handa (nh5757) – Review Assignment – bormashenko – (54880)
1
This printout should have 32 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
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
:l
i
m
x
→
0+
f
(
x
)
,L
2
i
m
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:
±or
x>
0thegrapho
F
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
≤
3
(
x

3)
2
,x
≥
3
.
±ind 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
)