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Nguyen (ln4328) – HW #4 – treisman – (54540)
1
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001
10.0 points
±ind the value oF
lim
x
→
1
√
10
−
x
−
3
√
5
−
x
−
2
iF the limit exists.
1.
limit =
2
3
correct
2.
limit =
−
2
3
3.
limit =
3
2
4.
limit does not exist
5.
limit =
−
3
2
Explanation:
By rationalization
√
10
−
x
−
3 =
(10
−
x
)
−
9
√
10
−
x
+ 3
=
−
p
x
−
1
√
10
−
x
+ 3
P
,
while
√
5
−
x
−
2 =
(5
−
x
)
−
4
√
5
−
x
+ 2
=
−
p
x
−
1
√
5
−
x
+ 2
P
.
Thus
√
10
−
x
−
3
√
5
−
x
−
2
=
√
5
−
x
+ 2
√
10
−
x
+ 3
when
x
n
= 1. But
lim
x
→
1
√
5
−
x
= 2
,
lim
x
→
1
√
10
−
x
= 3
.
Consequently, by the properties oF limits,
lim
x
→
1
√
10
−
x
−
3
√
5
−
x
−
2
=
2
3
.
002
10.0 points
Determine
lim
x
→
2
f
(
x
)
−
f
(2)
x
−
2
when
f
(
x
) = 5
x
2
−
2
x
+ 4
.
1.
limit = 21
2.
limit = 17
3.
limit does not exist
4.
limit = 19
5.
limit = 18
correct
6.
limit = 20
Explanation:
Since
f
(
x
)
−
f
(2) =
±
5
x
2
−
2
x
+ 4
²
−
20
= 5
x
2
−
2
x
−
16 = (5
x
+ 8)(
x
−
2)
,
we see that
f
(
x
)
−
f
(2)
x
−
2
= 5
x
+ 8
,
For
x
n
= 2. Consequently,
lim
x
→
2
f
(
x
)
−
f
(2)
x
−
2
= 18
.
Alternatively, iF we recognize that
lim
x
→
2
f
(
x
)
−
f
(2)
x
−
2
=
f
′
(2) = (10
x
−
2)

x
=2
,
we see again that
lim
x
→
2
f
(
x
)
−
f
(2)
x
−
2
= 18
.
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View Full DocumentNguyen (ln4328) – HW #4 – treisman – (54540)
2
003
10.0 points
Find the value of
b
,
b
≥
0, for which
lim
x
→
0
b
√
4
x
+
b
−
1
x
B
exists.
1.
b
= 2
2.
b
= 1
correct
3.
b
= 3
4.
b
= 4
5.
b
= 0
Explanation:
We are told that
lim
x
→
0
b
√
4
x
+
b
−
1
x
B
=
A
for some value
A
, but we aren’t told what the
particular value of
A
is. The question requires
us to see exactly what value
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 Fall '06
 McAdam
 Calculus, Limits

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