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Unformatted text preview: the limit from the right is Ý and the limit from the left is Ý.
To see why the f x
Ý as x 3 , suppose that w is any real number. Given x 3, the inequality
f x w says that
1 w.
x3
We can’t simply turn these expressions over because w need not be positive but we can make the
observation that the inequality
fx 216 1
x 3 w will certainly hold when
1
x w  1 3 which says that
1.
w  1
and observe that the condition f x w will hold whenever
3 x We therefore define w11

3 x 3.
To see why the f x
Ý as x
f x w says that 3 , suppose that w is any real number. Given x 3, the inequality
1
x 3 w which we can express as
1 w
3x
We can’t simply turn these expressions over because w need not be positive but we can make the
observation that the inequality
1 w
3x
will certainly hold when
1 w  1
x3
which says that
1.
3 x
w  1
We therefore define w11 and observe that the condition f x w will hold whenever

3 x 3.
4. Prove that
x3 8
x2 x 6 as x Ý.
We begin by observing that x 2 x Ý 6 0 whenever x 2. Given any number x 2 we have
8 x 2 2x 4 x 2 x.
x
x3
x2 x 6
Now suppose that w is any real number and define v to be the larger of the two numbers 2 and w.
The inequality
x3 8 w
2x
6
x
will hold whenver x v.
x3 5. Prove that
x 4 4x 3 x 2 x 7
x 3 2x 2 2x 3 Ý as x Ý.
Given any number x we have
x3 2x 2 2x 3 x 3 x2 x 1 2x 2 2x 3 will be positive whenever x 3. Now given any number x 3 we have
x 4 4x 3 x 2 x 7 x 4 4x 3 x 2 x 4 4x 3 x 3 x 5 x 5 .
x
3
x3
x3
x 3 2x 2 2x 3
Now suppose that w is any real number. We define v to be the larger of the two numbers 3 and and so x 3 217 w 5
3 and observe that the inequality
x 4 4x 3 x 2 x 7 w
x 3 2x 2 2x 3 will hold whenever x v.
6. Prove that
3x 2 x 1
5x 2 4 as x Ý.
Given any number x we have 3x 2 x 1
5x 2 4
Whenever x 17
5 5x 17
.
5 5x 2 4 we observe that
3x 2 x 1
5x 2 4 Now suppose that 3
5 3
5 5x
5 5x 2
5x 2
5 5x 3
5 0. As long as x 17
5 17
4
17
4 5x 1 .
5x
25x 2 , the inequality 3
x 1
5
5x 2 4
1
will hold whenever 51x which says that x 5 . We define v to be the larger of the two numbers
17
1
and 5 and oberve that the inequality
5
3x 2 holds whenever x v. 3x 2 x 1
5x 2 4 3
5 7. Given that
fx 1 if x is rational
0 if x is irrational explain why f does not have a limit from the right at 2. Solution: To obtain a contradiction, suppose that is a limit of the function f from the right at 2. Using the fact that 1/2 0 we choose a number 0 such that the inequality
f x  1
2
will hold for every number x
2, 2 . Choose a rational number x and an irrational number t in the
interval x, x . Then
1 f x f t  f x   f t  1 1 1
2
2
and we have reached the desired contradiction. 8. Suppose that a is an interior point of a set S of real numbers and that f : S R. Suppose that f x
0 as
x a and that f x
1 as x a . Prove that the fun...
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 Fall '08
 STAFF
 Math, Calculus

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