Taylor, Douglas – Exam 1 – Due: Oct 2 2007, 11:00 pm – Inst: JEGilbert
2
To see why the Squeeze theorem applies,
it’s a good idea to draw the graphs of
g
and
h
using, say, a graphing calculator. They look
like
g
:
h
:
(1
,
6)
(not drawn to scale), so the graphs of
g
and
h
‘touch’ at the point (1
,
6) while the graph of
f
is ‘sandwiched’ between these two graphs.
Thus again we see that
lim
x
→
1
f
(
x
) = 6
.
keywords: limit, squeeze theorem
003
(part 1 of 1) 10 points
Let
f
be the function defined by
f
(
x
) = 3
x
+ (
x

2 +

x

2

)
2
.
Determine if
lim
h
→
0
f
(4 +
h
)

f
(4)
h
exists, and if it does, find its value.
1.
limit = 15
2.
limit doesn’t exist
3.
limit = 16
4.
limit = 18
5.
limit = 17
6.
limit = 19
correct
Explanation:
Since

v

=
‰
v,
v
≥
0,

v,
v <
0,
we see that
x

2 +

x

2

=
‰
2(
x

2)
,
x
≥
2,
0
,
x <
2.
Thus
f
(
x
) =
‰
3
x,
x <
2,
3
x
+ 4(
x

2)
2
,
x
≥
2.
In particular, therefore,
lim
h
→
0
f
(4 +
h
)

f
(4)
h
=
d
dx
‡
3
x
+ 4(
x

2)
2
·fl
fl
fl
x
= 4
=
‡
3 + 8(
x

2)
·fl
fl
fl
x
= 4
because 4
,
4 +
h >
2 for all small
h
. Conse
quently,
limit = 19
.
keywords: limit, Newtonian quotient absolute
value function
004
(part 1 of 1) 10 points
Determine if the limit
lim
x
→
0
sin 5
x
6
x
exists, and if it does, find its value.
1.
limit = 6
2.
limit =
6
5
3.
limit =
5
6
correct
4.
limit = 5