Mahon, Kevin – Homework 3 – Due: Sep 15 2005, 3:00 am – Inst: Edward Odell
2
6
5
4
3
2
1
0
1
2
3
4
5
2
4

2

4
2
4

2

4
Use the graph to determine lim
x
→
2
f
(
x
).
1.
limit =

1
2.
limit = 1
3.
does not exist
4.
limit = 0
5.
limit =

2
correct
Explanation:
From the graph it is clear that the limit
lim
x
→
2

f
(
x
) =

2
,
from the left and the limit
lim
x
→
2+
f
(
x
) =

2
,
from the right exist and coincide in value.
Thus the twosided
lim
x
→
2
f
(
x
) =

2
.
keywords:
Stewart5e, limit, graph, limit at
removable discontinuity
004
(part 1 of 1) 10 points
Consider the function
f
(
x
) =
3

x,
x <

1
x,

1
≤
x <
3
(
x

1)
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.
(
∞
,
∞
)
2.
(
∞
,
3)
∪
(3
,
∞
)
3.
(
∞
,

1)
∪
(

1
,
∞
)
4.
(
∞
,

1]
∪
[3
,
∞
)
5.
(
∞
,

1)
∪
(

1
,
3)
∪
(3
,
∞
)
correct
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
) = 4
.
Thus neither of the limits
lim
x
→ 
1
f
(
x
)
,
lim
x
→
3
f
(
x
)
exists. Consequently, the limit exists only for
values of
a
in
(
∞
,

1)
∪
(

1
,
3)
∪
(3
,
∞
)
.