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(b)
[0, 24] by [0, 9]
This is continuous for all values of
x
in the domain
[0, 24] except for
x
5
0, 1, 2, 3, 4, 5, 6.
45. (a)
The function is defined when 1
1 }
1
x
}
.
0, that is, on
(
2‘
,
2
1)
<
(0,
‘
). (It can be argued that the domain
should also include certain values in the interval
(
2
1, 0), namely, those rational numbers that have odd
denominators when expressed in lowest terms.)
(b)
[
2
5, 5] by [
2
3, 10]
(c)
If we attempt to evaluate
f
(
x
) at these values, we obtain
f
(
2
1)
5
1
1
1 }
2
1
1
}
2
2
1
5
0
2
1
5 }
1
0
}
(undefined) and
f
(0)
5
1
1
1 }
1
0
}
2
0
(undefined). Since
f
is undefined at
these values due to division by zero, both values are
points of discontinuity.
(d)
The discontinuity at
x
5
0 is removable because the
righthand limit is 0. The discontinuity at
x
52
1 is not
removable because it is an infinite discontinuity.
(e)
[0, 20] by [0, 3]
The limit is about 2.718, or
e
.
46.
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 Spring '08
 GERMAN

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