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Unformatted text preview: number 3.
8
6
4 4 0 2 2x 4 2
4
6 The Graph y x2 9
x 3  We observe that
fx x
x3 3 if x 3
if x 3 To obtain a contradiction, suppose that the function f has a limit at 3. Choose 0 such that
1 and such that whenever x 3  and x 3 we have
f x  1.
Choose numbers x 1 and x 2 such that
3 x 1 3 x 2 3 .
We observe that
f x 2  f x 1
 f x 2
 1 1 2.
f x 1
On the other hand
f x1 x1 3 2 3 5
and 208 f x2 x2 3 6
and so
f x 1 f x2  6 5 2. This is the desired contradiction.
5. Given that S is a set of real numbers, that f : S R, that is a real number and that a is a limit point of S,
prove that the following conditions are equivalent:
a. f x as x a. b. For every number 0 there exists a number 0 such that the inequality f x
every number x in the set S
a that satisfies the inequality x a  .  3 holds for c. For every number 0 there exists a number 0 such that the inequality f x
every number x in the set S
a that satisfies the inequality x a  5.  3 holds for Solution: We shall provide the proof that condition c implies condition a. Suppose that condition c
holds. To prove that condition a holds, suppose that 0. Using the fact that the number /3 is positive
we now apply condition c to choose a number 0 such that the inequality
f x holds whenever x
x a  . S a and x  3 3
a  5. We see that f x  whenever x S a and 6. Given that S is a set of real numbers, that f : S R, that is a real number and that a is a limit point of S,
prove that the following conditions are equivalent:
a. f x as x a. b. For every number 0 there exists a neighborhood U of the number a such that the inequality
a.
f x  holds for every number x in the set U S
c. For every neighborhood V of the number there exists a number 0 such that the condition f x
holds for every number x in the set S
a that satisfies the inequality x a  . V To show that condition a implies condition b we assume that f x
as x a. Suppose that 0.
From condition a and the fact that the interval , is a neighborhood of we deduce that
there exists a neighborhood U of a such that the condition f x
, holds whenever
xUS
a.
To show that condition b implies condition a we assume that condition b holds. Suppose that V is a
neighborhood of . Choose a number 0 such that ,
V. Now, using condition b,
choose a neighborhood U of a such that the condition f x
, holds whenever
xUS
a . Then, whenever x U S
a we have f x
V.
To show that condition a implies condition c we assume that condition a holds. Suppose that V is a
neighborhood of and, using condition a, choose a neighborhood U of a such that the condition
fx
V will hold whenever x U S
a . Choose 0 such that a , a
U. We observe
that whenever x S
a and x a  , we must have x U S
a and so f x
V.
To show that condition c implies condition a we assume that condition c holds. Suppose that...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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