lim
x
→
a
f
(
x
)
exists (and =
f
(
a
)) for all
a
in (
∞
,

1).
Similarly, the graph of
f
on (

1
,
3) is a
straight line, so
lim
x
→
a
f
(
x
)
exists (and =
f
(
a
)) for all
a
in (

1
,
3).
On
(3
,
∞
), however, the graph of
f
is a parabola,
so
lim
x
→
a
f
(
x
)
still exists (and =
f
(
a
)) for all
a
in (3
,
∞
).
On the other hand,
lim
x
→ 
1

f
(
x
) = 4
,
lim
x
→ 
1+
f
(
x
) =

1
,
while
lim
x
→
3

f
(
x
) = 3
,
lim
x
→
3+
f
(
x
) = 0
.
Thus neither of the limits
lim
x
→ 
1
f
(
x
)
,
lim
x
→
3
f
(
x
)
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handa (nh5757) – Review Assignment – bormashenko – (54880)
2
exists. Consequently, the limit exists only for
values of
a
in
(
∞
,

1)
∪
(

1
,
3)
∪
(3
,
∞
)
.
003
10.0 points
Determine if
lim
x
→
0
x
2
cos
10
x
exists, and if it does, find its value.
1.
limit = 10
2.
limit = 0
correct
3.
limit = 1
4.
limit = 2
5.
limit does not exist
Explanation:
Since

cos
θ

≤
1 for all
θ
, we see that
cos
10
x
≤
1
,
holds for all
x
= 0. But then
x
2
cos
10
x
≤

x
2

,
for all
x
= 0; in other words, if we set
g
(
x
) =


x
2

,
h
(
x
) =

x
2

,
then the inequalities
g
(
x
)
≤
x
2
cos
10
x
≤
h
(
x
)
hold for all
x
= 0. But
lim
x
→
0
g
(
x
) =
lim
x
→
0
h
(
x
) = 0
.
Consequently, by the Squeeze Theorem, the
given limit
lim
x
→
0
x
2
cos
10
x
exists and
limit = 0
.
keywords:
limit,
trigonometric
function,
squeeze theorem
004
10.0 points
Determine if
lim
x
→
0
1
5
x
sin
x
exists, and if it does, find its value.
1.
limit =
1
5
2.
limit =

1
5
3.
limit =
∞
4.
none of the other answers
5.
limit =
∞
correct
Explanation:
Since
1
5
x
sin
x
>
0
for all small
x
= 0, both positive and negative,
while
lim
x
→
0
x
sin
x
= 0
,
we see that
lim
x
→
0
1
5
x
sin
x
=
∞
.
005
10.0 points
Determine the value of
lim
x
→
3
f
(
x
)
when
f
satisfies the inequalities
4
≤
f
(
x
)
≤
x
2

6
x
+ 13
on [

5
,
3)
∪
(3
,
5].
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 Spring '10
 Gualdini
 Limit, lim, Continuous function, Logarithm

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